Sum of primes in odd positions is prime: Difference between revisions
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<br>Let '''p(n)''' be a sequence of prime numbers. |
<br>Let '''p(n)''' be a sequence of prime numbers. |
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<br>Consider the '''p(1),p(3),p(5), ... ,p(i)''' for each ''' |
<br>Consider the '''p(1),p(3),p(5), ... ,p(i)''' for each '''p(i) < 1,000n''' and ''i''' is odd. |
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<br>Let '''sum''' be summarize of these primes. |
<br>Let '''sum''' be summarize of these primes. |
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<br>If '''sum''' is prime then print '''p(i)''' and '''sum''' |
<br>If '''sum''' is prime then print '''p(i)''' and '''sum''' |
Revision as of 13:42, 1 September 2021
Sum of primes in odd positions is prime 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
Let p(n) be a sequence of prime numbers.
Consider the p(1),p(3),p(5), ... ,p(i)' for each p(i) < 1,000n and i is odd.
Let sum be summarize of these primes.
If sum is prime then print p(i) and sum
Ring
<lang ring> load "stdlib.ring" see "working..." + nl see "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 "" + n + " " + sum + nl ok ok ok
next
see "done..." + nl </lang>
- Output:
working... p sum 2 2 5 7 31 89 103 659 149 1181 331 5021 467 9923 499 10909 523 11941 653 17959 823 26879 done...