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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 11. Sequence lists primes p for which s(p,k_1)=11.
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%I #12 Dec 07 2018 12:25:35

%S 2,5,7,13,23,29,31,43,47,53,59,79,83,97,137,139,173,191,227,239,241,

%T 257,263,281,317,331,337,349,353,359,373,383,421,439,443,449,461,463,

%U 467,479,499,509,523,547,557,563,569,593,599,607,619,641,643,653,659

%N 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 11. Sequence lists primes p for which s(p,k_1)=11.

%C For s(p,k_1)=5 and s(p,k_1)=7 see A251964 and A252280 respectively.

%t s[p_, k_] := Module[{s = Total[IntegerDigits[p^k]]}, s/2^IntegerExponent[s, 2]]; f11[p_] := Module[{k = 1}, While[! Divisible[s[p, k], 11], k++]; k]; ok11Q[p_] := s[p, f11[p]] == 11; Select[Range[1000], PrimeQ[#] && ok11Q[#] &] (* _Amiram Eldar_, Dec 07 2018 *)

%o (PARI) s(p,k) = my(s=sumdigits(p^k)); s >> valuation(s, 2);

%o f11(p) = my(k=1); while(s(p,k) % 11, k++); k;

%o isok11(p) = s(p, f11(p)) == 11;

%o lista11(nn) = forprime(p=2, nn, if (isok11(p), print1(p, ", "))); \\ _Michel Marcus_, Dec 07 2018

%Y Cf. A221858, A225039, A225093, A251964, A252280.

%K nonn,base

%O 1,1

%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Dec 16 2014