Left factorials
You are encouraged to solve this task according to the task description, using any language you may know.
Left factorials, !n, may refer to either subfactorials or to factorial sums;
the same notation can be confusingly seen used for the two different definitions.
Sometimes, subfactorials (also known as derangements) may use any of the notations:
- !n`
- !n
- n¡
(It may not be visually obvious, but the last example uses an upside-down exclamation mark.)
This Rosetta Code task will be using this formula for left factorial:
where
- Task
Display the left factorials for:
- zero through ten (inclusive)
- 20 through 110 (inclusive) by tens
Display the length (in decimal digits) of the left factorials for:
- 1,000, 2,000 through 10,000 (inclusive), by thousands.
- Also see
- The OEIS entry: A003422 left factorials
- The MathWorld entry: left factorial
- The MathWorld entry: factorial sums
- The MathWorld entry: subfactorial
- Related task
ALGOL 68
Uses the Algol 68G LONG LONG INT type which has programmer definable precision.
<lang algol68># set the precision of LONG LONG INT - large enough for !n up to ! 10 000 # PR precision 36000 PR
- stores left factorials in an array #
- we calculate the left factorials, storing their values in the "values" array #
- if step is <= 1, we store we store every left factorial, otherwise we store !x when x MOD step = 0 #
- note this means values[ 0 ] is always !0 #
PROC get left factorials = ( REF[]LONG LONG INT values, INT step )VOID:
BEGIN
INT store position := LWB values;
INT max values := UPB values;
LONG LONG INT result := 0;
LONG LONG INT factorial k := 1;
FOR k FROM 0
WHILE
IF IF step <= 1 THEN TRUE ELSE k MOD step = 0 FI THEN
values[ store position ] := result;
store position +:= 1
FI;
store position <= max values
DO
result +:= factorial k;
factorial k *:= ( k + 1 )
OD
END # get left factorials # ;
- returns the number of digits in n #
OP DIGITCOUNT = ( LONG LONG INT n )INT:
BEGIN
INT result := 1;
LONG LONG INT v := ABS n;
WHILE v > 100 000 000 DO
result +:= 8;
v OVERAB 100 000 000
OD;
WHILE v > 10 DO
result +:= 1;
v OVERAB 10
OD;
result
END # DIGITCOUNT # ;
BEGIN
print( ( "!n for n = 0(1)10", newline ) );
[ 0 : 10 ]LONG LONG INT v;
get left factorials( v, 1 );
FOR i FROM 0 TO UPB v DO
print( ( whole( v[ i ], 0 ), newline ) )
OD
END;
BEGIN
print( ( "!n for n = 20(10)110", newline ) );
[ 0 : 11 ]LONG LONG INT v;
get left factorials( v, 10 );
FOR i FROM 2 TO UPB v DO
print( ( whole( v[ i ], 0 ), newline ) )
OD
END;
BEGIN
print( ( "digit counts of !n for n = 1000(1000)10 000", newline ) );
[ 0 : 10 ]LONG LONG INT v;
get left factorials( v, 1 000 );
FOR i FROM 1 TO UPB v DO
print( ( whole( DIGITCOUNT v[ i ], 0 ), newline ) )
OD
END </lang>
- Output:
!n for n = 0(1)10 0 1 2 4 10 34 154 874 5914 46234 409114 !n for n = 20(10)110 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 digit counts of !n for n = 1000(1000)10 000 2565 5733 9128 12670 16322 20062 23875 27749 31678 35656
AWK
Old posix AWK doesn't support computing with large numbers. However modern gawk can use GMP if the flag -M is used. <lang AWK>
- !/usr/bin/gawk -Mf
function left_factorial(num) { result=0 adder=1 if (num==0) return(0) for (k = 1; k <=num; k++) { result = result + adder adder = adder * k } return(result) }
BEGIN { for (i = 0; i <= 10; i++) { print "!" i " = " left_factorial(i) } for (i = 20; i<= 110; i+=10) { print "!" i " = " left_factorial(i) } for (i = 1000; i<= 10000; i+=1000) { print "!" i " has " length(left_factorial(i)) " digits" }
}
</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
BBC BASIC
Use the 'Mapm' library. <lang bbcbasic> INSTALL @lib$+"BB4WMAPMLIB" : PROCMAPM_Init : MAPM_Dec%=200
Result$="0" : A$="1"
FOR I%=0 TO 10000
IF I% Result$=FNMAPM_Add(Result$,A$) : A$=FNMAPM_Multiply(A$,STR$I%)
IF I% < 111 IF I% MOD 10 = 0 OR I% < 11 PRINT "!";I% " = " FNMAPM_FormatDec(Result$,0)
IF I% > 999 IF I% MOD 1000 = 0 PRINT "!";I% " has " LENFNMAPM_FormatDec(Result$,0) " digits"
NEXT
END</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
Bracmat
<lang bracmat>( ( leftFact
= result factorial i
. 0:?result
& 1:?factorial
& 0:?i
& whl
' ( !i+1:~>!arg:?i
& !factorial+!result:?result
& !factorial*!i:?factorial
)
& !result
)
& ( iterate
= from to step c fun
. !arg:(?from.?to.?step.?fun)
& !from+-1*!step:?from
& !step:?c
& whl
' ( !step+!from:~>!to:?from
& !fun$(leftFact$!from)
)
&
)
& out$"First 11 left factorials:" & iterate$(0.10.1.out) & out$" 20 through 110 (inclusive) by tens:" & iterate$(20.110.10.out) & out$" Digits in 1,000 through 10,000 by thousands:" & iterate
$ ( 1000 . 10000 . 1000 . (=L.@(!arg:? [?L)&out$!L) )
)</lang>
- Output:
First 11 left factorials: 0 1 2 4 10 34 154 874 5914 46234 409114 20 through 110 (inclusive) by tens: 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 Digits in 1,000 through 10,000 by thousands: 2565 5733 9128 12670 16322 20062 23875 27749 31678 35656
C
<lang C>
- include <stdio.h>
- include <stdlib.h>
- include <string.h>
- include <gmp.h>
void mpz_left_fac_ui(mpz_t rop, unsigned long op) {
mpz_t t1; mpz_init_set_ui(t1, 1); mpz_set_ui(rop, 0);
size_t i;
for (i = 1; i <= op; ++i) {
mpz_add(rop, rop, t1);
mpz_mul_ui(t1, t1, i);
}
mpz_clear(t1);
}
size_t mpz_digitcount(mpz_t op) {
/* mpz_sizeinbase can not be trusted to give accurate base 10 length */ char *t = mpz_get_str(NULL, 10, op); size_t ret = strlen(t); free(t); return ret;
}
int main(void) {
mpz_t t; mpz_init(t); size_t i;
for (i = 0; i <= 110; ++i) {
if (i <= 10 || i % 10 == 0) {
mpz_left_fac_ui(t, i);
gmp_printf("!%u = %Zd\n", i, t);
}
}
for (i = 1000; i <= 10000; i += 1000) {
mpz_left_fac_ui(t, i);
printf("!%u has %u digits\n", i, mpz_digitcount(t));
}
mpz_clear(t); return 0;
} </lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
C++
<lang CPP>
- include <vector>
- include <string>
- include <algorithm>
- include <iostream>
- include <sstream>
using namespace std;
- if 1 // optimized for 64-bit architecture
typedef unsigned long usingle; typedef unsigned long long udouble; const int word_len = 32;
- else // optimized for 32-bit architecture
typedef unsigned short usingle; typedef unsigned long udouble; const int word_len = 16;
- endif
class bignum { private:
// rep_.size() == 0 if and only if the value is zero. // Otherwise, the word rep_[0] keeps the least significant bits. vector<usingle> rep_;
public:
explicit bignum(usingle n = 0) { if (n > 0) rep_.push_back(n); }
bool equals(usingle n) const {
if (n == 0) return rep_.empty();
if (rep_.size() > 1) return false;
return rep_[0] == n;
}
bignum add(usingle addend) const {
bignum result(0);
udouble sum = addend;
for (size_t i = 0; i < rep_.size(); ++i) {
sum += rep_[i];
result.rep_.push_back(sum & (((udouble)1 << word_len) - 1));
sum >>= word_len;
}
if (sum > 0) result.rep_.push_back((usingle)sum);
return result;
}
bignum add(const bignum& addend) const {
bignum result(0);
udouble sum = 0;
size_t sz1 = rep_.size();
size_t sz2 = addend.rep_.size();
for (size_t i = 0; i < max(sz1, sz2); ++i) {
if (i < sz1) sum += rep_[i];
if (i < sz2) sum += addend.rep_[i];
result.rep_.push_back(sum & (((udouble)1 << word_len) - 1));
sum >>= word_len;
}
if (sum > 0) result.rep_.push_back((usingle)sum);
return result;
}
bignum multiply(usingle factor) const {
bignum result(0);
udouble product = 0;
for (size_t i = 0; i < rep_.size(); ++i) {
product += (udouble)rep_[i] * factor;
result.rep_.push_back(product & (((udouble)1 << word_len) - 1));
product >>= word_len;
}
if (product > 0)
result.rep_.push_back((usingle)product);
return result;
}
void divide(usingle divisor, bignum& quotient, usingle& remainder) const {
quotient.rep_.resize(0);
udouble dividend = 0;
remainder = 0;
for (size_t i = rep_.size(); i > 0; --i) {
dividend = ((udouble)remainder << word_len) + rep_[i - 1];
usingle quo = (usingle)(dividend / divisor);
remainder = (usingle)(dividend % divisor);
if (quo > 0 || i < rep_.size())
quotient.rep_.push_back(quo);
}
reverse(quotient.rep_.begin(), quotient.rep_.end());
}
};
ostream& operator<<(ostream& os, const bignum& x);
ostream& operator<<(ostream& os, const bignum& x) {
string rep;
bignum dividend = x;
bignum quotient;
usingle remainder;
while (true) {
dividend.divide(10, quotient, remainder);
rep += (char)('0' + remainder);
if (quotient.equals(0)) break;
dividend = quotient;
}
reverse(rep.begin(), rep.end());
os << rep;
return os;
}
bignum lfact(usingle n);
bignum lfact(usingle n) {
bignum result(0);
bignum f(1);
for (usingle k = 1; k <= n; ++k) {
result = result.add(f);
f = f.multiply(k);
}
return result;
}
int main() {
for (usingle i = 0; i <= 10; ++i) {
cout << "!" << i << " = " << lfact(i) << endl;
}
for (usingle i = 20; i <= 110; i += 10) {
cout << "!" << i << " = " << lfact(i) << endl;
}
for (usingle i = 1000; i <= 10000; i += 1000) {
stringstream ss;
ss << lfact(i);
cout << "!" << i << " has " << ss.str().size()
<< " digits." << endl;
}
} </lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits. !2000 has 5733 digits. !3000 has 9128 digits. !4000 has 12670 digits. !5000 has 16322 digits. !6000 has 20062 digits. !7000 has 23875 digits. !8000 has 27749 digits. !9000 has 31678 digits. !10000 has 35656 digits.
C#
<lang csharp> using System; using System.Numerics;
namespace LeftFactorial {
class Program
{
static void Main(string[] args)
{
for (int i = 0; i <= 10; i++)
{
Console.WriteLine(string.Format("!{0} = {1}", i, LeftFactorial(i)));
}
for (int j = 20; j <= 110; j += 10)
{
Console.WriteLine(string.Format("!{0} = {1}", j, LeftFactorial(j)));
}
for (int k = 1000; k <= 10000; k += 1000)
{
Console.WriteLine(string.Format("!{0} has {1} digits", k, LeftFactorial(k).ToString().Length));
}
Console.ReadKey();
}
private static BigInteger Factorial(int number)
{
BigInteger accumulator = 1;
for (int factor = 1; factor <= number; factor++)
{
accumulator *= factor;
}
return accumulator;
}
private static BigInteger LeftFactorial(int n)
{
BigInteger result = 0;
for (int i = 0; i < n; i++)
{
result += Factorial(i);
}
return result;
}
}
} </lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
Faster Implementation
<lang csharp> using System; using System.Numerics;
namespace LeftFactorial {
class Program
{
static void Main(string[] args)
{
for (int i = 0; i <= 10; i++)
{
Console.WriteLine(string.Format("!{0} : {1}", i, LeftFactorial(i)));
}
for (int j = 20; j <= 110; j += 10)
{
Console.WriteLine(string.Format("!{0} : {1}", j, LeftFactorial(j)));
}
for (int k = 1000; k <= 10000; k += 1000)
{
Console.WriteLine(string.Format("!{0} : has {1} digits", k, LeftFactorial(k).ToString().Length));
}
Console.ReadKey();
}
private static BigInteger LeftFactorial(int n)
{
BigInteger result = 0;
BigInteger subResult = 1;
for (int i = 0; i < n; i++)
{
if (i == 0)
{
subResult = 1;
}
else
{
subResult *= i;
}
result += subResult;
}
return result;
}
}
} </lang>
Clojure
<lang lisp>(ns left-factorial
(:gen-class))
(defn left-factorial [n]
" Compute by updating the state [fact summ] for each k, where k equals 1 to n
Update is next state is [k*fact (summ+k)"
(second
(reduce (fn [[fact summ] k]
[(*' fact k) (+ summ fact)])
[1 0] (range 1 (inc n)))))
(doseq [n (range 11)]
(println (format "!%-3d = %5d" n (left-factorial n))))
(doseq [n (range 20 111 10)] (println (format "!%-3d = %5d" n (biginteger (left-factorial n)))))
(doseq [n (range 1000 10001 1000)]
(println (format "!%-5d has %5d digits" n (count (str (biginteger (left-factorial n)))))))
</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
Common Lisp
<lang lisp> (defun fact (n)
(reduce #'* (loop for i from 1 to n collect i)))
(defun left-fac (n)
(reduce #'+ (loop for i below n collect (fact i))))
(format t "0 -> 10~&") (format t "~a~&" (loop for i upto 10 collect (left-fac i))) (format t "20 -> 110 by 10~&") (format t "~{~a~&~}" (loop for i from 20 upto 110 by 10 collect (left-fac i))) (format t "1000 -> 10000 by 1000~&") (format t "~{~a digits~&~}" (loop for i from 1000 upto 10000 by 1000 collect (length (format nil "~a" (left-fac i))))) </lang>
- Output:
0 -> 10 (0 1 2 4 10 34 154 874 5914 46234 409114) 20 -> 110 by 10 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 1000 -> 10000 by 1000 2565 digits 5733 digits 9128 digits 12670 digits 16322 digits 20062 digits 23875 digits 27749 digits 31678 digits 35656 digits
D
<lang d>import std.stdio, std.bigint, std.range, std.algorithm, std.conv;
BigInt leftFact(in uint n) pure nothrow /*@safe*/ {
BigInt result = 0, factorial = 1;
foreach (immutable i; 1 .. n + 1) {
result += factorial;
factorial *= i;
}
return result;
}
void main() {
writeln("First 11 left factorials:\n", 11.iota.map!leftFact);
writefln("\n20 through 110 (inclusive) by tens:\n%(%s\n%)",
iota(20, 111, 10).map!leftFact);
writefln("\nDigits in 1,000 through 10,000 by thousands:\n%s",
iota(1_000, 10_001, 1_000).map!(i => i.leftFact.text.length));
}</lang>
- Output:
First 11 left factorials: [0, 1, 2, 4, 10, 34, 154, 874, 5914, 46234, 409114] 20 through 110 (inclusive) by tens: 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 Digits in 1,000 through 10,000 by thousands: [2565, 5733, 9128, 12670, 16322, 20062, 23875, 27749, 31678, 35656]
EchoLisp
We use the 'bigint' library and memoization : (remember 'function). <lang lisp> (lib 'bigint) (define (!n n) (if (zero? n) 0 (+ (!n (1- n)) (factorial (1- n))))) (remember '!n) </lang> Output: <lang lisp> (for ((n 11)) (printf "!n(%d) = %d" n (!n n))) (for ((n (in-range 20 120 10))) (printf "!n(%d) = %d" n (!n n))) !n(0) = 0 !n(1) = 1 !n(2) = 2 !n(3) = 4 !n(4) = 10 !n(5) = 34 !n(6) = 154 !n(7) = 874 !n(8) = 5914 !n(9) = 46234 !n(10) = 409114 !n(20) = 128425485935180314 !n(30) = 9157958657951075573395300940314 !n(40) = 20935051082417771847631371547939998232420940314 !n(50) = 620960027832821612639424806694551108812720525606160920420940314 !n(60) = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !n(70) = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !n(80) = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !n(90) = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !n(100) = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !n(110) = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
- Compute !n
- 5 seconds
(for ((n (in-range 1000 10001 500))) (!n n) (writeln n))
- Display results
- 12 seconds
(for ((n (in-range 1000 10001 1000))) (printf "Digits of !n(%d) = %d" n (number-length (!n n)))) Digits of !n(1000) = 2565 Digits of !n(2000) = 5733 Digits of !n(3000) = 9128 Digits of !n(4000) = 12670 Digits of !n(5000) = 16322 Digits of !n(6000) = 20062 Digits of !n(7000) = 23875 Digits of !n(8000) = 27749 Digits of !n(9000) = 31678 Digits of !n(10000) = 35656 </lang>
Elixir
<lang elixir>defmodule LeftFactorial do
def calc(0), do: 0
def calc(n) do
{result, _factorial} = Enum.reduce(1..n, {0, 1}, fn i,{res, fact} ->
{res + fact, fact * i}
end)
result
end
end
Enum.each(0..10, fn i ->
IO.puts "!#{i} = #{LeftFactorial.calc(i)}"
end) Enum.each(Enum.take_every(20..110, 10), fn i ->
IO.puts "!#{i} = #{LeftFactorial.calc(i)}"
end) Enum.each(Enum.take_every(1000..10000, 1000), fn i ->
digits = LeftFactorial.calc(i) |> to_char_list |> length
IO.puts "!#{i} has #{digits} digits"
end)</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
F#
The Functıon
<lang fsharp> // Generate Sequence of Left Factorials: Nigel Galloway, March 5th., 2019. let LF=Seq.unfold(fun (Σ,n,g)->Some(Σ,(Σ+n,n*g,g+1I))) (0I,1I,1I) </lang>
The Tasks
- Display LF 0..10
<lang fsharp> LF |> Seq.take 11|>Seq.iter(printfn "%A") </lang>
- Output:
0 1 2 4 10 34 154 874 5914 46234 409114
- Display LF 20..110 in steps of 10
<lang fsharp> LF |> Seq.skip 20 |> Seq.take 91 |> Seq.iteri(fun n g->if n%10=0 then printfn "%A" g) </lang>
- Output:
128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
- Display the length (in decimal digits) of LF 1000 .. 10000 in steps of 1000
<lang fsharp> LF |> Seq.skip 1000 |> Seq.take 9001 |> Seq.iteri(fun n g->if n%1000=0 then printfn "%d" (string g).Length) </lang>
- Output:
2565 5733 9128 12670 16322 20062 23875 27749 31678 35656
Factor
<lang>USING: formatting fry io kernel math math.factorials math.functions math.parser math.ranges sequences ; IN: rosetta-code.left-factorials
- left-factorial ( n -- m ) <iota> [ n! ] map-sum ;
- print-left-factorials ( seq quot -- )
'[
dup left-factorial @
[ number>string "!" prepend ] dip
"%6s %-6d\n" printf
] each nl ; inline
- digit-count ( n -- count ) log10 >integer 1 + ;
- part1 ( -- ) 11 <iota> [ ] print-left-factorials ;
- part2 ( -- ) 20 110 10 <range> [ ] print-left-factorials ;
- part3 ( -- )
"Number of digits for" print 1,000 10,000 1,000 <range> [ digit-count ] print-left-factorials ;
- main ( -- ) part1 part2 part3 ;
MAIN: main</lang>
- Output:
!0 0
!1 1
!2 2
!3 4
!4 10
!5 34
!6 154
!7 874
!8 5914
!9 46234
!10 409114
!20 128425485935180314
!30 9157958657951075573395300940314
!40 20935051082417771847631371547939998232420940314
!50 620960027832821612639424806694551108812720525606160920420940314
!60 141074930726669571000530822087000522211656242116439949000980378746128920420940314
!70 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
!80 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
!90 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
!100 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
!110 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
Number of digits for
!1000 2565
!2000 5733
!3000 9128
!4000 12670
!5000 16322
!6000 20062
!7000 23875
!8000 27749
!9000 31678
!10000 35656
Fōrmulæ
In this page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Forth
This solution inspired by the Fortran one. <lang Forth>36000 CONSTANT #DIGITS \ Enough for !10000 CREATE S #DIGITS ALLOT S #DIGITS ERASE VARIABLE S# CREATE F #DIGITS ALLOT F #DIGITS ERASE VARIABLE F# 1 F C! 1 F# ! \ F = 1 = 0!
\ "Bignums": represented by two cells on the stack: \ 1) An address pointing to the least-significant unit \ 2) An integer size representing the number of character-size units
- mod/ /mod swap ;
- B+ ( addr u addr' u' -- u) \ Add the second "bignum" into the first
over + >R -rot over + >R ( addr' addr R:end' R:end)
swap >R 0 over R> ( addr 0 addr addr' R:end' R:end)
\ 0: Assume second has equal or more digits, as in our problem
BEGIN over R@ < WHILE \ 1: add all digits from S
dup >R C@ swap dup >R C@ ( addr c a a' R:end' R:end R:addr'* R:addr*)
+ + 10 mod/ R@ C! R> 1+ R> 1+
REPEAT R> drop ( addr c addr* addr'* R:end')
BEGIN dup R@ < WHILE \ 2: add any remaining digits from F
dup >R C@ swap >R ( addr c a' R:end' R:addr'* R:addr*)
+ 10 mod/ R@ C! R> 1+ R> 1+
REPEAT R> drop drop ( addr c addr*)
BEGIN over WHILE \ 3: add any carry digits
>R 10 mod/ ( addr m d R:addr*) R@ C! R> 1+
REPEAT rot - nip ; \ calculate travel distance, discard 0 carry
- B* ( addr u u' -- u) \ Multiply "bignum" inplace by U'
0 2swap over >R dup >R bounds ( u' 0 addr+u addr R:addr R:u)
DO ( u' c) over I C@ * + 10 mod/ I C! LOOP
nip R> BEGIN ( c u) over WHILE \ insert carry, may have multiple digits
>R 10 mod/ R@ swap R> R@ + ( m u d addr+u R:addr) C! 1+
REPEAT nip R> ( u addr) drop ;
- .B ( addr u) over + BEGIN 1- \ print bignum
dup C@ [char] 0 + EMIT over over >= UNTIL drop drop ;
- .!n 0 <# #s [char] ! hold #> 6 over - spaces type space ;
- REPORT ( n)
dup 10 <= over dup 20 111 within swap 10 mod 0= and or
IF .!n [char] = emit space S S# @ .B cr
ELSE dup 1000 mod 0=
IF .!n ." has " S# @ . ." digits" cr
ELSE drop THEN
THEN ;
- GO 0 REPORT
1 BEGIN dup 10000 <=
WHILE
S S# @ F F# @ B+ S# !
dup REPORT
dup F F# @ rot B* F# !
1+ REPEAT drop ;</lang>
- Output:
$ gforth left-factorials.fs -e 'GO bye'
!0 = 0
!1 = 1
!2 = 2
!3 = 4
!4 = 10
!5 = 34
!6 = 154
!7 = 874
!8 = 5914
!9 = 46234
!10 = 409114
!20 = 128425485935180314
!30 = 9157958657951075573395300940314
!40 = 20935051082417771847631371547939998232420940314
!50 = 620960027832821612639424806694551108812720525606160920420940314
!60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314
!70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314
!80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314
!90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
!100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314
!110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
!1000 has 2565 digits
!2000 has 5733 digits
!3000 has 9128 digits
!4000 has 12670 digits
!5000 has 16322 digits
!6000 has 20062 digits
!7000 has 23875 digits
!8000 has 27749 digits
!9000 has 31678 digits
!10000 has 35656 digits
Fortran
First, to see how far INTEGER*8 arithmetic can reach. This is the largest size likely available, even though the syntax could easily allow INTEGER*2400 or the like. The F90 MODULE protocol is used simply to avoid the tedium of declaring the type of the function FACT(n) in all routines invoking it, though at the cost of typing out the required blather. Otherwise, this would be acceptable to older Fortran compilers, except for the appending of name information on END statements.
The function names are used as ordinary variables within the function while building the result; earlier compilers often did not allow such usage or produced incorrect code in certain cases. Ordinary integer variables are used since it is obvious that their range will not be exercised before the function result overflows, even with sixty-four bit integers for them. With two's complement arithmetic, negative numbers can appear in spite of the mathematics involved only being able to generate positive numbers, but this relies on the "sign bit" happening to become set and cannot be regarded as a definite check. Only a proper test such as the IF OVERFLOW found in First Fortran (1958) will do, but the modernisers have long abandoned this detail.
Because this calculation won't get far, no attempt is made to save intermediate results (such as the factorial numbers) nor develop the results progressively even though they are to be produced in sequence. Each result is computed from the start, as per the specified formulae.
For output, to have the exclamation mark precede the number without a gap, format sequence "!",I0 will do, the I0 format code being standardised in F90. However, this produces varying-length digit sequences, which will mean that the following output changes position likewise. Rather than use say I20 for the result and have a wide gap, code I0 will do, and to start each such number in the same place, the code T6 will start it in column six, far enough along not to clash with the first number on the line, given that it will not be large. <lang Fortran> MODULE LAIROTCAF !Calculates "left factorials".
CONTAINS !The usual suspects.
INTEGER*8 FUNCTION FACT(N) !Factorial, the ordinary.
INTEGER N !The number won't ever get far.
INTEGER I !The stepper.
FACT = 1 !Here we go.
DO I = 2,N !Does nothing for N < 2.
FACT = FACT*I !Perhaps this overflows.
IF (FACT.LE.0) STOP "Factorial: Overflow!" !Two's complement arithmetic.
END DO !No longer any IF OVERFLOW tests.
END FUNCTION FACT !Simple enough.
INTEGER*8 FUNCTION LFACT(N) !Left factorial.
INTEGER N !This number won't get far either.
INTEGER K !A stepper.
LFACT = 0 !Here we go.
DO K = 0,N - 1 !Apply the definition.
LFACT = LFACT + FACT(K) !Perhaps this overflows.
IF (LFACT.LE.0) STOP "Lfact: Overflow!" !Unreliable test.
END DO !On to the next step in the summation.
END FUNCTION LFACT !No attempts at saving effort.
END MODULE LAIROTCAF !Just the minimum.
PROGRAM POKE
USE LAIROTCAF
INTEGER I
WRITE (6,*) "Left factorials, from 0 to 10..."
DO I = 0,10
WRITE (6,1) I,LFACT(I)
1 FORMAT ("!",I0,T6,I0)
END DO
WRITE (6,*) "Left factorials, from 20 to 110 by tens..."
DO I = 20,110,10
WRITE (6,1) I,LFACT(I)
END DO
END</lang>
Output:
Left factorials, from 0 to 10... !0 0 !1 1 !2 2 !3 4 !4 10 !5 34 !6 154 !7 874 !8 5914 !9 46234 !10 409114 Left factorials, from 20 to 110 by tens... !20 128425485935180314 Factorial: Overflow!
Obviously, one could proceed using the services of some collection of "bignum" routines, and then the code would merely depict their uses for this problem. Since the task is to produce consecutive values, all that need be done is to maintain a S value holding the accumulated sum, and a F value for the successive factorials to be added into S. The only difficulty is to arrange the proper phasing of the starting values so that the calculation will work. Since only one multiply and one addition is needed per step, explicit code might as well be used, as follows: <lang Fortran>Calculates "left factorials", in sequence, and shows some.
INTEGER ENUFF,BASE !Some parameters.
PARAMETER (BASE = 10, ENUFF = 40000) !This should do.
INTEGER LF,F(ENUFF),LS,S(ENUFF) !Big numbers in digits F(1:LF), S(1:LS)
INTEGER N !A stepper.
INTEGER L !Locates digits.
INTEGER C !A carry for arithmetic.
INTEGER MSG !I/O unit number.
MSG = 6 !Standard output.
LF = 1; F(1) = 1 !Set F = 1 = 0!
LS = 1; S(1) = 0 !Set S = 0 = !0
WRITE (MSG,1) 0,0 !Pre-emptive first result.
1 FORMAT ("!",I0,T6,666I1) !This will do for reasonable sizes.
10 DO N = 1,10000 !Step away.
Commence the addition of F to S.
20 C = 0 !Clear the carry.
DO L = 1,MIN(LF,LS) !First, both S and F have low-order digits.
C = S(L) + F(L) + C !So, a three-part addition.
S(L) = MOD(C,BASE) !Place the digit.
C = C/BASE !Carry to the next digit up.
END DO !Ends with L and C important.
Careful. L fingers the next digit up, and C is to carry in to that digit.
IF (LF.GT.LS) THEN !Has F more digits than S?
DO L = L,LF !Yes. Continue adding, with leading zero digits from S.
C = F(L) + C !Thus.
LS = LS + 1 !Another digit for S.
S(LS) = MOD(C,BASE) !Place.
C = C/BASE !Carry to the next digit up.
END DO !Continue to the end of F.
END IF !Either way, F has been added in.
Continue carrying, with C for digit L.
DO WHILE(C .GT. 0) !Extend the carry into S.
IF (L.LE.LS) THEN !If F had fewer digits than S,
C = C + S(L) !S digits await.
ELSE !Otherwise,
LS = LS + 1 !Extend S.
END IF !C is ready.
S(L) = MOD(C,BASE) !Place it.
C = C/BASE !The carry for the next digit up.
L = L + 1 !Locate it.
END DO !Perhaps a multi-digit carry.
Contemplate what to do with the current S.
IF (N.LE.10) THEN !First selection: !N for 0 to 10.
WRITE (MSG,1) N,S(LS:1:-1) !Show the value. Digits from the high-order end down.
ELSE IF (20.LE.N .AND. N.LE.110) THEN !Second selection: for 20 to 110,
IF (MOD(N,10).EQ.0) WRITE (MSG,1) N,S(LS:1:-1) !Show only every tenth.
ELSE !Third selection
IF (MOD(N,1000).EQ.0) WRITE (MSG,21) N,LS !Show only the number of digits.
21 FORMAT ("!",I0," has ",I0," digits.") !Which is why BASE is only 10.
END IF !So much for the selection of output.
Calculate the next factorial, ready for the next one up.
C = 0 !Start a multiply.
DO L = 1,LF !Step up the digits to produce N! in F.
C = F(L)*N + C !A digit.
F(L) = MOD(C,BASE) !Place.
C = C/BASE !Extract the carry.
END DO !On to the next digit.
DO WHILE(C .GT. 0) !While any carry remains,
LF = LF + 1 !Add another digit to F.
IF (LF.GT.ENUFF) STOP "F overflow!" !Perhaps not.
F(LF) = MOD(C,BASE) !The digit.
C = C/BASE !Carry to the next digit up.
END DO !If there is one, as when N > BASE.
END DO !On to the next result.
END !Ends with a new factorial that won't be used.</lang>
Output: achieved in a few seconds. A larger BASE would give a faster calculation, but would complicate the digit count.
!0 0 !1 1 !2 2 !3 4 !4 10 !5 34 !6 154 !7 874 !8 5914 !9 46234 !10 409114 !20 128425485935180314 !30 9157958657951075573395300940314 !40 20935051082417771847631371547939998232420940314 !50 620960027832821612639424806694551108812720525606160920420940314 !60 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits. !2000 has 5733 digits. !3000 has 9128 digits. !4000 has 12670 digits. !5000 has 16322 digits. !6000 has 20062 digits. !7000 has 23875 digits. !8000 has 27749 digits. !9000 has 31678 digits. !10000 has 35656 digits.
FreeBASIC
<lang freebasic>' FB 1.05.0 Win64
- include "gmp.bi"
Sub leftFactorial(rop As __mpz_struct, op As ULong)
Dim As __mpz_struct t1 mpz_init_set_ui(@t1, 1) mpz_set_ui(@rop, 0) For i As ULong = 1 To op mpz_add(@rop, @rop, @t1) mpz_mul_ui(@t1, @t1, i) Next mpz_clear(@t1)
End Sub
Function digitCount(op As __mpz_struct) As ULong
Dim As ZString Ptr t = mpz_get_str(0, 10, @op) Dim As ULong ret = Len(*t) Deallocate(t) Return ret
End Function
Dim As __mpz_struct t mpz_init(@t)
For i As ULong = 0 To 110
If i <= 10 OrElse i Mod 10 = 0 Then leftFactorial(t, i) gmp_printf(!"!%u = %Zd\n", i, @t) End If
Next
For i As ULong = 1000 To 10000 Step 1000
leftFactorial(t, i) Print "!"; Str(i); " has "; digitCount(t); " digits"
Next
mpz_clear(@t) Print Print "Press any key to quit" Sleep</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
Go
<lang go>package main
import (
"fmt" "math/big"
)
func main() {
fmt.Print("!0 through !10: 0")
one := big.NewInt(1)
n := big.NewInt(1)
f := big.NewInt(1)
l := big.NewInt(1)
next := func() { f.Mul(f, n); l.Add(l, f); n.Add(n, one) }
for ; ; next() {
fmt.Print(" ", l)
if n.Int64() == 10 {
break
}
}
fmt.Println()
for {
for i := 0; i < 10; i++ {
next()
}
fmt.Printf("!%d: %d\n", n, l)
if n.Int64() == 110 {
break
}
}
fmt.Println("Lengths of !1000 through !10000 by thousands:")
for i := 110; i < 1000; i++ {
next()
}
for {
fmt.Print(" ", len(l.String()))
if n.Int64() == 10000 {
break
}
for i := 0; i < 1000; i++ {
next()
}
}
fmt.Println()
}</lang>
- Output:
!0 through !10: 0 1 2 4 10 34 154 874 5914 46234 409114 !20: 128425485935180314 !30: 9157958657951075573395300940314 !40: 20935051082417771847631371547939998232420940314 !50: 620960027832821612639424806694551108812720525606160920420940314 !60: 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70: 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80: 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90: 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100: 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110: 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 Lengths of !1000 through !10000 by thousands: 2565 5733 9128 12670 16322 20062 23875 27749 31678 35656
Haskell
<lang haskell>fact :: [Integer] fact = scanl (*) 1 [1 ..]
leftFact :: [Integer] leftFact = scanl (+) 0 fact
main :: IO () main =
mapM_ putStrLn [ "0 ~ 10:" , show $ (leftFact !!) <$> [0 .. 10] , "" , "20 ~ 110 by tens:" , unlines $ show . (leftFact !!) <$> [20,30 .. 110] , "" , "length of 1,000 ~ 10,000 by thousands:" , show $ (length . show . (leftFact !!)) <$> [1000,2000 .. 10000] , "" ]</lang>
- Output:
0 ~ 10: [0,1,2,4,10,34,154,874,5914,46234,409114] 20 ~ 110 by tens: 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 length of 1,000 ~ 10,000 by thousands: [2565,5733,9128,12670,16322,20062,23875,27749,31678,35656]
Icon and Unicon
The following works in both languages: <lang>procedure main()
every writes(lfact(0 | !10)," ") write() write() every write(lfact(20 to 110 by 10)) write() every writes(*lfact(1000 to 10000 by 1000)," ") write()
end
procedure lfact(n)
r := 0 f := 1 every (i := !n, r +:= .f, f *:= .i) return r
end</lang>
- Output:
->lfact 0 1 2 4 10 34 154 874 5914 46234 409114 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 2565 5733 9128 12670 16322 20062 23875 27749 31678 35656 ->
J
This could be made more efficient (in terms of machine time), is there a practical application for this? The more efficient machine approach would require a more specialized interface or memory dedicated to caching.
<lang J>leftFact=: +/@:!@i."0</lang>
Task examples:
<lang J> (,. leftFact) i.11
0 0 1 1 2 2 3 4 4 10 5 34 6 154 7 874 8 5914 9 46234
10 409114
(,. leftFact) 10*2+i.10x 20 128425485935180314 30 9157958657951075573395300940314 40 20935051082417771847631371547939998232420940314 50 620960027832821612639424806694551108812720525606160920420940314 60 141074930726669571000530822087000522211656242116439949000980378746128920420940314 70 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 80 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 90 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314
100 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 110 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
(,. #@":@leftFact) 1000*1+i.10x 1000 2565 2000 5733 3000 9128 4000 12670 5000 16322 6000 20062 7000 23875 8000 27749 9000 31678
10000 35656</lang>
Java
<lang java>import java.math.BigInteger;
public class LeftFac{ public static BigInteger factorial(BigInteger n){ BigInteger ans = BigInteger.ONE; for(BigInteger x = BigInteger.ONE; x.compareTo(n) <= 0; x = x.add(BigInteger.ONE)){ ans = ans.multiply(x); } return ans; }
public static BigInteger leftFact(BigInteger n){ BigInteger ans = BigInteger.ZERO; for(BigInteger k = BigInteger.ZERO; k.compareTo(n.subtract(BigInteger.ONE)) <= 0; k = k.add(BigInteger.ONE)){ ans = ans.add(factorial(k)); } return ans; }
public static void main(String[] args){ for(int i = 0; i <= 10; i++){ System.out.println("!" + i + " = " + leftFact(BigInteger.valueOf(i))); }
for(int i = 20; i <= 110; i += 10){ System.out.println("!" + i + " = " + leftFact(BigInteger.valueOf(i))); }
for(int i = 1000; i <= 10000; i += 1000){ System.out.println("!" + i + " has " + leftFact(BigInteger.valueOf(i)).toString().length() + " digits"); } } }</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
jq
jq currently only has builtin support for IEEE 64-bit numbers, so in this section we will first present the algorithm using the builtin arithmetic operators and then adapt it for use with the BigInt library at https://gist.github.com/pkoppstein/d06a123f30c033195841
Using builtin arithmetic: <lang jq>def left_factorial:
reduce range(1; .+1) as $i # state: [i!, !i] ([1,0]; .[1] += .[0] | .[0] *= $i) | .[1];</lang>
Using BigInt.jq:
The BigInt library can be used with jq 1.4, but we will take this opportunity to showcase jq 1.5's support for importing such libraries as modules. If your jq does not have support for modules, add the BigInt.jq file, remove the 'import' statement and strip off the "BigInt::" prefix.
To compute the lengths of the decimal representation without having to recompute !n, we also define left_factorial_lengths(gap) to emit [n, ( !n|length) ] when n % gap == 0. <lang jq>import "BigInt" as BigInt;
- integer input
def long_left_factorial:
reduce range(1; .+1) as $i
# state: [i!, !i]
( ["1", "0"];
.[1] = BigInt::long_add(.[0]; .[1])
| .[0] = BigInt::long_multiply(.[0]; $i | tostring) )
| .[1];
- input and gap should be integers
def long_left_factorial_lengths(gap):
reduce range(1; .+1) as $i # state: [i!, !i, gap] (["1", "0", []]; .[1] = BigInt::long_add(.[0]; .[1]) | .[0] = BigInt::long_multiply(.[0]; $i|tostring) | (.[1] | tostring | length) as $lf | if $i % gap == 0 then .[2] += $i, $lf else . end) | .[2];</lang>
The specific tasks: <lang sh>((range(0;11), (range(2; 12) * 10)) | "\(.): \(long_left_factorial)"),
(10000 | long_left_factorial_lengths(1000) | .[] | "\(.[0]): length is \(.[1])")</lang>
- Output:
(scrollable)
<lang sh>$ jq -r -n -L . -f Long_left_factorial.jq 0: 0 1: 1 2: 2 3: 4 4: 10 5: 34 6: 154 7: 874 8: 5914 9: 46234 10: 409114 20: 128425485935180314 30: 9157958657951075573395300940314 40: 20935051082417771847631371547939998232420940314 50: 620960027832821612639424806694551108812720525606160920420940314 60: 141074930726669571000530822087000522211656242116439949000980378746128920420940314 70: 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 80: 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 90: 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 100: 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 110: 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 1000: length is 2565 2000: length is 5733 3000: length is 9128 4000: length is 12670 5000: length is 16322 6000: length is 20062 7000: length is 23875 8000: length is 27749 9000: length is 31678 10000: length is 35656</lang>
Julia
<lang julia>leftfactorial(n::Integer) = n ≤ 0 ? zero(n) : sum(factorial, 0:n-1)
@show leftfactorial.(0:10) @show ndigits.(leftfactorial.(big.(1000:1000:10_000)))</lang>
- Output:
leftfactorial.(0:10) = [0, 1, 2, 4, 10, 34, 154, 874, 5914, 46234, 409114] ndigits.(leftfactorial.(big.(1000:1000:10000))) = [2565, 5733, 9128, 12670, 16322, 20062, 23875, 27749, 31678, 35656]
Kotlin
<lang scala>// version 1.0.6
import java.math.BigInteger
fun leftFactorial(n: Int): BigInteger {
if (n == 0) return BigInteger.ZERO
var fact = BigInteger.ONE
var sum = fact
for (i in 1 until n) {
fact *= BigInteger.valueOf(i.toLong())
sum += fact
}
return sum
}
fun main(args: Array<String>) {
for (i in 0..110)
if (i <= 10 || (i % 10) == 0)
println("!${i.toString().padEnd(3)} = ${leftFactorial(i)}")
println("\nLength of the following left factorials:")
for (i in 1000..10000 step 1000)
println("!${i.toString().padEnd(5)} has ${leftFactorial(i).toString().length} digits")
}</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 Length of the following left factorials: !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
Lambdatalk
The code can be tested in this wiki page: http://lambdaway.free.fr/lambdawalks/?view=left_factorial <lang scheme> 1) defining !n
{def !n // the main function's name
{def !n.mem {A.new 0 1 2}} // initializing a global array
// memorizing the computed values of !n
{def !n.set // computing and storing the values
{lambda {:n}
{A.set! :n // assign at n
{long_mult :n {A.get {- :n 1} {!n.mem}} } // product of n and computed value at n-1
{!n.mem}}}} // in the global array
{def !n.get // getting a value
{lambda {:n}
{A.get :n {if {W.equal? {A.get :n {!n.mem}} undefined} // if it doesn't exist
then {!n.set :n} // then compute it
else {!n.mem}}}}} // else get it from the global array
{lambda {:n} // the main function's body
{if {< :n 2} // if n=0 and n=1
then :n // then return 0 or 1
else {S.reduce long_add // apply add_long to
1 {S.map !n.get {S.serie 1 {- :n 1}}}}}}} // the sequence of computed values
2) the task
A) computing !n from 2 to 10 takes about 4ms
{S.map {lambda {:n} {br}!n(:n) = {!n :n}}
{S.serie 0 10}}
-> !n(0) = 0 !n(1) = 1 !n(2) = 2 !n(3) = 4 !n(4) = 10 !n(5) = 34 !n(6) = 154 !n(7) = 874 !n(8) = 5914 !n(9) = 46234 !n(10) = 409114
B) computing !n from 20 to 110 with step 10 takes about 25ms
{S.map {lambda {:n} {br}!n(:n) = {!n :n}}
{S.serie 20 110 10}}
-> !n(20) = 128425485935180314 !n(30) = 9157958657951075573395300940314 !n(40) = 20935051082417771847631371547939998232420940314 !n(50) = 620960027832821612639424806694551108812720525606160920420940314 !n(60) = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !n(70) = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !n(80) = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !n(90) = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !n(100) = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !n(110) = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
C) computing !n from 1000 to 1000 with step 1000 takes about 87 seconds
1) compute n! from 1 to 10000 and store in MEM // about 86 seconds {def foo {!n 10000}} -> foo
2) take 10 values from MEM // about 1ms {S.map {lambda {:n}
{br}Digits of !n(:n) = {W.length {A.get {- :n 1} {MEM}}}}
{S.serie 1000 10000 1000}}
-> Digits of !n(1000) = 2565 Digits of !n(2000) = 5733 Digits of !n(3000) = 9128 Digits of !n(4000) = 12670 Digits of !n(5000) = 16322 Digits of !n(6000) = 20062 Digits of !n(7000) = 23875 Digits of !n(8000) = 27749 Digits of !n(9000) = 31678 Digits of !n(10000) = 35656
</lang>
Lua
Takes about five seconds... <lang Lua>-- Lua bindings for GNU bc require("bc")
-- Return table of factorials from 0 to n function facsUpTo (n)
local f, fList = bc.number(1), {}
fList[0] = 1
for i = 1, n do
f = bc.mul(f, i)
fList[i] = f
end
return fList
end
-- Return left factorial of n function leftFac (n)
local sum = bc.number(0) for k = 0, n - 1 do sum = bc.add(sum, facList[k]) end return bc.tostring(sum)
end
-- Main procedure facList = facsUpTo(10000) for i = 0, 10 do print("!" .. i .. " = " .. leftFac(i)) end for i = 20, 110, 10 do print("!" .. i .. " = " .. leftFac(i)) end for i = 1000, 10000, 1000 do
print("!" .. i .. " contains " .. #leftFac(i) .. " digits")
end</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 contains 2565 digits !2000 contains 5733 digits !3000 contains 9128 digits !4000 contains 12670 digits !5000 contains 16322 digits !6000 contains 20062 digits !7000 contains 23875 digits !8000 contains 27749 digits !9000 contains 31678 digits !10000 contains 35656 digits
Maple
<lang Maple>left_factorial := n -> sum(k!, k = 1 .. n - 1); seq(left_factorial(i), i = 1 .. 10); seq(left_factorial(i), i = 20 .. 110, 10); seq(length(left_factorial(i)), i = 1000 .. 10000, 1000);</lang>
- Output:
0, 1, 3, 9, 33, 153, 873, 5913, 46233, 409113 128425485935180313, 9157958657951075573395300940313, 20935051082417771847631371547939998232420940313, 620960027832821612639424806694551108812720525606160920420940313, 141074930726669571000530822087000522211656242116439949000980378746128920420940313, 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940313, 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940313, 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940313, 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940313, 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940313 2565, 5733, 9128, 12670, 16322, 20062, 23875, 27749, 31678, 35656
Mathematica
<lang Mathematica>left[n_] := left[n] = Sum[k!, {k, 0, n - 1}] Print["left factorials 0 through 10:"] Print[left /@ Range[0, 10] // TableForm] Print["left factorials 20 through 110, by tens:"] Print[left /@ Range[20, 110, 10] // TableForm] Print["Digits in left factorials 1,000 through 10,000, by thousands:"] Print[Length[IntegerDigits[left[#]]] & /@ Range[1000, 10000, 1000] // TableForm]</lang>
- Output:
left factorials 0 through 10: 0 1 2 4 10 34 154 874 5914 46234 409114 left factorials 20 through 110, by tens: 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 Digits in left factorials 1,000 through 10,000, by thousands: 2565 5733 9128 12670 16322 20062 23875 27749 31678 35656
Nim
<lang nim>import iterutils, bigints
proc lfact: iterator: BigInt =
result = iterator: BigInt =
yield 0.initBigInt
var
fact = 1.initBigInt
sum = 0.initBigInt
n = 1.initBigInt
while true:
sum += fact
fact *= n
n += 1
yield sum
echo "first 11:\n " for i in lfact().slice(last = 10):
echo " ", i
echo "20 through 110 (inclusive) by tens:" for i in lfact().slice(20, 110, 10):
echo " ", i
echo "Digits in 1,000 through 10,000 (inclusive) by thousands:" for i in lfact().slice(1_000, 10_000, 1_000):
echo " ", ($i).len</lang>
- Output:
first 11: 0 1 2 4 10 34 154 874 5914 46234 409114 20 through 110 (inclusive) by tens: 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 Digits in 1,000 through 10,000 (inclusive) by thousands: 2565 5733 9128 12670 16322 20062 23875 27749 31678 35656
Oforth
<lang Oforth>: leftFact | i | 0 1 rot loop: i [ tuck + swap i * ] drop ;</lang>
- Output:
>seqFrom(0, 10) map(#leftFact) println [0, 1, 2, 4, 10, 34, 154, 874, 5914, 46234, 409114]
>seqFrom(2, 11) apply(#[ 10 * leftFact println ]) 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
>seq(10) map(#[ 1000 * leftFact asString size ]) println [2565, 5733, 9128, 12670, 16322, 20062, 23875, 27749, 31678, 35656]
PARI/GP
<lang parigp>lf(n)=sum(k=0,n-1,k!); apply(lf, [0..10]) apply(lf, 10*[2..11]) forstep(n=1000,1e4,1000,print1(#digits(lf(n))", "))</lang>
- Output:
%1 = [0, 1, 2, 4, 10, 34, 154, 874, 5914, 46234, 409114] %2 = [128425485935180314, 9157958657951075573395300940314, 20935051082417771847631371547939998232420940314, 620960027832821612639424806694551108812720525606160920420940314, 141074930726669571000530822087000522211656242116439949000980378746128920420940314, 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314, 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314, 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314, 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314, 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314] 2565, 5733, 9128, 12670, 16322, 20062, 23875, 27749, 31678, 35656,
Perl
By caching the last used factorial and left factorial values, I avoid needless recomputation. By only retaining the most recently used values, instead of all past values, I avoid the need to store twenty thousand enormous numbers.
If performance is a concern, this will run over 100x faster by replacing the line "use bigint" with "use Math::GMP qw/:constant/" (after installing that module).
<lang perl>#!perl use 5.010; use strict; use warnings; use bigint;
sub leftfact { my ($n) = @_; state $cached = 0; state $factorial = 1; state $leftfact = 0; if( $n < $cached ) { ($cached, $factorial, $leftfact) = (0, 1, 0); } while( $n > $cached ) { $leftfact += $factorial; $factorial *= ++$cached; } return $leftfact; }
printf "!%d = %s\n", $_, leftfact($_) for 0 .. 10, map $_*10, 2..11; printf "!%d has %d digits.\n", $_, length leftfact($_) for map $_*1000, 1..10;
</lang>
Since I copied the printf format strings from the perl6 implementation, the output from the code above is identical to the output of the perl6 code.
Perl 6
Implement left factorial as a prefix !. Note that this redefines the core prefix ! (not) function.
<lang perl6>sub prefix:<!> ($k) { (constant l = 0, |[\+] 1, (|[\*] 1..*))[$k] }
$ = !10000; # Pre-initialize
.say for ( 0 … 10, 20 … 110 ).hyper(:4batch).map: { sprintf "!%d = %s", $_, !$_ }; .say for (1000, 2000 … 10000).hyper(:4batch).map: { sprintf "!%d has %d digits.", $_, chars !$_ };</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits. !2000 has 5733 digits. !3000 has 9128 digits. !4000 has 12670 digits. !5000 has 16322 digits. !6000 has 20062 digits. !7000 has 23875 digits. !8000 has 27749 digits. !9000 has 31678 digits. !10000 has 35656 digits.
If you would rather not override prefix ! operator and you can live with just defining lazy lists and indexing into them, this should suffice; (and is in fact very slightly faster than the first example since it avoids routine dispatch overhead): <lang perl6>constant leftfact = 0, |[\+] 1, (|[\*] 1..*);
$ = leftfact[10000]; # Pre-initialize
.say for ( 0 … 10, 20 … 110 ).hyper(:4batch).map: { sprintf "!%d = %s", $_, leftfact[$_] }; .say for (1000, 2000 … 10000).hyper(:4batch).map: { sprintf "!%d has %d digits.", $_, chars leftfact[$_] };</lang>
Same output.
Phix
(now over 1500 times faster than the previous bigatom version.) <lang Phix>include mpfr.e
sequence lf_list
procedure init(integer n)
mpz f = mpz_init(1)
lf_list = repeat(f,n+1)
for i=1 to n do
f = mpz_init_set(f)
mpz_mul_si(f,f,i)
lf_list[i+1] = f
end for
end procedure
function lf(integer n, bool len=false) -- Returns left factorial of n, or it's length, as a string
mpz sumf = mpz_init(0)
for k=1 to n do mpz_add(sumf,sumf,lf_list[k]) end for
return iff(len?sprintf("%d",mpz_sizeinbase(sumf,10))
:mpz_get_str(sumf))
end function
-- Main procedure atom t0 = time() init(10000) for i=0 to 10 do printf(1,"!%d = %s\n",{i,lf(i)}) end for for i=20 to 110 by 10 do printf(1,"!%d = %s\n",{i,lf(i)}) end for for i=1000 to 10000 by 1000 do printf(1,"!%d contains %s digits\n",{i,lf(i,true)}) end for printf(1,"complete (%3.2fs)\n",{time()-t0})</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 contains 2565 digits !2000 contains 5733 digits !3000 contains 9128 digits !4000 contains 12670 digits !5000 contains 16323 digits !6000 contains 20062 digits !7000 contains 23875 digits !8000 contains 27749 digits !9000 contains 31678 digits !10000 contains 35656 digits complete (0.25s)
PicoLisp
<lang PicoLisp>(de n! (N)
(cache '(NIL) N
(if (> 2 N) 1
(* N (n! (dec N))))))
(de !n (Num)
(if (= Num 0) 1
(sum n! (range 0 (dec Num)))))
(de pril (List) (mapcar 'println List))
(prinl "0-10") (pril (mapcar '!n (range 0 10))) (prinl "20 - 110") (pril (mapcar '!n (range 20 110 10))) (prinl "length of 1000 - 10000") (pril (mapcar 'length (mapcar '!n (range 1000 10000 1000)))) </lang>
- Output:
<lang>0-10 1 1 2 4 10 34 154 874 5914 46234 409114 20 - 110 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 1000 - 10000 2565 5733 9128 12670 16322 20062 23875 27749 31678 35656 </lang>
PL/I
In PL/I the biggest integer type is fixed decimal(31) i.e. 31 digits. To the best of my knowledge, no big integers exist. Results are shown for the first 11 integers, as required; then for the integers from 20 through 30 only, because factorials for n = 40 and larger are not possible. <lang pli>lf: procedure (n) returns (fixed decimal (31) );
declare n fixed binary; declare (s, f) fixed (31); declare (i, j) fixed;
s = 0;
do i = n-1 to 0 by -1;
f = 1;
do j = i to 1 by -1;
f = f * j;
end;
s = s + f;
end;
return (s);
end lf;
declare n fixed binary;
do n = 0 to 10, 20 to 30;
put skip list ('Left factorial of ' || n || '=' || lf(n) );
end;
end left_factorials;</lang>
- Output:
Left factorial of 0= 0 Left factorial of 1= 1 Left factorial of 2= 2 Left factorial of 3= 4 Left factorial of 4= 10 Left factorial of 5= 34 Left factorial of 6= 154 Left factorial of 7= 874 Left factorial of 8= 5914 Left factorial of 9= 46234 Left factorial of 10= 409114 Left factorial of 20= 128425485935180314 Left factorial of 21= 2561327494111820314 Left factorial of 22= 53652269665821260314 Left factorial of 23= 1177652997443428940314 Left factorial of 24= 27029669736328405580314 Left factorial of 25= 647478071469567844940314 Left factorial of 26= 16158688114800553828940314 Left factorial of 27= 419450149241406189412940314 Left factorial of 28= 11308319599659758350180940314 Left factorial of 29= 316196664211373618851684940314 Left factorial of 30= 9157958657951075573395300940314
PowerShell
<lang PowerShell> function left-factorial ([BigInt]$n) {
[BigInt]$k, [BigInt]$fact = ([BigInt]::Zero), ([BigInt]::One)
[BigInt]$lfact = ([BigInt]::Zero)
while($k -lt $n){
if($k -gt ([BigInt]::Zero)) {
$fact = [BigInt]::Multiply($fact, $k)
$lfact = [BigInt]::Add($lfact, $fact)
} else {
$lfact = ([BigInt]::One)
}
$k = [BigInt]::Add($k, [BigInt]::One)
}
$lfact
} 0..9 | foreach{
"!$_ = $(left-factorial $_)"
} for($i = 10; $i -le 110; $i += 10) {
"!$i = $(left-factorial $i)"
} for($i = 1000; $i -le 10000; $i += 1000) {
$digits = [BigInt]::Log10($(left-factorial $i))
$digits = [Math]::Floor($digits) + 1
if($digits -gt 1) {"!$i has $digits digits"}
else {"!$i has $digits digit"}
} </lang> Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 9060895879876953465345168046502906376940248300119563651843276746197520942896963148820085319918409223365289204 20940314 !90 = 1669557007262421076703416768839462336073351516357586413634591033592403996240486951022572307223584266878750799 3136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203 520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337 422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
Python
<lang python>from itertools import islice
def lfact():
yield 0
fact, summ, n = 1, 0, 1
while 1:
fact, summ, n = fact*n, summ + fact, n + 1
yield summ
print('first 11:\n %r' % [lf for i, lf in zip(range(11), lfact())]) print('20 through 110 (inclusive) by tens:') for lf in islice(lfact(), 20, 111, 10):
print(lf)
print('Digits in 1,000 through 10,000 (inclusive) by thousands:\n %r'
% [len(str(lf)) for lf in islice(lfact(), 1000, 10001, 1000)] )</lang>
- Output:
first 11: [0, 1, 2, 4, 10, 34, 154, 874, 5914, 46234, 409114] 20 through 110 (inclusive) by tens: 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 Digits in 1,000 through 10,000 (inclusive) by thousands: [2565, 5733, 9128, 12670, 16322, 20062, 23875, 27749, 31678, 35656]
Or, sidestepping the use of while and yield, we can directly define left factorials in terms of the scanl abstraction
(a fold or catamorphism, like functools.reduce, but one which returns the whole accumulation of intermediate values – see, for example, The Algebra of Programming, Bird and de Moor, 1997).
scanl in turn, has a natural definition in terms of the itertools functions accumulate and chain.
<lang python>"""Left factorials"""
from itertools import (accumulate, chain, count, islice) from operator import (mul, add)
- leftFact :: [Integer]
def leftFact():
Left factorial series defined in terms of the factorial series
return scanl(add)(0)(
fact()
)
- fact :: [Integer]
def fact():
Factorial series – a non-finite list
return scanl(mul)(1)(
enumFrom(1)
)
- TEST ----------------------------------------------------
- main :: IO ()
def main():
Tests
print(
'Terms 0 thru 10 inclusive:\n %r'
% take(11)(leftFact())
)
print('\nTerms 20 thru 110 (inclusive) by tens:')
for x in takeFromThenTo(20)(30)(110)(leftFact()):
print(x)
print(
'\n\nDigit counts for terms 1k through 10k (inclusive) by k:\n %r'
% list(map(
compose(len)(str),
takeFromThenTo(1000)(2000)(10000)(
leftFact()
)
))
)
- GENERIC -------------------------------------------------
- compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
Function composition. return lambda f: lambda x: g(f(x))
- enumFrom :: Enum a => a -> [a]
def enumFrom(x):
A non-finite stream of enumerable values,
starting from the given value.
return count(x) if isinstance(x, int) else (
map(chr, count(ord(x)))
)
- scanl :: (b -> a -> b) -> b -> [a] -> [b]
def scanl(f):
scanl is like reduce, but returns a succession of
intermediate values, building from the left.
return lambda a: lambda xs: (
accumulate(chain([a], xs), f)
)
- take :: Int -> [a] -> [a]
- take :: Int -> String -> String
def take(n):
The prefix of xs of length n,
or xs itself if n > length xs
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else list(islice(xs, n))
)
- takeFromThenTo :: Int -> Int -> Int -> [a] -> [a]
def takeFromThenTo(a):
Values drawn from a series betweens positions a and b
at intervals of size z
return lambda b: lambda z: lambda xs: islice(
xs, a, 1 + z, b - a
)
if __name__ == '__main__':
main()</lang>
- Output:
Terms 0 thru 10 inclusive: [0, 1, 2, 4, 10, 34, 154, 874, 5914, 46234, 409114] Terms 20 thru 110 (inclusive) by tens: 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 Digit counts for terms 1k through 10k (inclusive) by k: [2565, 5733, 9128, 12670, 16322, 20062, 23875, 27749, 31678, 35656]
R
<lang rsplus> library(gmp)
left_factorial <- function(n) { if (n == 0) return(0) result <- as.bigz(0)
adder <- as.bigz(1) for (k in 1:n) { result <- result + adder adder <- adder * k } result }
digit_count <- function(n) { nchar(as.character(n)) }
for (n in 0:10) { cat("!",n," = ",sep = "") cat(as.character(left_factorial(n))) cat("\n") } for (n in seq(20,110,10)) { cat("!",n," = ",sep = "") cat(as.character(left_factorial(n))) cat("\n") } for (n in seq(1000,10000,1000)) {
cat("!",n," has ",digit_count(left_factorial(n))," digits\n", sep = "")
} </lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
Racket
<lang racket>#lang racket (define ! (let ((rv# (make-hash))) (λ (n) (hash-ref! rv# n (λ () (if (= n 0) 1 (* n (! (- n 1)))))))))
(define (!n n)
;; note that in-range n is from 0 to n-1 inclusive (for/sum ((k (in-range n))) (! k)))
(define (dnl. s) (for-each displayln s)) (dnl
"Display the left factorials for:" "zero through ten (inclusive)" (pretty-format (for/list ((i (in-range 0 (add1 10)))) (!n i))) "20 through 110 (inclusive) by tens" (pretty-format (for/list ((i (in-range 20 (add1 110) 10))) (!n i))) "Display the length (in decimal digits) of the left factorials for:" "1,000, 2,000 through 10,000 (inclusive), by thousands." (pretty-format (for/list ((i (in-range 1000 10001 1000))) (add1 (order-of-magnitude (!n i))))))</lang>
- Output:
Display the left factorials for: zero through ten (inclusive) '(0 1 2 4 10 34 154 874 5914 46234 409114) 20 through 110 (inclusive) by tens '(128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314) Display the length (in decimal digits) of the left factorials for: 1,000, 2,000 through 10,000 (inclusive), by thousands. '(2565 5733 9128 12670 16322 20062 23875 27749 31678 35656)
REXX
<lang rexx>/*REXX program computes/display the left factorial (or its width) of N (or range). */ parse arg bot top inc . /*obtain optional argumenst from the CL*/ if bot== | bot=="," then bot= 1 /*Not specified: Then use the default.*/ if top== | top=="," then top=bot /* " " " " " " */ if inc= | inc=="," then inc= 1 /* " " " " " " */ tellDigs= (bot<0) /*if BOT < 0, only show # of digits. */ bot=abs(bot) /*use the │bot│ for the DO loop. */ @= 'left ! of ' /*a handy literal used in the display. */ w=length(H) /*width of the largest number request. */
do j=bot to top by inc /*traipse through the numbers requested*/
if tellDigs then say @ right(j,w) " ───► " length(L!(j)) ' digits'
else say @ right(j,w) " ───► " L!(j)
end /*j*/ /* [↑] show either L! or # of digits*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ L!: procedure; parse arg x .; if x<3 then return x; s=4 /*some shortcuts. */ !=2; do f=3 to x-1 /*compute L! for all numbers ─── ► X.*/
!=!*f /*compute intermediate factorial. */
if pos(.,!)\==0 then numeric digits digits()*1.5%1 /*bump decimal digits.*/
s=s+! /*add the factorial ───► L! sum. */
end /*f*/ /* [↑] handles gihugeic numbers. */
return s /*return the sum (L!) to the invoker.*/</lang> output when using the input: 0 10
left ! of 0 ───► 0 left ! of 1 ───► 1 left ! of 2 ───► 2 left ! of 3 ───► 4 left ! of 4 ───► 10 left ! of 5 ───► 34 left ! of 6 ───► 154 left ! of 7 ───► 874 left ! of 8 ───► 5914 left ! of 9 ───► 46234 left ! of 10 ───► 409114
output when using the input: 20 110 10
left ! of 20 ───► 128425485935180314 left ! of 30 ───► 9157958657951075573395300940314 left ! of 40 ───► 20935051082417771847631371547939998232420940314 left ! of 50 ───► 620960027832821612639424806694551108812720525606160920420940314 left ! of 60 ───► 141074930726669571000530822087000522211656242116439949000980378746128920420940314 left ! of 70 ───► 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 left ! of 80 ───► 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 left ! of 90 ───► 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 left ! of 100 ───► 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 left ! of 110 ───► 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314
output when using the input: -1000 10000 1000
left ! of 1000 ───► 2565 digits left ! of 2000 ───► 5733 digits left ! of 3000 ───► 9128 digits left ! of 4000 ───► 12670 digits left ! of 5000 ───► 16322 digits left ! of 6000 ───► 20062 digits left ! of 7000 ───► 23875 digits left ! of 8000 ───► 27749 digits left ! of 9000 ───► 31678 digits left ! of 10000 ───► 35656 digits
Ring
<lang ring> a = leftFact(0,10,1) see "" + a + nl
func leftFact f,t,s
see "------ From " + f + " --To -> " + t +" Step " + s + " -------" + nl
for i = f to t step s
leftFact = 1
fct = 1
for j = 1 to i - 1
fct = fct * j
leftFact = leftFact + fct
next
if i >= 1000 see "" + i + " " + len(string(leftFact)) + " digits" + nl
else see "" + i + " " + leftFact + nl ok
next
</lang>
Ruby
<lang ruby>left_fact = Enumerator.new do |y|
f, lf = 1, 0 1.step do |n| y << lf #yield left_factorial lf += f f *= n end
end</lang> Test: <lang ruby>tens = 20.step(110, 10) thousands = 1000.step(10_000, 1000)
10001.times do |n|
lf = left_fact.next
case n
when 0..10, *tens
puts "!#{n} = #{lf}"
when *thousands
puts "!#{n} has #{lf.to_s.size} digits"
end
end</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
Run BASIC
<lang Runbasic>a = lftFct(0,10,1) a = lftFct(20,110,10) a = lftFct(1000,10000,1000)
function lftFct(f,t,s) print :print "------ From ";f;" --To-> ";t;" Step ";s;" -------" for i = f to t step s lftFct = 1 fct = 1 for j = 1 to i-1 fct = fct * j lftFct = lftFct + fct next j if i >= 1000 then print i;" ";len(str$(lftFct));" "digits" else print i;" ";lftFct end if next i end function</lang>Output:
------ From 0 --To-> 10 Step 1 ------- 0 1 1 1 2 2 3 4 4 10 5 34 6 154 7 874 8 5914 9 46234 10 409114 ------ From 20 --To-> 110 Step 10 ------- 20 128425485935180314 30 9157958657951075573395300940314 40 20935051082417771847631371547939998232420940314 50 620960027832821612639424806694551108812720525606160920420940314 60 141074930726669571000530822087000522211656242116439949000980378746128920420940314 70 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 80 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 90 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 100 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 110 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 ------ From 1000 --To-> 10000 Step 1000 ------- 1000 2565 digits 2000 5733 digits 3000 9128 digits 4000 12670 digits 5000 16322 digits 6000 20062 digits 7000 23875 digits 8000 27749 digits 9000 31678 digits 10000 35656 digits
Rust
<lang Rust>
- [cfg(target_pointer_width = "64")]
type USingle = u32;
- [cfg(target_pointer_width = "64")]
type UDouble = u64;
- [cfg(target_pointer_width = "64")]
const WORD_LEN: i32 = 32;
- [cfg(not(target_pointer_width = "64"))]
type USingle = u16;
- [cfg(not(target_pointer_width = "64"))]
type UDouble = u32;
- [cfg(not(target_pointer_width = "64"))]
const WORD_LEN: i32 = 16;
use std::cmp;
- [derive(Debug,Clone)]
struct BigNum {
// rep_.size() == 0 if and only if the value is zero. // Otherwise, the word rep_[0] keeps the least significant bits. rep_: Vec<USingle>,
}
impl BigNum {
pub fn new(n: USingle) -> BigNum {
let mut result = BigNum { rep_: vec![] };
if n > 0 { result.rep_.push(n); }
result
}
pub fn equals(&self, n: USingle) -> bool {
if n == 0 { return self.rep_.is_empty() }
if self.rep_.len() > 1 { return false }
self.rep_[0] == n
}
pub fn add_big(&self, addend: &BigNum) -> BigNum {
let mut result = BigNum::new(0);
let mut sum = 0 as UDouble;
let sz1 = self.rep_.len();
let sz2 = addend.rep_.len();
for i in 0..cmp::max(sz1, sz2) {
if i < sz1 { sum += self.rep_[i] as UDouble }
if i < sz2 { sum += addend.rep_[i] as UDouble }
result.rep_.push(sum as USingle);
sum >>= WORD_LEN;
}
if sum > 0 { result.rep_.push(sum as USingle) }
result
}
pub fn multiply(&self, factor: USingle) -> BigNum {
let mut result = BigNum::new(0);
let mut product = 0 as UDouble;
for i in 0..self.rep_.len() {
product += self.rep_[i] as UDouble * factor as UDouble;
result.rep_.push(product as USingle);
product >>= WORD_LEN;
}
if product > 0 {
result.rep_.push(product as USingle);
}
result
}
pub fn divide(&self, divisor: USingle, quotient: &mut BigNum,
remainder: &mut USingle) {
quotient.rep_.truncate(0);
let mut dividend: UDouble;
*remainder = 0;
for i in 0..self.rep_.len() {
let j = self.rep_.len() - 1 - i;
dividend = ((*remainder as UDouble) << WORD_LEN)
+ self.rep_[j] as UDouble;
let quo = (dividend / divisor as UDouble) as USingle;
*remainder = (dividend % divisor as UDouble) as USingle;
if quo > 0 || j < self.rep_.len() - 1 {
quotient.rep_.push(quo);
}
}
quotient.rep_.reverse();
}
fn to_string(&self) -> String {
let mut rep = String::new();
let mut dividend = (*self).clone();
let mut remainder = 0 as USingle;
let mut quotient = BigNum::new(0);
loop {
dividend.divide(10, &mut quotient, &mut remainder);
rep.push(('0' as USingle + remainder) as u8 as char);
if quotient.equals(0) { break; }
dividend = quotient.clone();
}
rep.chars().rev().collect::<String>()
}
}
use std::fmt; impl fmt::Display for BigNum {
fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
write!(f, "{}", self.to_string())
}
}
fn lfact(n: USingle) -> BigNum {
let mut result = BigNum::new(0);
let mut f = BigNum::new(1);
for k in 1 as USingle..n + 1 {
result = result.add_big(&f);
f = f.multiply(k);
}
result
}
fn main() {
for i in 0..11 {
println!("!{} = {}", i, lfact(i));
}
for i in 2..12 {
let j = i * 10;
println!("!{} = {}", j, lfact(j));
}
for i in 1..11 {
let j = i * 1000;
println!("!{} has {} digits.", j, lfact(j).to_string().len());
}
} </lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits. !2000 has 5733 digits. !3000 has 9128 digits. !4000 has 12670 digits. !5000 has 16322 digits. !6000 has 20062 digits. !7000 has 23875 digits. !8000 has 27749 digits. !9000 has 31678 digits. !10000 has 35656 digits.
Scala
<lang scala>object LeftFactorial extends App {
// this part isn't really necessary, it just shows off Scala's ability
// to match the mathematical syntax: !n
implicit class RichInt(n:Int) {
def unary_!() = factorial.take(n).sum
}
val factorial: Stream[BigInt] = 1 #:: factorial.zip(Stream.from(1)).map(n => n._2 * factorial(n._2 - 1))
for (n <- (0 to 10) ++
(20 to 110 by 10);
value = !n) {
println(s"!${n} = ${value}")
}
for (n <- 1000 to 10000 by 1000;
length = (!n).toString.length) {
println(s"length !${n} = ${length}")
}
} </lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 length !1000 = 2565 length !2000 = 5733 length !3000 = 9128 length !4000 = 12670 length !5000 = 16322 length !6000 = 20062 length !7000 = 23875 length !8000 = 27749 length !9000 = 31678 length !10000 = 35656
Scheme
This version uses the iota method in the standard lists library. iota takes three values, a count, an optional start value (defaults to 0), and an optional step value (defaults to 1) so (iota 5) produces a list (0 1 2 3 4) and (iota 5 100 2) produces a list (100 102 104 106 108)
<lang scheme> (import (scheme base) ;; library imports in R7RS style
(scheme write)
(srfi 1 lists))
(define (factorial n)
(fold * 1 (iota n 1)))
(define (left-factorial n)
(fold + 0 (map factorial (iota n))))
(define (show i r) ; to pretty print the results
(display "!") (display i) (display " ") (display r) (newline))
- show left factorials for zero through ten (inclusive)
(for-each
(lambda (i) (show i (left-factorial i))) (iota 11))
- show left factorials for 20 through 110 (inclusive) by tens
(for-each
(lambda (i) (show i (left-factorial i))) (iota 10 20 10))
- number of digits in 1000 through 10000 by thousands
(for-each
(lambda (i) (show i (string-length (number->string (left-factorial i))))) (iota 10 1000 1000))
</lang>
Seed7
<lang seed7>$ include "seed7_05.s7i";
include "bigint.s7i";
const func bigInteger: leftFact (in integer: n) is func
result
var bigInteger: leftFact is 0_;
local
var bigInteger: factorial is 1_;
var integer: i is 0;
begin
for i range 1 to n do
leftFact +:= factorial;
factorial *:= bigInteger conv i;
end for;
end func;
const proc: main is func
local
var integer: n is 0;
begin
writeln("First 11 left factorials:");
for n range 0 to 10 do
write(" " <& leftFact(n));
end for;
writeln;
writeln("20 through 110 (inclusive) by tens:");
for n range 20 to 110 step 10 do
writeln(leftFact(n));
end for;
writeln;
writeln("Digits in 1,000 through 10,000 by thousands:");
for n range 1000 to 10000 step 1000 do
writeln(length(str(leftFact(n))));
end for;
writeln;
end func;</lang>
- Output:
First 11 left factorials: 0 1 2 4 10 34 154 874 5914 46234 409114 20 through 110 (inclusive) by tens: 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 Digits in 1,000 through 10,000 by thousands: 2565 5733 9128 12670 16322 20062 23875 27749 31678 35656
Sidef
Built-in: <lang ruby>say 20.of { .left_factorial }</lang>
Straightforward: <lang ruby>func left_factorial(n) {
^n -> sum { _! }
}</lang>
Alternatively, using Range.reduce(): <lang ruby>func left_factorial(n) {
^n -> reduce({ |a,b| a + b! }, 0)
}</lang>
A faster approach: <lang ruby>func left_factorial(n) {
static cached = 0 static factorial = 1 static leftfact = 0
if (n < cached) {
cached = 0
factorial = 1
leftfact = 0
}
while (n > cached) {
leftfact += factorial
factorial *= ++cached
}
leftfact
}</lang>
Completing the task: <lang ruby>for n in (0..10, 20..110 `by` 10) {
printf("!%d = %s\n", n, left_factorial(n))
}
for n in (1000..10000 `by` 1000) {
printf("!%d has %d digits.\n", n, left_factorial(n).len)
}</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits. !2000 has 5733 digits. !3000 has 9128 digits. !4000 has 12670 digits. !5000 has 16322 digits. !6000 has 20062 digits. !7000 has 23875 digits. !8000 has 27749 digits. !9000 has 31678 digits. !10000 has 35656 digits.
Swift
<lang swift>import BigInt
func factorial<T: BinaryInteger>(_ n: T) -> T {
guard n != 0 else {
return 1
}
return stride(from: n, to: 0, by: -1).reduce(1, *)
}
prefix func ! <T: BinaryInteger>(n: T) -> T {
guard n != 0 else {
return 0
}
return stride(from: 0, to: n, by: 1).lazy.map(factorial).reduce(0, +)
}
for i in 0...10 {
print("!\(i) = \(!i)")
}
print()
for i in stride(from: BigInt(20), through: 110, by: 10) {
print("!\(i) = \(!i)")
}
print()
print("!1000 = \((!BigInt(1000)).description.count) digit number")
print()
for i in stride(from: BigInt(2000), through: 10_000, by: 1000) {
print("!\(i) = \((!i).description.count) digit number")
}</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 = 2565 digit number !2000 = 5733 digit number !3000 = 9128 digit number !4000 = 12670 digit number !5000 = 16322 digit number !6000 = 20062 digit number !7000 = 23875 digit number !8000 = 27749 digit number !9000 = 31678 digit number !10000 = 35656 digit number
Standard ML
<lang sml> (* reuse earlier factorial calculations in dfac, apply to listed arguments in cumlfac *) (* example: left factorial n, is #3 (dfac (0,n-1,1,1) ) *) (* output list contains (number, factorial, left factorial) *) (* tested in PolyML *)
val store = ref 0;
val rec dfac = fn
(from,to,acc,cm) => if from = to then (from,acc,cm) else (store:=(from+1)*acc;dfac (from+1,to,!store,!store+cm ) );
val rec cumlfac = fn
(x::y::rm) => x :: cumlfac ( dfac (#1 x, #1 y, #2 x, #3 x) :: rm ) |
rm =>rm ;
val arguments = List.tabulate (10,fn 0=>(0,1,1)|i=>(i,0,0)) @
List.tabulate (10,fn i=> (10*i+19,0,0) ) @
List.tabulate ( 10,fn i=> (1000*i+999,0,0));
val result = (~1,0,0)::(cumlfac arguments);
(* done *) (* display: *)
List.app (fn triple :int*int*int =>
print(Int.toString (1+ #1 triple ) ^ " : " ^ Int.fmt StringCvt.DEC (#3 triple ) ^" \n" )
) (List.take(result,21) ) ;
List.app (fn triple :int*int*int =>
print( Int.toString (1+ #1 triple ) ^ " : " ^ Int.toString (size(Int.toString (#3 triple ))) ^" \n" ) ) (List.drop(result,21) );
</lang>
- Output:
time poly --script thisscript 0 : 0 1 : 1 2 : 2 3 : 4 4 : 10 5 : 34 6 : 154 7 : 874 8 : 5914 9 : 46234 10 : 409114 20 : 128425485935180314 30 : 9157958657951075573395300940314 40 : 20935051082417771847631371547939998232420940314 50 : 620960027832821612639424806694551108812720525606160920420940314 60 : 141074930726669571000530822087000522211656242116439949000980378746128920420940314 70 : 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 80 : 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 90 : 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 100 : 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 110 : 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 1000 : 2565 2000 : 5733 3000 : 9128 4000 : 12670 5000 : 16322 6000 : 20062 7000 : 23875 8000 : 27749 9000 : 31678 10000 : 35656 (CPU 2.1Ghz:) 0.36 real 0.29 user 0.08 sys
Tcl
<lang tcl>proc leftfact {n} {
set s 0
for {set i [set f 1]} {$i <= $n} {incr i} {
incr s $f set f [expr {$f * $i}]
} return $s
}
for {set i 0} {$i <= 110} {incr i [expr {$i>9?10:1}]} {
puts "!$i = [leftfact $i]"
} for {set i 1000} {$i <= 10000} {incr i 1000} {
puts "!$i has [string length [leftfact $i]] digits"
}</lang>
- Output:
!0 = 0 !1 = 1 !2 = 2 !3 = 4 !4 = 10 !5 = 34 !6 = 154 !7 = 874 !8 = 5914 !9 = 46234 !10 = 409114 !20 = 128425485935180314 !30 = 9157958657951075573395300940314 !40 = 20935051082417771847631371547939998232420940314 !50 = 620960027832821612639424806694551108812720525606160920420940314 !60 = 141074930726669571000530822087000522211656242116439949000980378746128920420940314 !70 = 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 !80 = 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 !90 = 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 !100 = 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 !110 = 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 !1000 has 2565 digits !2000 has 5733 digits !3000 has 9128 digits !4000 has 12670 digits !5000 has 16322 digits !6000 has 20062 digits !7000 has 23875 digits !8000 has 27749 digits !9000 has 31678 digits !10000 has 35656 digits
zkl
<lang zkl>var BN=Import("zklBigNum");
fcn leftFact(n){
[1..n].reduce(fcn(p,n,rf){ p+=rf.value; rf.set(rf.value*n); p },
BN(0),Ref(BN(1)));
}</lang> <lang zkl>println("First 11 left factorials:\n", [0..10].apply(leftFact)); lfs:=[20..111,10].apply(leftFact); println(("\n20 through 110 (inclusive) by tens:\n" + "%d\n"*lfs.len()).fmt(lfs.xplode()));
println("Digits in 1,000 through 10,000 by thousands:\n",
[0d1_000..0d10_000, 1000].pump(List,fcn(n){leftFact(n).toString().len()}));</lang>
- Output:
First 11 left factorials: L(0,1,2,4,10,34,154,874,5914,46234,409114) 20 through 110 (inclusive) by tens: 128425485935180314 9157958657951075573395300940314 20935051082417771847631371547939998232420940314 620960027832821612639424806694551108812720525606160920420940314 141074930726669571000530822087000522211656242116439949000980378746128920420940314 173639511802987526699717162409282876065556519849603157850853034644815111221599509216528920420940314 906089587987695346534516804650290637694024830011956365184327674619752094289696314882008531991840922336528920420940314 16695570072624210767034167688394623360733515163575864136345910335924039962404869510225723072235842668787507993136908442336528920420940314 942786239765826579160595268206839381354754349601050974345395410407078230249590414458830117442618180732911203520208889371641659121356556442336528920420940314 145722981061585297004706728001906071948635199234860720988658042536179281328615541936083296163475394237524337422204397431927131629058103519228197429698252556442336528920420940314 Digits in 1,000 through 10,000 by thousands: L(2565,5733,9128,12670,16322,20062,23875,27749,31678,35656)
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