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A251964 For a prime p, denote by s(p,k) the odd part of the digital sum of p^k. Let k_1 be the smallest k such that s(p,k) is divisible by 5. Sequence lists primes p for which s(p,k_1)=5.

%I A251964 #22 Dec 07 2018 12:25:22
%S A251964 2,5,7,11,19,23,37,41,61,71,73,101,109,113,127,131,163,179,181,211,
%T A251964 229,241,251,271,307,311,313,383,389,401,421,433,449,479,521,523,541,
%U A251964 557,569,571,587,601,613,631,659,677,751,811,827,839,857,929,947,971,977
%N A251964 For a prime p, denote by s(p,k) the odd part of the digital sum of p^k. Let k_1 be the smallest k such that s(p,k) is divisible by 5. Sequence lists primes p for which s(p,k_1)=5.
%C A251964 Let p be a prime other than 3. If p is not in the sequence, then either s(p,k_1) >= 25 or k_1 does not exist. We conjecture that k_1=k_1(p) exists for every prime p.
%e A251964 For p=7, s(p,1) = 7, s(p,2) = 4+9 = 13, s(p,3) = (3+4+3)/2 = 5. So 7 is a term.
%e A251964 For p=13, s(p,1) = 1, s(p,2) = 1, s(p,3) = 19, s(p,4) = 11, s(p,5) = 25. So 13 is not in the sequence.
%t A251964 s[p_, k_] := Module[{s = Total[IntegerDigits[p^k]]}, s/2^IntegerExponent[s, 2]]; f5[p_] := Module[{k = 1}, While[! Divisible[s[p, k], 5], k++]; k]; ok5Q[p_] := s[p, f5[p]] == 5; Select[Range[1000], PrimeQ[#] && ok5Q[#] &] (* _Amiram Eldar_, Dec 07 2018 *)
%o A251964 (PARI) s(p,k) = my(s=sumdigits(p^k)); s >> valuation(s, 2);
%o A251964 f5(p) = my(k=1); while(s(p,k) % 5, k++); k;
%o A251964 isok5(p) = s(p, f5(p)) == 5;
%o A251964 lista5(nn) = forprime(p=2, nn, if (isok5(p), print1(p, ", "))); \\ _Michel Marcus_, Dec 07 2018
%Y A251964 Cf. A221858, A225039, A225093.
%K A251964 nonn,base
%O A251964 1,1
%A A251964 _Vladimir Shevelev_, Dec 11 2014
%E A251964 More terms from _Peter J. C. Moses_, Dec 11 2014