Sum of two adjacent numbers are primes
Show on this page the first 20 numbers and sum of two adjacent numbers which sum is prime.
Sum of two adjacent numbers are primes 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.
- Task
- Extra credit
Show the ten millionth such prime sum.
ALGOL 68
BEGIN # find the first 20 primes which are n + ( n + 1 ) for some n #
PR read "primes.incl.a68" PR # include prime utilities #
[]BOOL prime = PRIMESIEVE 200; # should be enough primes #
INT p count := 0;
FOR n WHILE p count < 20 DO
INT n1 = n + 1;
INT p = n + n1;
IF prime[ p ] THEN
p count +:= 1;
print( ( whole( n, -2 ), " + ", whole( n1, -2 ), " = ", whole( p, -3 ), newline ) )
FI
OD
END- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
AWK
# syntax: GAWK -f SUM_OF_TWO_ADJACENT_NUMBERS_ARE_PRIMES.AWK
BEGIN {
n = 1
stop = 20
printf("The first %d pairs of numbers whose sum is prime:\n",stop)
while (count < stop) {
if (is_prime(2*n + 1)) {
printf("%2d + %2d = %2d\n",n,n+1,2*n+1)
count++
}
n++
}
exit(0)
}
function is_prime(n, d) {
d = 5
if (n < 2) { return(0) }
if (n % 2 == 0) { return(n == 2) }
if (n % 3 == 0) { return(n == 3) }
while (d*d <= n) {
if (n % d == 0) { return(0) }
d += 2
if (n % d == 0) { return(0) }
d += 4
}
return(1)
}
- Output:
The first 20 pairs of numbers whose sum is prime: 1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
BASIC
BASIC256
function isPrime(v)
if v < 2 then return False
if v mod 2 = 0 then return v = 2
if v mod 3 = 0 then return v = 3
d = 5
while d * d <= v
if v mod d = 0 then return False else d += 2
end while
return True
end function
n = 0
num = 0
print "The first 20 pairs of numbers whose sum is prime:"
while True
n += 1
suma = 2*n+1
if isPrime(suma) then
num += 1
if num < 21 then
print n; " + "; (n+1); " = "; suma
else
exit while
end if
end if
end while
end- Output:
Igual que la entrada de FreeBASIC.
FreeBASIC
Function isPrime(Byval ValorEval As Integer) As Boolean
If ValorEval <= 1 Then Return False
For i As Integer = 2 To Int(Sqr(ValorEval))
If ValorEval Mod i = 0 Then Return False
Next i
Return True
End Function
Dim As Integer n = 0, num = 0, suma
print "The first 20 pairs of numbers whose sum is prime:"
Do
n += 1
suma = 2*n+1
If isPrime(suma) Then
num += 1
If num < 21 Then
Print using "## + ## = ##"; n; (n+1); suma
Else
Exit Do
End If
End If
Loop
Sleep- Output:
The first 20 pairs of numbers whose sum is prime: 1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
PureBasic
Procedure isPrime(v.i)
If v <= 1 : ProcedureReturn #False
ElseIf v < 4 : ProcedureReturn #True
ElseIf v % 2 = 0 : ProcedureReturn #False
ElseIf v < 9 : ProcedureReturn #True
ElseIf v % 3 = 0 : ProcedureReturn #False
Else
Protected r = Round(Sqr(v), #PB_Round_Down)
Protected f = 5
While f <= r
If v % f = 0 Or v % (f + 2) = 0
ProcedureReturn #False
EndIf
f + 6
Wend
EndIf
ProcedureReturn #True
EndProcedure
If OpenConsole()
Define n.i = 0, num.i = 0, suma.i
PrintN("The first 20 pairs of numbers whose sum is prime:")
Repeat
n + 1
suma = 2*n+1
If isPrime(suma)
num + 1
If num < 21
PrintN(Str(n) + " + " + Str(n+1) + " = " + Str(suma))
Else
Break
EndIf
EndIf
ForEver
CloseConsole()
EndIf- Output:
Igual que la entrada de FreeBASIC.
QBasic
FUNCTION isPrime (ValorEval)
IF ValorEval <= 1 THEN isPrime = 0
FOR i = 2 TO INT(SQR(ValorEval))
IF ValorEval MOD i = 0 THEN isPrime = 0
NEXT i
isPrime = 1
END FUNCTION
n = 0
num = 0
PRINT "The first 20 pairs of numbers whose sum is prime:"
DO
n = n + 1
suma = 2 * n + 1
IF isPrime(suma) THEN
num = num + 1
IF num < 21 THEN
PRINT USING "## + ## = ##"; n; (n + 1); suma
ELSE
EXIT DO
END IF
END IF
LOOP
END
- Output:
Igual que la entrada de FreeBASIC.
True BASIC
FUNCTION isPrime (ValorEval)
IF ValorEval <= 1 THEN LET isPrime = 0
FOR i = 2 TO INT(SQR(ValorEval))
IF MOD(ValorEval, i) = 0 THEN LET isPrime = 0
NEXT i
LET isPrime = 1
END FUNCTION
LET n = 0
LET num = 0
PRINT "The first 20 pairs of numbers whose sum is prime:"
DO
LET n = n + 1
LET suma = 2 * n + 1
IF isPrime(suma) = 1 THEN
LET num = num + 1
IF num < 21 THEN
PRINT USING "##": n;
PRINT " + ";
PRINT USING "##": n+1;
PRINT " = ";
PRINT USING "##": suma
ELSE
EXIT DO
END IF
END IF
LOOP
END
- Output:
Igual que la entrada de FreeBASIC.
Yabasic
sub isPrime(v)
if v < 2 then return False : fi
if mod(v, 2) = 0 then return v = 2 : fi
if mod(v, 3) = 0 then return v = 3 : fi
d = 5
while d * d <= v
if mod(v, d) = 0 then return False else d = d + 2 : fi
wend
return True
end sub
n = 0
num = 0
print "The first 20 pairs of numbers whose sum is prime:"
do
n = n + 1
suma = 2*n+1
if isPrime(suma) then
num = num + 1
if num < 21 then
print n using "##", " + ", (n+1) using "##", " = ", suma using "##"
else
break
end if
end if
loop
end- Output:
Igual que la entrada de FreeBASIC.
C
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define TRUE 1
#define FALSE 0
typedef int bool;
void primeSieve(int *c, int limit, bool processEven, bool primesOnly) {
int i, ix, p, p2;
limit++;
c[0] = TRUE;
c[1] = TRUE;
if (processEven) {
for (i = 4; i < limit; i +=2) c[i] = TRUE;
}
p = 3;
while (TRUE) {
p2 = p * p;
if (p2 >= limit) break;
for (i = p2; i < limit; i += 2*p) c[i] = TRUE;
while (TRUE) {
p += 2;
if (!c[p]) break;
}
}
if (primesOnly) {
/* move the actual primes to the front of the array */
c[0] = 2;
for (i = 3, ix = 1; i < limit; i += 2) {
if (!c[i]) c[ix++] = i;
}
}
}
int main() {
int i, p, hp, n = 10000000;
int limit = (int)(log(n) * (double)n * 1.2); /* should be more than enough */
int *primes = (int *)calloc(limit, sizeof(int));
primeSieve(primes, limit-1, FALSE, TRUE);
printf("The first 20 pairs of natural numbers whose sum is prime are:\n");
for (i = 1; i <= 20; ++i) {
p = primes[i];
hp = p / 2;
printf("%2d + %2d = %2d\n", hp, hp+1, p);
}
printf("\nThe 10 millionth such pair is:\n");
p = primes[n];
hp = p / 2;
printf("%2d + %2d = %2d\n", hp, hp+1, p);
free(primes);
return 0;
}
- Output:
The first 20 pairs of natural numbers whose sum is prime are: 1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73 The 10 millionth such pair is: 89712345 + 89712346 = 179424691
F#
This task uses Extensible Prime Generator (F#)
// 2n+1 is prime. Nigel Galloway: Januuary 22nd., 2022
primes32()|>Seq.skip 1|>Seq.take 20|>Seq.map(fun n->n/2)|>Seq.iteri(fun n g->printfn "%2d: %2d + %2d=%d" (n+1) g (g+1) (g+g+1))
- Output:
1: 1 + 2=3 2: 2 + 3=5 3: 3 + 4=7 4: 5 + 6=11 5: 6 + 7=13 6: 8 + 9=17 7: 9 + 10=19 8: 11 + 12=23 9: 14 + 15=29 10: 15 + 16=31 11: 18 + 19=37 12: 20 + 21=41 13: 21 + 22=43 14: 23 + 24=47 15: 26 + 27=53 16: 29 + 30=59 17: 30 + 31=61 18: 33 + 34=67 19: 35 + 36=71 20: 36 + 37=73
Factor
USING: arrays formatting kernel lists lists.lazy math
math.primes.lists sequences ;
20 lprimes cdr [ 2/ dup 1 + 2array ] lmap-lazy ltake
[ dup sum suffix "%d + %d = %d\n" vprintf ] leach
- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
Go
package main
import (
"fmt"
"math"
"rcu"
)
func main() {
limit := int(math.Log(1e7) * 1e7 * 1.2) // should be more than enough
primes := rcu.Primes(limit)
fmt.Println("The first 20 pairs of natural numbers whose sum is prime are:")
for i := 1; i <= 20; i++ {
p := primes[i]
hp := p / 2
fmt.Printf("%2d + %2d = %2d\n", hp, hp+1, p)
}
fmt.Println("\nThe 10 millionth such pair is:")
p := primes[1e7]
hp := p / 2
fmt.Printf("%2d + %2d = %2d\n", hp, hp+1, p)
}
- Output:
The first 20 pairs of natural numbers whose sum is prime are: 1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73 The 10 millionth such pair is: 89712345 + 89712346 = 179424691
J
Here, we generate the first 20 odd primes, divide by 2, drop the fractional part and add 0 and 1 to that value. Then we format that pair of numbers with their sum and with decorations indicating the relevant arithmetic:
(+/,&":'=',{.,&":'+',&":{:)"#. 0 1+/~<.-: p:1+i.20
3=1+2
5=2+3
7=3+4
11=5+6
13=6+7
17=8+9
19=9+10
23=11+12
29=14+15
31=15+16
37=18+19
41=20+21
43=21+22
47=23+24
53=26+27
59=29+30
61=30+31
67=33+34
71=35+36
73=36+37
jq
Works with gojq, the Go implementation of jq
This entry uses a memory-less approach to computing `is_prime`, and so a naive approach to computing the first 20 numbers satisfying the condition is appropriate.
def is_prime:
. as $n
| if ($n < 2) then false
elif ($n % 2 == 0) then $n == 2
elif ($n % 3 == 0) then $n == 3
else 5
| until( . <= 0;
if .*. > $n then -1
elif ($n % . == 0) then 0
else . + 2
| if ($n % . == 0) then 0
else . + 4
end
end)
| . == -1
end;
def naive:
limit(20;
range(0; infinite) as $i
| (2*$i + 1) as $q
| select($q | is_prime)
| "\($i) + \($i + 1) = \($q)" );
naive- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
Julia
using Lazy
using Primes
seq = @>> Lazy.range() filter(n -> isprime(2n + 1))
for n in take(20, seq)
println("$n + $(n + 1) = $(n + n + 1)")
end
let
i, cnt = 0, 0
while cnt < 10_000_000
i += 1
if isprime(2i + 1)
cnt += 1
end
end
println("Ten millionth: $i + $(i + 1) = $(i + i + 1)")
end
- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73 Ten millionth: 89712345 + 89712346 = 179424691
Mathematica/Wolfram Language
Column[Row[{Floor[#/2], " + ", Ceiling[#/2], " = ", #}] & /@ Prime[1 + Range[20]]]
Row[{Floor[#/2], " + ", Ceiling[#/2], " = ", #}] &[Prime[10^7 + 1]]
- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73 89712345 + 89712346 = 179424691
Perl
use strict;
use warnings;
use ntheory 'is_prime';
my($n,$c);
while () { is_prime(1 + 2*++$n) and printf "%2d + %2d = %2d\n", $n, $n+1, 1+2*$n and ++$c == 20 and last }
- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
Phix
Every prime p greater than 2 is odd, hence p/2 is k.5 and the adjacent numbers needed are k and k+1. DOH.
with javascript_semantics procedure doh(integer p) printf(1,"%d + %d = %d\n",{floor(p/2),ceil(p/2),p}) end procedure papply(get_primes(-21)[2..$],doh) doh(get_prime(1e7+1))
- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73 89712345 + 89712346 = 179424691
Python
#!/usr/bin/python
def isPrime(n):
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
return False
return True
if __name__ == "__main__":
n = 0
num = 0
print('The first 20 pairs of numbers whose sum is prime:')
while True:
n += 1
suma = 2*n+1
if isPrime(suma):
num += 1
if num < 21:
print('{:2}'.format(n), "+", '{:2}'.format(n+1), "=", '{:2}'.format(suma))
else:
break
- Output:
The first 20 pairs of numbers whose sum is prime: 1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
Raku
my @n-n1-triangular = map { $_, $_ + 1, $_ + ($_ + 1) }, ^Inf;
my @wanted = @n-n1-triangular.grep: *.[2].is-prime;
printf "%2d + %2d = %2d\n", |.list for @wanted.head(20);
- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
Ring
load "stdlibcore.ring"
see "working..." + nl
n = 0
num = 0
while true
n++
sum = 2*n+1
if isprime(sum)
num++
if num < 21
? "" + n + " + " + (n+1) + " = " + sum
else
exit
ok
ok
end
see "done..." + nl- Output:
working... 1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73 done...
Sidef
var wanted = (1..Inf -> lazy.map {|n| [n, n+1, n+(n+1)] }\
.grep { .tail.is_prime })
wanted.first(20).each_2d {|a,b,c|
printf("%2d + %2d = %2d\n", a,b,c)
}
- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
Wren
import "./math" for Int
import "./fmt" for Fmt
var limit = (1e7.log * 1e7 * 1.2).floor // should be more than enough
var primes = Int.primeSieve(limit)
System.print("The first 20 pairs of natural numbers whose sum is prime are:")
for (i in 1..20) {
var p = primes[i]
var hp = (p/2).floor
Fmt.print("$2d + $2d = $2d", hp, hp + 1, p)
}
System.print("\nThe 10 millionth such pair is:")
var p = primes[1e7]
var hp = (p/2).floor
Fmt.print("$2d + $2d = $2d", hp, hp + 1, p)- Output:
The first 20 pairs of natural numbers whose sum is prime are: 1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73 The 10 millionth such pair is: 89712345 + 89712346 = 179424691
XPL0
include xpllib;
int N, Num, Sum;
[Text(0, "Working...^M^J");
N:= 0;
Num:= 0;
loop
[N:= N+1;
Sum:= 2*N + 1;
if IsPrime(Sum) then
[Num:= Num+1;
if Num < 21 then
[Text(0,"N = "); IntOut(0,N); Text(0," Sum = "); IntOut(0,Sum); CrLf(0)]
else
quit
]
];
Text(0, "Done...^M^J");
]- Output:
Working... N = 1 Sum = 3 N = 2 Sum = 5 N = 3 Sum = 7 N = 5 Sum = 11 N = 6 Sum = 13 N = 8 Sum = 17 N = 9 Sum = 19 N = 11 Sum = 23 N = 14 Sum = 29 N = 15 Sum = 31 N = 18 Sum = 37 N = 20 Sum = 41 N = 21 Sum = 43 N = 23 Sum = 47 N = 26 Sum = 53 N = 29 Sum = 59 N = 30 Sum = 61 N = 33 Sum = 67 N = 35 Sum = 71 N = 36 Sum = 73 Done...