De Polignac numbers
Alphonse de Polignac, a French mathematician in the 1800s, conjectured that every positive odd integer could be formed from the sum of a power of 2 and a prime number.
De Polignac numbers 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.
He was subsequently proved incorrect.
The numbers that fail this condition are now known as de Polignac numbers.
Technically 1 is a de Polignac number, as there is no prime and power of 2 that sum to 1. De Polignac was aware but thought that 1 was a special case. However.
127 is also fails that condition, as there is no prime and power of 2 that sum to 127.
As it turns out, de Polignac numbers are not uncommon, in fact, there are an infinite number of them.
- Task
- Find and display the first fifty de Polignac numbers.
- Stretch
- Find and display the one thousandth de Polignac number.
- Find and display the ten thousandth de Polignac number.
- See also
Action!
which is based on the ALGOL 68 sample.
;;; Find some De Polignac numbers: positive odd numbers that can't be
;;; written as p + 2**n for some prime p and integer n
INCLUDE "H6:SIEVE.ACT"
PROC showDePolignac( CARD dpNumber )
Put(' )
IF dpNumber < 10 THEN Put(' ) FI
IF dpNumber < 100 THEN Put(' ) FI
IF dpNumber < 1000 THEN Put(' ) FI
PrintC( dpNumber )
RETURN
PROC Main()
DEFINE MAX_DP = "4000", MAX_POWER_OF_2 = "11"
BYTE ARRAY primes(MAX_DP+1)
CARD ARRAY powersOf2(MAX_POWER_OF_2+1) =
[1 2 4 8 16 32 64 128 256 512 1024 2048]
CARD i, p, count
BYTE found
Sieve(primes,MAX_DP+1)
; the numbers must be odd and not of the form p + 2**n
; either p is odd and 2**n is even and hence n > 0 and p > 2
; or 2**n is odd and p is even and hence n = 0 and p = 2
; n = 0, p = 2 - the only possibility is 3
FOR i = 1 TO 3 STEP 2 DO
p = 2
IF p + 1 <> i THEN
count ==+ 1
showDePolignac( i )
FI
OD
; n > 0, p > 2
i = 3
WHILE i < MAX_DP AND count < 50 DO
i ==+ 2
found = 0
p = 1
WHILE found = 0 AND p <= MAX_POWER_OF_2 AND i > powersOf2( p ) DO
found = primes( i - powersOf2( p ) )
p ==+ 1
OD
IF found = 0 THEN
count ==+ 1
showDePolignac( i )
IF count MOD 10 = 0 THEN PutE() FI
FI
OD
RETURN- Output:
1 127 149 251 331 337 373 509 599 701 757 809 877 905 907 959 977 997 1019 1087 1199 1207 1211 1243 1259 1271 1477 1529 1541 1549 1589 1597 1619 1649 1657 1719 1759 1777 1783 1807 1829 1859 1867 1927 1969 1973 1985 2171 2203 2213
ALGOL 68
BEGIN # find some De Polignac Numbers - positive odd numbers that can't be #
# written as p + 2^n for some prime p and integer n #
INT max number = 500 000; # maximum number we will consider #
# sieve the primes to max number #
[ 0 : max number ]BOOL prime;
prime[ 0 ] := prime[ 1 ] := FALSE;
prime[ 2 ] := TRUE;
FOR i FROM 3 BY 2 TO UPB prime DO prime[ i ] := TRUE OD;
FOR i FROM 4 BY 2 TO UPB prime DO prime[ i ] := FALSE OD;
FOR i FROM 3 BY 2 TO ENTIER sqrt( UPB prime ) DO
IF prime[ i ] THEN
FOR s FROM i * i BY i + i TO UPB prime DO prime[ s ] := FALSE OD
FI
OD;
# table of powers of 2 greater than 2^0 ( up to around 2 000 000 ) #
# increase the table size if max number > 2 000 000 #
[ 1 : 20 ]INT powers of 2;
BEGIN
INT p2 := 1;
FOR i TO UPB powers of 2 DO
powers of 2[ i ] := ( p2 *:= 2 )
OD
END;
# the numbers must be odd and not of the form p + 2^n #
# either p is odd and 2^n is even and hence n > 0 and p > 2 #
# or 2^n is odd and p is even and hence n = 0 and p = 2 #
INT dp count := 0;
# n = 0, p = 2 - the only possibility is 3 #
FOR i BY 2 TO 3 DO
INT p = 2;
IF p + 1 /= i THEN
print( ( whole( i, -5 ) ) );
dp count +:= 1
FI
OD;
# n > 0, p > 2 #
FOR i FROM 5 BY 2 TO max number DO
BOOL found := FALSE;
FOR p TO UPB powers of 2 WHILE NOT found AND i > powers of 2[ p ] DO
found := prime[ i - powers of 2[ p ] ]
OD;
IF NOT found THEN
IF ( dp count +:= 1 ) <= 50 THEN
print( ( whole( i, -5 ) ) );
IF dp count MOD 10 = 0 THEN print( ( newline ) ) FI
ELIF dp count = 1 000 OR dp count = 10 000 THEN
print( ( "The ", whole( dp count, -5 )
, "th De Polignac number is ", whole( i, -7 )
, newline
)
)
FI
FI
OD;
print( ( "Found ", whole( dp count, 0 )
, " De Polignac numbers up to ", whole( max number, 0 )
, newline
)
)
END- Output:
1 127 149 251 331 337 373 509 599 701 757 809 877 905 907 959 977 997 1019 1087 1199 1207 1211 1243 1259 1271 1477 1529 1541 1549 1589 1597 1619 1649 1657 1719 1759 1777 1783 1807 1829 1859 1867 1927 1969 1973 1985 2171 2203 2213 The 1000th De Polignac number is 31941 The 10000th De Polignac number is 273421 Found 19075 De Polignac numbers up to 500000
ALGOL W
begin % find some De Polignac numbers - positive odd numbers that can't be %
% written as p + 2^n for some prime p and integer n %
integer MAX_NUMBER; % maximum number we will consider %
integer MAX_POWER; % maximum powerOf2 < MAX_NUMBER %
MAX_NUMBER := 500000;
MAX_POWER := 20;
begin
logical array prime ( 0 :: MAX_NUMBER );
integer array powersOf2 ( 1 :: MAX_POWER );
integer dpCount;
% sieve the primes to MAX_NUMBER %
begin
integer rootMaxNumber;
rootMaxNumber:= entier( sqrt( MAX_NUMBER ) );
prime( 0 ) := prime( 1 ) := false;
prime( 2 ) := true;
for i := 3 step 2 until MAX_NUMBER do prime( i ) := true;
for i := 4 step 2 until MAX_NUMBER do prime( i ) := false;
for i := 3 step 2 until rootMaxNumber do begin
if prime( i ) then begin
integer i2;
i2 := i + i;
for s := i * i step i2 until MAX_NUMBER do prime( s ) := false
end if_prime_i_and_i_lt_rootMaxNumber
end for_i
end;
% table of powers of 2 greater than 2^0 ( up to around 2 000 000 ) %
% increase the table size if MAX_NUMBER > 2 000 000 %
begin
integer p2;
p2 := 1;
for i := 1 until MAX_POWER do begin
p2 := p2 * 2;
powersOf2( i ) := p2
end for_i
end;
% the numbers must be odd and not of the form p + 2^n %
% either p is odd and 2^n is even and hence n > 0 and p > 2 %
% or 2^n is odd and p is even and hence n = 0 and p = 2 %
% n = 0, p = 2 - the only possibility is 3 %
dpCount := 0;
for i := 1 step 2 until 3 do begin
integer p;
p := 2;
if p + 1 not = i then begin
dpCount := dpCount + 1;
writeon( i_w := 5, i )
end if_p_plus_1_ne_i
end for_i ;
% n > 0, p > 2 %
for i := 5 step 2 until MAX_NUMBER do begin
logical found;
integer p;
found := false;
p := 1;
while p <= MAX_POWER and not found and i > powersOf2( p ) do begin
found := prime( i - powersOf2( p ) );
p := p + 1
end while_not_found_and_have_a_suitable_power ;
if not found then begin
dpCount := dpCount + 1;
if dpCount <= 50 then begin
writeon( i_w := 5, i );
if dpCount rem 10 = 0 then write()
end
else if dpCount = 1000 or dpCount = 10000 then begin
write( i_w := 5, s_w := 0, "The ", dpCount
, "th De Polignac number is ", i
)
end if_various_DePolignac_numbers
end if_not_found
end for_i;
write( i_w := 1, s_w := 0
, "Found ", dpCount, " De Polignac numbers up to ", MAX_NUMBER
)
end
end.- Output:
1 127 149 251 331 337 373 509 599 701 757 809 877 905 907 959 977 997 1019 1087 1199 1207 1211 1243 1259 1271 1477 1529 1541 1549 1589 1597 1619 1649 1657 1719 1759 1777 1783 1807 1829 1859 1867 1927 1969 1973 1985 2171 2203 2213 The 1000th De Polignac number is 31941 The 10000th De Polignac number is 273421 Found 19075 De Polignac numbers up to 500000
C++
#include <iomanip>
#include <iostream>
bool is_prime(int n) {
if (n < 2)
return false;
if (n % 2 == 0)
return n == 2;
if (n % 3 == 0)
return n == 3;
for (int p = 5; p * p <= n; p += 4) {
if (n % p == 0)
return false;
p += 2;
if (n % p == 0)
return false;
}
return true;
}
bool is_depolignac_number(int n) {
for (int p = 1; p < n; p <<= 1) {
if (is_prime(n - p))
return false;
}
return true;
}
int main() {
std::cout.imbue(std::locale(""));
std::cout << "First 50 de Polignac numbers:\n";
for (int n = 1, count = 0; count < 10000; n += 2) {
if (is_depolignac_number(n)) {
++count;
if (count <= 50)
std::cout << std::setw(5) << n
<< (count % 10 == 0 ? '\n' : ' ');
else if (count == 1000)
std::cout << "\nOne thousandth: " << n << '\n';
else if (count == 10000)
std::cout << "\nTen thousandth: " << n << '\n';
}
}
}
- Output:
First 50 de Polignac numbers:
1 127 149 251 331 337 373 509 599 701
757 809 877 905 907 959 977 997 1,019 1,087
1,199 1,207 1,211 1,243 1,259 1,271 1,477 1,529 1,541 1,549
1,589 1,597 1,619 1,649 1,657 1,719 1,759 1,777 1,783 1,807
1,829 1,859 1,867 1,927 1,969 1,973 1,985 2,171 2,203 2,213
One thousandth: 31,941
Ten thousandth: 273,421
FreeBASIC
' find some De Polignac numbers - positive odd numbers that can't be
' written as p + 2^n for some prime p and integer n
Const maxNumber = 500000 ' maximum number we will consider
Const maxPower = 20 ' maximum powerOf2 < maxNumber
Dim As Boolean prime(0 To maxNumber)
Dim As Uinteger powersOf2(1 To maxPower)
' sieve the primes to maxNumber
Dim As Integer rootMaxNumber = Sqr(maxNumber)
prime(0) = false
prime(1) = false
prime(2) = true
Dim As Integer i, s
For i = 3 To maxNumber Step 2
prime(i) = true
Next i
For i = 4 To maxNumber Step 2
prime(i) = false
Next i
For i = 3 To rootMaxNumber Step 2
If prime(i) Then
Dim As Integer i2 = i + i
For s = i * i To maxNumber Step i2
prime(s) = false
Next s
End If
Next i
' table of powers of 2 greater than 2^0 (up to around 2 000 000)
' increase the table size if maxNumber > 2 000 000
Dim As Integer p2 = 1
For i = 1 To maxPower
p2 *= 2
powersOf2(i) = p2
Next i
' the numbers must be odd and not of the form p + 2^n
' either p is odd and 2^n is even and hence n > 0 and p > 2
' or 2^n is odd and p is even and hence n = 0 and p = 2
' n = 0, p = 2 - the only possibility is 3
Dim As Uinteger dpCount = 0
For i = 1 To 3 Step 2
Dim As Integer p = 2
If p + 1 <> i Then
dpCount += 1
Print Using "#####"; i;
End If
Next i
' n > 0, p > 2
For i = 5 To maxNumber Step 2
Dim As Boolean found = false
Dim As Integer p = 1
While p <= maxPower And Not found And i > powersOf2(p)
found = prime(i - powersOf2(p))
p += 1
Wend
If Not found Then
dpCount += 1
If dpCount <= 50 Then
Print Using "#####"; i;
If dpCount Mod 10 = 0 Then Print
Elseif dpCount = 1000 Or dpCount = 10000 Then
Print Using "The #####th De Polignac number is ######"; dpCount; i
End If
End If
Next
Print "Found"; dpCount; " De Polignac numbers up to"; maxNumber
Sleep- Output:
Same as ALGOL W entry.
J
Implementation:
depolignac=: (1+i.@>.&.-:) (-.,) i.@>.&.(2&^.) +/ i.&.(p:inv)
Task example:
5 10$depolignac 3000
1 127 149 251 331 337 373 509 599 701
757 809 877 905 907 959 977 997 1019 1087
1199 1207 1211 1243 1259 1271 1477 1529 1541 1549
1589 1597 1619 1649 1657 1719 1759 1777 1783 1807
1829 1859 1867 1927 1969 1973 1985 2171 2203 2213
Stretch:
T=: depolignac 3e5
999 { T
31941
9999 { T
273421
(We use 999 here for the 1000th number because 0 is the first J index.)
Phix
with javascript_semantics constant MAX_VALUE = 1_000_000, all_primes = get_primes_le(MAX_VALUE), powers_of_2 = sq_power(2,tagset(floor(log2(MAX_VALUE)),0)) sequence allvalues = repeat(true,MAX_VALUE), dePolignac = {} for i in all_primes do for j in powers_of_2 do if i + j < MAX_VALUE then allvalues[i + j] = false end if end for end for for n=1 to MAX_VALUE by 2 do if allvalues[n] then dePolignac &= n end if end for printf(1,"First fifty de Polignac numbers:\n%s\n",{join_by(dePolignac[1..50],1,10,"",fmt:="%5d")}) printf(1,"One thousandth: %,d\n",dePolignac[1000]) printf(1,"Ten thousandth: %,d\n",dePolignac[10000])
- Output:
First fifty de Polignac numbers:
1 127 149 251 331 337 373 509 599 701
757 809 877 905 907 959 977 997 1019 1087
1199 1207 1211 1243 1259 1271 1477 1529 1541 1549
1589 1597 1619 1649 1657 1719 1759 1777 1783 1807
1829 1859 1867 1927 1969 1973 1985 2171 2203 2213
One thousandth: 31,941
Ten thousandth: 273,421
alternative
No magic constant or arbitrary limit, but about seven times slower. Based on Wren/Raku, same output as above.
with javascript_semantics sequence depolignac = {1} function dePolignac(integer n) -- return the nth depolignac number integer t = depolignac[$]+2 while length(depolignac)<n do integer p2 = 1 bool found = false while p2<t do if is_prime(t-p2) then found = true exit end if p2 *= 2 end while if not found then depolignac &= t end if t += 2 end while return depolignac[n] end function printf(1,"First fifty de Polignac numbers:\n%s\n",{join_by(apply(tagset(50),dePolignac),1,10,"",fmt:="%5d")}) printf(1,"One thousandth: %,d\n",dePolignac(1000)) printf(1,"Ten thousandth: %,d\n",dePolignac(10000))
PL/M
Based on the ALGOL 68 sample.
... under CP/M (or an emulator)
100H: /* FIND SOME DE POLIGNAC NUMBERS - POSITIVE ODD NUMBERS THAT CAN'T BE */
/* WRITTEN AS P + 2**N FOR SOME PRIME P AND INTEGER N */
DECLARE FALSE LITERALLY '0', TRUE LITERALLY '0FFH';
/* CP/M SYSTEM CALL AND I/O ROUTINES */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR$NL: PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
PR$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH */
DECLARE N ADDRESS;
DECLARE V ADDRESS, N$STR ( 6 )BYTE, W BYTE;
V = N;
W = LAST( N$STR );
N$STR( W ) = '$';
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR$STRING( .N$STR( W ) );
END PR$NUMBER;
/* END SYSTEM CALL AND I/O ROUTINES */
DECLARE MAX$DP LITERALLY '4000', /* MAXIMUM NUMBER TO CONSIDER */
MAX$DP$PLUS$1 LITERALLY '4001'; /* MAX$DP + 1 FOR ARRAY BOUNDS */
/* SIEVE THE PRIMES TO MAX$DP */
DECLARE PRIME ( MAX$DP$PLUS$1 )BYTE;
DO;
DECLARE ( I, S ) ADDRESS;
PRIME( 0 ), PRIME( 1 ) = FALSE;
PRIME( 2 ) = TRUE;
DO I = 3 TO LAST( PRIME ) BY 2; PRIME( I ) = TRUE; END;
DO I = 4 TO LAST( PRIME ) BY 2; PRIME( I ) = FALSE; END;
DO I = 3 TO LAST( PRIME ) / 2 BY 2;
IF PRIME( I ) THEN DO;
DO S = I + I TO LAST( PRIME ) BY I; PRIME( S ) = FALSE; END;
END;
END;
END;
/* TABLE OF POWERS OF 2 UP TO MAX$DP */
DECLARE POWERS$OF$2 ( 12 )ADDRESS
INITIAL( 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048 );
/* DISPLAYS A DE POLIGNAC NUMBER, IF NECESSARY */
SHOW$DE$POLIGNAC: PROCEDURE( DP$NUMBER );
DECLARE ( DP$NUMBER ) ADDRESS;
CALL PR$CHAR( ' ' );
IF DP$NUMBER < 10 THEN CALL PR$CHAR( ' ' );
IF DP$NUMBER < 100 THEN CALL PR$CHAR( ' ' );
IF DP$NUMBER < 1000 THEN CALL PR$CHAR( ' ' );
CALL PR$NUMBER( DP$NUMBER );
END SHOW$DE$POLIGNAC;
DECLARE ( I, P, COUNT ) ADDRESS;
DECLARE FOUND BYTE;
/* THE NUMBERS MUST BE ODD AND NOT OF THE FORM P + 2**N */
/* EITHER P IS ODD AND 2**N IS EVEN AND HENCE N > 0 AND P > 2 */
/* OR 2**N IS ODD AND P IS EVEN AND HENCE N = 0 AND P = 2 */
/* N = 0, P = 2 - THE ONLY POSSIBILITY IS 3 */
DO I = 1 TO 3 BY 2;
P = 2;
IF P + 1 <> I THEN DO;
COUNT = COUNT + 1;
CALL SHOW$DE$POLIGNAC( I );
END;
END;
/* N > 0, P > 2 */
I = 3;
DO WHILE I < MAX$DP AND COUNT < 50;
I = I + 2;
FOUND = FALSE;
P = 1;
DO WHILE NOT FOUND
AND P <= LAST( POWERS$OF$2 )
AND I > POWERS$OF$2( P );
FOUND = PRIME( I - POWERS$OF$2( P ) );
P = P + 1;
END;
IF NOT FOUND THEN DO;
CALL SHOW$DE$POLIGNAC( I );
IF ( COUNT := COUNT + 1 ) MOD 10 = 0 THEN CALL PR$NL;
END;
END;
EOF- Output:
1 127 149 251 331 337 373 509 599 701 757 809 877 905 907 959 977 997 1019 1087 1199 1207 1211 1243 1259 1271 1477 1529 1541 1549 1589 1597 1619 1649 1657 1719 1759 1777 1783 1807 1829 1859 1867 1927 1969 1973 1985 2171 2203 2213
Python
''' Rosetta code rosettacode.org/wiki/De_Polignac_numbers '''
from sympy import isprime
from math import log
from numpy import ndarray
max_value = 1_000_000
all_primes = [i for i in range(max_value + 1) if isprime(i)]
powers_of_2 = [2**i for i in range(int(log(max_value, 2)))]
allvalues = ndarray(max_value, dtype=bool)
allvalues[:] = True
for i in all_primes:
for j in powers_of_2:
if i + j < max_value:
allvalues[i + j] = False
dePolignac = [n for n in range(1, max_value) if n & 1 == 1 and allvalues[n]]
print('First fifty de Polignac numbers:')
for i, n in enumerate(dePolignac[:50]):
print(f'{n:5}', end='\n' if (i + 1) % 10 == 0 else '')
print(f'\nOne thousandth: {dePolignac[999]:,}')
print(f'\nTen thousandth: {dePolignac[9999]:,}')
- Output:
First fifty de Polignac numbers:
1 127 149 251 331 337 373 509 599 701
757 809 877 905 907 959 977 997 1019 1087
1199 1207 1211 1243 1259 1271 1477 1529 1541 1549
1589 1597 1619 1649 1657 1719 1759 1777 1783 1807
1829 1859 1867 1927 1969 1973 1985 2171 2203 2213
One thousandth: 31,941
Ten thousandth: 273,421
Raku
use List::Divvy;
use Lingua::EN::Numbers;
constant @po2 = (1..∞).map: 2 ** *;
my @dePolignac = lazy 1, |(2..∞).hyper.map(* × 2 + 1).grep: -> $n { all @po2.&upto($n).map: { !is-prime $n - $_ } };
say "First fifty de Polignac numbers:\n" ~ @dePolignac[^50]».&comma».fmt("%5s").batch(10).join: "\n";
say "\nOne thousandth: " ~ comma @dePolignac[999];
say "\nTen thousandth: " ~ comma @dePolignac[9999];
- Output:
First fifty de Polignac numbers:
1 127 149 251 331 337 373 509 599 701
757 809 877 905 907 959 977 997 1,019 1,087
1,199 1,207 1,211 1,243 1,259 1,271 1,477 1,529 1,541 1,549
1,589 1,597 1,619 1,649 1,657 1,719 1,759 1,777 1,783 1,807
1,829 1,859 1,867 1,927 1,969 1,973 1,985 2,171 2,203 2,213
One thousandth: 31,941
Ten thousandth: 273,421
Wren
import "./math" for Int
import "./fmt" for Fmt
var pows2 = (0..19).map { |i| 1 << i }.toList
var dp = [1]
var dp1000
var dp10000
var count = 1
var n = 3
while (true) {
var found = false
for (pow in pows2) {
if (pow > n) break
if (Int.isPrime(n-pow)) {
found = true
break
}
}
if (!found) {
count = count + 1
if (count <= 50) {
dp.add(n)
} else if (count == 1000) {
dp1000 = n
} else if (count == 10000) {
dp10000 = n
break
}
}
n = n + 2
}
System.print("First 50 De Polignac numbers:")
Fmt.tprint("$,5d", dp, 10)
Fmt.print("\nOne thousandth: $,d", dp1000)
Fmt.print("\nTen thousandth: $,d", dp10000)- Output:
First 50 De Polignac numbers:
1 127 149 251 331 337 373 509 599 701
757 809 877 905 907 959 977 997 1,019 1,087
1,199 1,207 1,211 1,243 1,259 1,271 1,477 1,529 1,541 1,549
1,589 1,597 1,619 1,649 1,657 1,719 1,759 1,777 1,783 1,807
1,829 1,859 1,867 1,927 1,969 1,973 1,985 2,171 2,203 2,213
One thousandth: 31,941
Ten thousandth: 273,421
XPL0
Slow. Takes 225 seconds on Pi4.
func IsPrime(N); \Return 'true' if N is prime
int N, I;
[if N <= 2 then return N = 2;
if (N&1) = 0 then \even >2\ return false;
for I:= 3 to sqrt(N) do
[if rem(N/I) = 0 then return false;
I:= I+1;
];
return true;
];
func IsDePolignac(N); \Return 'true' if N is a de Polignac number
int N, P, T;
[for P:= N-1 downto 2 do
if IsPrime(P) then
[T:= 1;
repeat if T+P = N then return false;
T:= T<<1;
until T+P > N;
];
return true;
];
int N, C;
[Format(5, 0);
N:= 1; C:= 0;
loop [if IsDePolignac(N) then
[C:= C+1;
if C<=50 or C=1000 or C=10000 then
[RlOut(0, float(N));
if rem(C/10) = 0 then CrLf(0);
if C = 10000 then quit;
];
];
N:= N+2;
];
]- Output:
1 127 149 251 331 337 373 509 599 701 757 809 877 905 907 959 977 997 1019 1087 1199 1207 1211 1243 1259 1271 1477 1529 1541 1549 1589 1597 1619 1649 1657 1719 1759 1777 1783 1807 1829 1859 1867 1927 1969 1973 1985 2171 2203 2213 31941 273421