Sum of primes in odd positions is prime
- Task
Let p(i) be a sequence of prime numbers.
Consider the p(1),p(3),p(5), ... ,p(i), for each p(i) < 1,000 and i is odd.
Let sum be the sum of these primes.
If sum is prime then print i, p(i) and sum.
Factor
<lang factor>USING: assocs assocs.extras kernel math.primes math.statistics prettyprint sequences.extras ;
1000 primes-upto <evens> dup cum-sum zip [ prime? ] filter-values .</lang>
- Output:
{
{ 2 2 }
{ 5 7 }
{ 31 89 }
{ 103 659 }
{ 149 1181 }
{ 331 5021 }
{ 467 9923 }
{ 499 10909 }
{ 523 11941 }
{ 653 17959 }
{ 823 26879 }
}
Go
<lang go>package main
import (
"fmt" "rcu"
)
func main() {
primes := rcu.Primes(999)
sum := 0
fmt.Println(" i p[i] Σp[i]")
fmt.Println("----------------")
for i := 0; i < len(primes); i += 2 {
sum += primes[i]
if rcu.IsPrime(sum) {
fmt.Printf("%3d %3d %6s\n", i+1, primes[i], rcu.Commatize(sum))
}
}
}</lang>
- Output:
i p[i] Σp[i] ---------------- 1 2 2 3 5 7 11 31 89 27 103 659 35 149 1,181 67 331 5,021 91 467 9,923 95 499 10,909 99 523 11,941 119 653 17,959 143 823 26,879
Nim
<lang Nim>import strformat
template isOdd(n: Natural): bool = (n and 1) != 0 template isEven(n: Natural): bool = (n and 1) == 0
func isPrime(n: Positive): bool =
if n == 1: return false if n.isEven: return n == 2 if n mod 3 == 0: return n == 3 var d = 5 while d * d <= n: if n mod d == 0: return false inc d, 2 if n mod d == 0: return false inc d, 4 result = true
- Compute the sums of primes at odd position.
echo " i p(i) sum" var idx = 0 var sum = 0 var p = 1 while p < 1000:
inc p
if p.isPrime:
inc idx
if idx.isOdd:
inc sum, p
if sum.isPrime:
echo &"{idx:3} {p:3} {sum:5}"</lang>
- Output:
i p(i) sum 1 2 2 3 5 7 11 31 89 27 103 659 35 149 1181 67 331 5021 91 467 9923 95 499 10909 99 523 11941 119 653 17959 143 823 26879
Phix
with javascript_semantics sequence primes = get_primes_le(1000) integer total = 0 printf(1," i p sum\n") printf(1,"----------------\n") for i=1 to length(primes) by 2 do total += primes[i] if is_prime(total) then printf(1,"%3d %3d %,6d\n", {i, primes[i], total}) end if end for
- Output:
i p sum ---------------- 1 2 2 3 5 7 11 31 89 27 103 659 35 149 1,181 67 331 5,021 91 467 9,923 95 499 10,909 99 523 11,941 119 653 17,959 143 823 26,879
Raku
<lang perl6>my @odd = grep { ++$ !%% 2 }, grep &is-prime, 2 ..^ 1000; my @sums = [\+] @odd;
say .fmt('%5d') for grep { .[1].is-prime }, ( @odd Z @sums );</lang>
- Output:
2 2 5 7 31 89 103 659 149 1181 331 5021 467 9923 499 10909 523 11941 653 17959 823 26879
Ring
<lang ring> load "stdlib.ring" see "working..." + nl see "i p sum" + nl
nr = 0 sum = 0 limit = 1000
for n = 2 to limit
if isprime(n)
nr++
if nr%2 = 1
sum += n
if isprime(sum)
see "" + nr + " " + n + " " + sum + nl
ok
ok
ok
next
see "done..." + nl </lang>
- Output:
working... i p sum 1 2 2 3 5 7 11 31 89 27 103 659 35 149 1181 67 331 5021 91 467 9923 95 499 10909 99 523 11941 119 653 17959 143 823 26879 done...
Wren
<lang ecmascript>import "/math" for Int import "/trait" for Indexed import "/fmt" for Fmt
var primes = Int.primeSieve(999) var sum = 0 System.print(" i p[i] Σp[i]") System.print("----------------") for (se in Indexed.new(primes, 2)) {
sum = sum + se.value
if (Int.isPrime(sum)) Fmt.print("$3d $3d $,6d", se.index + 1, se.value, sum)
}</lang>
- Output:
i p[i] Σp[i] ---------------- 1 2 2 3 5 7 11 31 89 27 103 659 35 149 1,181 67 331 5,021 91 467 9,923 95 499 10,909 99 523 11,941 119 653 17,959 143 823 26,879
XPL0
<lang XPL0>func IsPrime(N); \Return 'true' if N is a prime number int N, I; [if N <= 1 then return false; for I:= 2 to sqrt(N) do
if rem(N/I) = 0 then return false;
return true; ];
int I, Sum, N; [Text(0, "p(n) sum^m^j"); Sum:= 0; I:= 0; for N:= 2 to 1000-1 do
[if IsPrime(N) then
[I:= I+1;
if I&1 then \odd
[Sum:= Sum + N;
if IsPrime(Sum) then
[IntOut(0, N); ChOut(0, ^ ); IntOut(0, Sum); CrLf(0)];
];
];
];
]</lang>
- Output:
p(n) sum 2 2 5 7 31 89 103 659 149 1181 331 5021 467 9923 499 10909 523 11941 653 17959 823 26879