Sum of two adjacent numbers are primes: Difference between revisions
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=={{header|Factor}}== |
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{{works with|Factor|0.99 2021-06-02}} |
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<lang factor>USING: arrays formatting kernel lists lists.lazy math |
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math.primes.lists sequences ; |
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20 lprimes cdr [ 2/ dup 1 + 2array ] lmap-lazy ltake |
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[ dup sum suffix "%d + %d = %d\n" vprintf ] leach</lang> |
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{{out}} |
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<pre> |
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1 + 2 = 3 |
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2 + 3 = 5 |
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3 + 4 = 7 |
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5 + 6 = 11 |
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6 + 7 = 13 |
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8 + 9 = 17 |
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9 + 10 = 19 |
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11 + 12 = 23 |
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14 + 15 = 29 |
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15 + 16 = 31 |
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18 + 19 = 37 |
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20 + 21 = 41 |
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21 + 22 = 43 |
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23 + 24 = 47 |
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26 + 27 = 53 |
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29 + 30 = 59 |
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30 + 31 = 61 |
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33 + 34 = 67 |
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35 + 36 = 71 |
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36 + 37 = 73 |
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</pre> |
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=={{header|J}}== |
=={{header|J}}== |
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Revision as of 14:32, 25 January 2022
- Task
Show on this page the first 20 numbers and sum of two adjacent numbers which sum is prime.
- Extra credit
Show the ten millionth such prime sum.
ALGOL 68
<lang algol68>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</lang>
- 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
C
<lang c>#include <stdio.h>
- define TRUE 1
- define FALSE 0
int isPrime(int n) {
if (n < 2) return FALSE; if (!(n%2)) return n == 2; if (!(n%3)) return n == 3; int d = 5; while (d*d <= n) { if (!(n%d)) return FALSE; d += 2; if (!(n%d)) return FALSE; d += 4; } return TRUE;
}
int main() {
int count = 0, n = 1; printf("The first 20 pairs of natural numbers whose sum is prime are:\n"); while (count < 20) { if (isPrime(2*n + 1)) { printf("%2d + %2d = %2d\n", n, n + 1, 2*n + 1); count++; } n++; } return 0;
}</lang>
- 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
F#
This task uses Extensible Prime Generator (F#) <lang fsharp> // 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)) </lang>
- 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
<lang 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</lang>
- 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
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:
<lang J> (+/,&":'=',{.,&":'+',&":{:)"#. 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</lang>
Julia
<lang 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) = $(i + i + 1)")
end
</lang>
- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73 Ten millionth: 89712345 + 89712346 = 179424691
Perl
<lang perl>use strict; use warnings; use ntheory 'is_prime';
my($n,$c); while () { is_prime(1 + 2*++$n) and printf "%2d + %2d = %2d\n", $n, $n+1, 1+2*$n and ++$c == 20 and last }</lang>
- Output:
1 + 2 = 3 2 + 3 = 5 3 + 4 = 7 5 + 6 = 11 6 + 7 = 13 8 + 9 = 17 9 + 10 = 19 11 + 12 = 23 14 + 15 = 29 15 + 16 = 31 18 + 19 = 37 20 + 21 = 41 21 + 22 = 43 23 + 24 = 47 26 + 27 = 53 29 + 30 = 59 30 + 31 = 61 33 + 34 = 67 35 + 36 = 71 36 + 37 = 73
Phix
Every prime p greater than 2 is odd, hence p/2 is k.5 and the adjacent numbers needed are k and k+1. DOH.
with javascript_semantics procedure doh(integer p) printf(1,"%d + %d = %d\n",{floor(p/2),ceil(p/2),p}) end procedure papply(get_primes(-21)[2..$],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
Raku
<lang perl6>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);</lang>
- 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
<lang 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 </lang>
- 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...
Wren
<lang ecmascript>import "./math" for Int import "./fmt" for Fmt
System.print("The first 20 pairs of natural numbers whose sum is prime are:") var count = 0 var n = 1 while (count < 20) {
if (Int.isPrime(2*n + 1)) { Fmt.print("$2d + $2d = $2d", n, n + 1, 2*n + 1) count = count + 1 } n = n + 1
}</lang>
- 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
XPL0
<lang 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"); ]</lang>
- 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...