Greatest prime dividing the n-th cubefree number
A cubefree number is a positive integer whose prime factorization does not contain any third (or higher) power factors. If follows that all primes are trivially cubefree and the first cubefree number is 1 because it has no prime factors.
Greatest prime dividing the n-th cubefree number is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
- Definitions
Let a[n] be the greatest prime dividing the n-th cubefree number for n >= 2. By convention, let a[1] = 1 even though the first cubefree number, 1, has no prime factors.
- Examples
a[2] is clearly 2 because it is the second cubefree number and also prime. The fourth cubefree number is 4 and it's highest prime factor is 2, hence a[4] = 2.
- Task
Compute and show on this page the first 100 terms of a[n]. Also compute and show the 1,000th, 10,000th and 100,000th members of the sequence.
- Stretch
Compute and show the 1 millionth and 10 millionth terms of the sequence.
This may take a while for interpreted languages.
- Reference
Dart
import 'dart:math';
void main() {
List<int> res = [1];
int count = 1;
int i = 2;
int lim1 = 100;
int lim2 = 1000;
double max = 1e7;
var t0 = DateTime.now();
while (count < max) {
bool cubeFree = false;
List<int> factors = primeFactors(i);
if (factors.length < 3) {
cubeFree = true;
} else {
cubeFree = true;
for (int i = 2; i < factors.length; i++) {
if (factors[i - 2] == factors[i - 1] && factors[i - 1] == factors[i]) {
cubeFree = false;
break;
}
}
}
if (cubeFree) {
if (count < lim1) res.add(factors.last);
count += 1;
if (count == lim1) {
print("First $lim1 terms of a[n]:");
print(res.take(lim1).join(', '));
print("");
} else if (count == lim2) {
print("The $count term of a[n] is ${factors.last}");
lim2 *= 10;
}
}
i += 1;
}
print("${DateTime.now().difference(t0).inSeconds} sec.");
}
List<int> primeFactors(int n) {
List<int> factors = [];
while (n % 2 == 0) {
factors.add(2);
n ~/= 2;
}
for (int i = 3; i <= sqrt(n); i += 2) {
while (n % i == 0) {
factors.add(i);
n ~/= i;
}
}
if (n > 2) {
factors.add(n);
}
return factors;
}
- Output:
First 100 terms of a[n]: 1, 2, 3, 2, 5, 3, 7, 3, 5, 11, 3, 13, 7, 5, 17, 3, 19, 5, 7, 11, 23, 5, 13, 7, 29, 5, 31, 11, 17, 7, 3, 37, 19, 13, 41, 7, 43, 11, 5, 23, 47, 7, 5, 17, 13, 53, 11, 19, 29, 59, 5, 61, 31, 7, 13, 11, 67, 17, 23, 7, 71, 73, 37, 5, 19, 11, 13, 79, 41, 83, 7, 17, 43, 29, 89, 5, 13, 23, 31, 47, 19, 97, 7, 11, 5, 101, 17, 103, 7, 53, 107, 109, 11, 37, 113, 19, 23, 29, 13, 59 The 1000 term of a[n] is 109 The 10000 term of a[n] is 101 The 100000 term of a[n] is 1693 The 1000000 term of a[n] is 1202057 The 10000000 term of a[n] is 1202057
FreeBASIC
Without using external libraries, it takes about 68 seconds to run on my system (Core i5).
Dim Shared As Integer factors()
Dim As Integer res(101)
res(0) = 1
Dim As Integer count = 1
Dim As Integer j, i = 2
Dim As Integer lim1 = 100
Dim As Integer lim2 = 1000
Dim As Integer max = 1e7
Dim As Integer cubeFree = 0
Sub primeFactors(n As Integer, factors() As Integer)
Dim As Integer i = 2, cont = 2
While (i * i <= n)
While (n Mod i = 0)
n /= i
cont += 1
Redim Preserve factors(1 To cont)
factors(cont) = i
Wend
i += 1
Wend
If (n > 1) Then
cont += 1
Redim Preserve factors(1 To cont)
factors(cont) = n
End If
End Sub
While (count < max)
primeFactors(i, factors())
If (Ubound(factors) < 3) Then
cubeFree = 1
Else
cubeFree = 1
For j = 2 To Ubound(factors)
If (factors(j-2) = factors(j-1) And factors(j-1) = factors(j)) Then
cubeFree = 0
Exit For
End If
Next j
End If
If (cubeFree = 1) Then
If (count < lim1) Then
res(count) = factors(Ubound(factors))
End If
count += 1
If (count = lim1) Then
Print "First "; lim1; " terms of a[n]:"
For j = 1 To lim1
Print Using "####"; res(j-1);
If (j Mod 10 = 0) Then Print
Next j
Print
Elseif (count = lim2) Then
Print "The"; count; " term of a[n] is"; factors(Ubound(factors))
lim2 *= 10
End If
End If
i += 1
Wend
Sleep
- Output:
First 100 terms of a[n]: 1 2 3 2 5 3 7 3 5 11 3 13 7 5 17 3 19 5 7 11 23 5 13 7 29 5 31 11 17 7 3 37 19 13 41 7 43 11 5 23 47 7 5 17 13 53 11 19 29 59 5 61 31 7 13 11 67 17 23 7 71 73 37 5 19 11 13 79 41 83 7 17 43 29 89 5 13 23 31 47 19 97 7 11 5 101 17 103 7 53 107 109 11 37 113 19 23 29 13 59 The 1000th term of a[n] is 109 The 10000th term of a[n] is 101 The 100000th term of a[n] is 1693 The 1000000th term of a[n] is 1202057 The 10000000th term of a[n] is 1202057
Pascal
Free Pascal
Uses sieving with cube powers of primes.
Nearly linear runtime. 0.8s per Billion. Highest Prime for 1E9 is 997
program CubeFree2;
{$IFDEF FPC}
{$MODE DELPHI}
{$OPTIMIZATION ON,ALL}
//{$CODEALIGN proc=16,loop=8} //TIO.RUN (Intel Xeon 2.3 Ghz) loves it
{$COPERATORS ON}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
uses
sysutils
{$IFDEF WINDOWS},Windows{$ENDIF}
;
const
SizeCube235 =4* (2*2*2* 3*3*3 *5*5*5);//2*27000 <= 64kb level I
type
tPrimeIdx = 0..65535;
tPrimes = array[tPrimeIdx] of Uint32;
tSv235IDx = 0..SizeCube235-1;
tSieve235 = array[tSv235IDx] of boolean;
tDl3 = record
dlPr3 : UInt64;
dlSivMod,
dlSivNum : Uint32;
end;
tDelCube = array[tPrimeIdx] of tDl3;
var
{$ALIGN 8}
SmallPrimes: tPrimes;
Sieve235,
Sieve : tSieve235;
DelCube : tDelCube;
procedure InitSmallPrimes;
//get primes. #0..65535.Sieving only odd numbers
const
MAXLIMIT = (821641-1) shr 1;
var
pr : array[0..MAXLIMIT] of byte;
p,j,d,flipflop :NativeUInt;
Begin
SmallPrimes[0] := 2;
fillchar(pr[0],SizeOf(pr),#0);
p := 0;
repeat
repeat
p +=1
until pr[p]= 0;
j := (p+1)*p*2;
if j>MAXLIMIT then
BREAK;
d := 2*p+1;
repeat
pr[j] := 1;
j += d;
until j>MAXLIMIT;
until false;
SmallPrimes[1] := 3;
SmallPrimes[2] := 5;
j := 3;
d := 7;
flipflop := (2+1)-1;//7+2*2,11+2*1,13,17,19,23
p := 3;
repeat
if pr[p] = 0 then
begin
SmallPrimes[j] := d;
inc(j);
end;
d += 2*flipflop;
p+=flipflop;
flipflop := 3-flipflop;
until (p > MAXLIMIT) OR (j>High(SmallPrimes));
end;
procedure Init235(var Sieve235:tSieve235);
var
i,j,k : NativeInt;
begin
fillchar(Sieve235,SizeOf(Sieve235),Ord(true));
Sieve235[0] := false;
for k in [2,3,5] do
Begin
j := k*k*k;
i := j;
while i < SizeCube235 do
begin
Sieve235[i] := false;
inc(i,j);
end;
end;
end;
procedure InitDelCube(var DC:tDelCube);
var
i,q,r : Uint64;
begin
for i in tPrimeIdx do
begin
q := SmallPrimes[i];
q *= sqr(q);
with DC[i] do
begin
dlPr3 := q;
r := q div SizeCube235;
dlSivNum := r;
dlSivMod := q-r*SizeCube235;
end;
end;
end;
function Numb2USA(n:Uint64):Ansistring;
var
pI :pChar;
i,j : NativeInt;
Begin
str(n,result);
i := length(result);
//extend s by the count of comma to be inserted
j := i+ (i-1) div 3;
if i<> j then
Begin
setlength(result,j);
pI := @result[1];
dec(pI);
while i > 3 do
Begin
//copy 3 digits
pI[j] := pI[i];pI[j-1] := pI[i-1];pI[j-2] := pI[i-2];
// insert comma
pI[j-3] := ',';
dec(i,3);
dec(j,4);
end;
end;
end;
function highestDiv(n: uint64):Uint64;
//can test til 821641^2 ~ 6,75E11
var
pr : Uint64;
i : integer;
begin
result := n;
i := 0;
repeat
pr := Smallprimes[i];
if pr*pr>result then
BREAK;
while (result > pr) AND (result MOD pr = 0) do
result := result DIV pr;
inc(i);
until i > High(SmallPrimes);
end;
procedure OutNum(lmt,n:Uint64);
begin
writeln(Numb2Usa(lmt):18,Numb2Usa(n):19,Numb2Usa(highestDiv(n)):18);
end;
var
sieveNr,minIdx,maxIdx : Uint32;
procedure SieveOneSieve;
var
j : Uint64;
i : Uint32;
begin
// sieve with previous primes
Sieve := Sieve235;
For i := minIdx to MaxIdx do
with DelCube[i] do
if dlSivNum = sievenr then
begin
j := dlSivMod;
repeat
sieve[j] := false;
inc(j,dlPr3);
until j >= SizeCube235;
dlSivMod := j Mod SizeCube235;
dlSivNum += j div SizeCube235;
end;
//sieve with new primes
while DelCube[maxIdx+1].dlSivNum = sieveNr do
begin
inc(maxIdx);
with DelCube[maxIdx] do
begin
j := dlSivMod;
repeat
sieve[j] := false;
inc(j,dlPr3);
until j >= SizeCube235;
dlSivMod := j Mod SizeCube235;
dlSivNum := sieveNr + j div SizeCube235;
end;
end;
end;
var
T0:Int64;
cnt,lmt : Uint64;
i : integer;
Begin
T0 := GetTickCount64;
InitSmallPrimes;
Init235(Sieve235);
InitDelCube(DelCube);
sieveNr := 0;
minIdx := low(tPrimeIdx);
while Smallprimes[minIdx]<=5 do
inc(minIdx);
MaxIdx := minIdx;
while DelCube[maxIdx].dlSivNum <= sieveNr do
inc(maxIdx);
SieveOneSieve;
i := 1;
cnt := 0;
lmt := 100;
repeat
if sieve[i] then
begin
inc(cnt);
write(highestDiv(i):4);
if cnt mod 10 = 0 then
Writeln;
end;
inc(i);
until cnt = lmt;
Writeln;
cnt := 0;
lmt *=10;
repeat
For i in tSv235IDx do
begin
if sieve[i] then
begin
inc(cnt);
if cnt = lmt then
Begin
OutNum(lmt,i+sieveNr*SizeCube235);
lmt*= 10;
end;
end;
end;
inc(sieveNr);
SieveOneSieve;
until lmt > 1*1000*1000*1000;
T0 := GetTickCount64-T0;
writeln('runtime ',T0/1000:0:3,' s');
end.
- @home:
1 2 3 2 5 3 7 3 5 11
3 13 7 5 17 3 19 5 7 11
23 5 13 7 29 5 31 11 17 7
3 37 19 13 41 7 43 11 5 23
47 7 5 17 13 53 11 19 29 59
5 61 31 7 13 11 67 17 23 7
71 73 37 5 19 11 13 79 41 83
7 17 43 29 89 5 13 23 31 47
19 97 7 11 5 101 17 103 7 53
107 109 11 37 113 19 23 29 13 59
1,000 1,199 109
10,000 12,019 101
100,000 120,203 1,693
1,000,000 1,202,057 1,202,057
10,000,000 12,020,570 1,202,057
100,000,000 120,205,685 20,743
1,000,000,000 1,202,056,919 215,461 // TIO.RUN 2.4 s
10,000,000,000 12,020,569,022 1,322,977
100,000,000,000 120,205,690,298 145,823
1,000,000,000,000 1,202,056,903,137 400,685,634,379
runtime 805.139 s
real 13m25,140s
Phix
Quite possibly flawed, but it does at least complete to 1e9 in the blink of an eye, and 93s for 1e11.
with javascript_semantics
-- Note this routine had a major hiccup at 27000 = 2^3*3^3*5^3 and another was predicted at 9261000 = 27000*7^3.
-- The "insufficient kludge" noted below makes it match Wren over than limit, but quite possibly only by chance.
-- (it has o/c passed every test I can think of, but the logic of combinations/permutes behind it all eludes me)
function cubes_before(atom n)
-- nb: if n is /not/ cube-free it /is/ included in the result.
-- nth := n-cubes_before(n) means n is the nth cube-free integer,
-- but only if cubes_before(n-1)==cubes_before(n),
-- otherwise cubes_before(cubicate) isn't really very meaningful.
atom r = 0
bool xpm = true -- extend prior multiples
sequence pm = {}
for i=1 to floor(power(n,1/3)) do
atom p3 = power(get_prime(i),3)
if p3>n then exit end if
integer k = floor(n/p3)
for mask=1 to power(2,length(pm))-1 do
integer m = mask, mi = 0, bc = count_bits(mask)
atom kpm = p3
while m do
mi += 1
if odd(m) then
kpm *= pm[mi]
end if
m = floor(m/2)
end while
if kpm>n then
if bc=1 then
xpm = false
pm = pm[1..mi-1]
exit
end if
else
-- subtract? already counted multiples..
integer l = floor(n/kpm)
-- DEV insufficient kludge... (probably)
--if bc=1 then
if odd(bc) then
k -= l
else
k += l
end if
end if
end for
r += k
if xpm then
pm &= p3
end if
end for
return r
end function
function cube_free(atom nth)
-- get the nth cube-free integer
atom lo = nth, hi = lo*2, mid, cb, k
while hi-cubes_before(hi)<nth do
lo = hi
hi = lo*2
end while
-- bisect until we have a possible...
atom t1 = time()+1
while true do
mid = floor((lo+hi)/2)
cb = cubes_before(mid)
k = mid-cb
if k=nth then
-- skip omissions
while cubes_before(mid-1)!=cb do
mid -= 1
cb -= 1
end while
exit
elsif k<nth then
lo = mid
else
hi = mid
end if
if time()>t1 then
progress("bisecting %,d..%,d...\r",{lo,hi})
t1 = time()+1
end if
end while
return mid
end function
function A370833(atom nth)
if nth=1 then return {1,1,1} end if
atom n = cube_free(nth)
sequence f = prime_powers(n)
return {nth,n,f[$][1]}
end function
atom t0 = time()
sequence f100 = vslice(apply(tagset(100),A370833),3)
printf(1,"First 100 terms of a[n]:\n%s\n",join_by(f100,1,10," ",fmt:="%3d"))
for n in sq_power(10,tagset(11,3)) do
printf(1,"The %,dth term of a[n] is %,d with highest divisor %,d.\n",A370833(n))
end for
?elapsed(time()-t0)
- Output:
First 100 terms of a[n]: 1 2 3 2 5 3 7 3 5 11 3 13 7 5 17 3 19 5 7 11 23 5 13 7 29 5 31 11 17 7 3 37 19 13 41 7 43 11 5 23 47 7 5 17 13 53 11 19 29 59 5 61 31 7 13 11 67 17 23 7 71 73 37 5 19 11 13 79 41 83 7 17 43 29 89 5 13 23 31 47 19 97 7 11 5 101 17 103 7 53 107 109 11 37 113 19 23 29 13 59 The 1,000th term of a[n] is 1,199 with highest divisor 109. The 10,000th term of a[n] is 12,019 with highest divisor 101. The 100,000th term of a[n] is 120,203 with highest divisor 1,693. The 1,000,000th term of a[n] is 1,202,057 with highest divisor 1,202,057. The 10,000,000th term of a[n] is 12,020,570 with highest divisor 1,202,057. The 100,000,000th term of a[n] is 120,205,685 with highest divisor 20,743. The 1,000,000,000th term of a[n] is 1,202,056,919 with highest divisor 215,461. The 10,000,000,000th term of a[n] is 12,020,569,022 with highest divisor 1,322,977. The 100,000,000,000th term of a[n] is 120,205,690,298 with highest divisor 145,823. "1 minute and 33s"
Python
This takes under 12 minutes to run on my system (Core i5).
#!/usr/bin/python
from sympy import primefactors
res = [1]
count = 1
i = 2
lim1 = 100
lim2 = 1000
max = 1e7
while count < max:
cubeFree = False
factors = primefactors(i)
if len(factors) < 3:
cubeFree = True
else:
cubeFree = True
for j in range(2, len(factors)):
if factors[j-2] == factors[j-1] and factors[j-1] == factors[j]:
cubeFree = False
break
if cubeFree:
if count < lim1:
res.append(factors[-1])
count += 1
if count == lim1:
print("First {} terms of a[n]:".format(lim1))
for k in range(0, len(res), 10):
print(" ".join(map(str, res[k:k+10])))
print("")
elif count == lim2:
print("The {} term of a[n] is {}".format(count, factors[-1]))
lim2 *= 10
i += 1
- Output:
Similar to FreeBASIC entry.
RPL
I confirm it does take a while for an interpreted language like RPL. Getting the 100,000th term is likely to be a question of hours, even on an emulator.
« 1 { } DUP2 + → n res2 res1
« 2
WHILE 'n' INCR 10000 ≤ REPEAT
WHILE DUP FACTORS DUP 1 « 3 < NSUB 2 MOD NOT OR » DOSUBS ΠLIST NOT
REPEAT DROP 1 + END
HEAD
CASE
n 100 ≤ THEN 'res1' OVER STO+ END
n LOG FP NOT THEN 'res2' OVER STO+ END
END
DROP 1 +
END
DROP res1 res2
» » 'TASK' STO
- Output:
2: { 1 2 3 2 5 3 7 3 5 11 3 13 7 5 17 3 19 5 7 11 23 5 13 7 29 5 31 11 17 7 3 37 19 13 41 7 43 11 5 23 47 7 5 17 13 53 11 19 29 59 5 61 31 7 13 11 67 17 23 7 71 73 37 5 19 11 13 79 41 83 7 17 43 29 89 5 13 23 31 47 19 97 7 11 5 101 17 103 7 53 107 109 11 37 113 19 23 29 13 59 }
1: { 109 101 }
Wren
Version 1
Simple brute force approach though skipping numbers which are multiples of 8 or 27 as these can't be cubefree.
Runtime about 3 minutes 36 seconds on my system (Core i7).
import "./math" for Int
import "./fmt" for Fmt
var res = [1]
var count = 1
var i = 2
var lim1 = 100
var lim2 = 1000
var max = 1e7
while (count < max) {
var cubeFree = false
var factors = Int.primeFactors(i)
if (factors.count < 3) {
cubeFree = true
} else {
cubeFree = true
for (i in 2...factors.count) {
if (factors[i-2] == factors[i-1] && factors[i-1] == factors[i]) {
cubeFree = false
break
}
}
}
if (cubeFree) {
if (count < lim1) res.add(factors[-1])
count = count + 1
if (count == lim1) {
System.print("First %(lim1) terms of a[n]:")
Fmt.tprint("$3d", res, 10)
} else if (count == lim2) {
Fmt.print("\nThe $,r term of a[n] is $,d.", count, factors[-1])
lim2 = lim2 * 10
}
}
i = i + 1
if (i % 8 == 0 || i % 27 == 0) i = i + 1
}
- Output:
First 100 terms of a[n]: 1 2 3 2 5 3 7 3 5 11 3 13 7 5 17 3 19 5 7 11 23 5 13 7 29 5 31 11 17 7 3 37 19 13 41 7 43 11 5 23 47 7 5 17 13 53 11 19 29 59 5 61 31 7 13 11 67 17 23 7 71 73 37 5 19 11 13 79 41 83 7 17 43 29 89 5 13 23 31 47 19 97 7 11 5 101 17 103 7 53 107 109 11 37 113 19 23 29 13 59 The 1,000th term of a[n] is 109. The 10,000th term of a[n] is 101. The 100,000th term of a[n] is 1,693. The 1,000,000th term of a[n] is 1,202,057. The 10,000,000th term of a[n] is 1,202,057.
Version 2
Astonishing speed-up, now taking only 0.04 seconds to reach 10 million and 36.5 seconds to reach 10 billion.
import "./math" for Int
import "./fmt" for Conv, Fmt
import "./iterate" for Stepped
var start = System.clock
var primes = Int.primeSieve(3000)
var countBits = Fn.new { |n| Conv.bin(n).count { |c| c == "1" } }
var cubesBefore = Fn.new { |n|
var r = 0
var xpm = true
var pm = []
for (i in 1..n.cbrt.floor) {
var p3 = primes[i-1].pow(3)
if (p3 > n) break
var k = (n/p3).floor
for (mask in Stepped.ascend(1..2.pow(pm.count)-1)) {
var m = mask
var mi = 0
var bc = countBits.call(mask)
var kpm = p3
while (m != 0) {
mi = mi + 1
if (m % 2 == 1) kpm = kpm * pm[mi-1]
m = (m/2).floor
}
if (kpm > n) {
if (bc == 1) {
xpm = false
pm = pm[0...mi-1]
break
}
} else {
var l = (n/kpm).floor
if (bc % 2 == 1) {
k = k - l
} else {
k = k + l
}
}
}
r = r + k
if (xpm) pm.add(p3)
}
return r
}
var cubeFree = Fn.new { |nth|
var lo = nth
var hi = lo * 2
var mid
var cb
var k
while (hi - cubesBefore.call(hi) < nth) {
lo = hi
hi = lo * 2
}
while (true) {
mid = ((lo + hi)/2).floor
cb = cubesBefore.call(mid)
k = mid - cb
if (k == nth) {
while (cubesBefore.call(mid-1) != cb) {
mid = mid - 1
cb = cb - 1
}
break
} else if (k < nth) {
lo = mid
} else {
hi = mid
}
}
return mid
}
var a370833 = Fn.new { |n|
if (n == 1) return [1, 1]
var nth = cubeFree.call(n)
return [Int.primeFactors(nth)[-1], nth]
}
System.print("First 100 terms of a[n]:")
Fmt.tprint("$3d", (1..100).map { |i| a370833.call(i)[0] }, 10)
System.print()
var n = 1000
while (n <= 1e10) {
var res = a370833.call(n)
Fmt.print("The $,r term of a[n] is $,d for cubefree number $,d.", n, res[0], res[1])
n = n * 10
}
System.print("\nTook %(System.clock - start) seconds.")
- Output:
First 100 terms of a[n]: 1 2 3 2 5 3 7 3 5 11 3 13 7 5 17 3 19 5 7 11 23 5 13 7 29 5 31 11 17 7 3 37 19 13 41 7 43 11 5 23 47 7 5 17 13 53 11 19 29 59 5 61 31 7 13 11 67 17 23 7 71 73 37 5 19 11 13 79 41 83 7 17 43 29 89 5 13 23 31 47 19 97 7 11 5 101 17 103 7 53 107 109 11 37 113 19 23 29 13 59 The 1,000th term of a[n] is 109 for cubefree number 1,199. The 10,000th term of a[n] is 101 for cubefree number 12,019. The 100,000th term of a[n] is 1,693 for cubefree number 120,203. The 1,000,000th term of a[n] is 1,202,057 for cubefree number 1,202,057. The 10,000,000th term of a[n] is 1,202,057 for cubefree number 12,020,570. The 100,000,000th term of a[n] is 20,743 for cubefree number 120,205,685. The 1,000,000,000th term of a[n] is 215,461 for cubefree number 1,202,056,919. The 10,000,000,000th term of a[n] is 1,322,977 for cubefree number 12,020,569,022. Took 36.458647 seconds.