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%I #29 Oct 16 2019 13:24:35
%S 7,7,7,7,421,2311,43321,59730109,537052693
%N In view of their definitions, let us refer to A251964 as sequence "5", A252280 as sequence "7", and similarly define sequence "prime(n)"; a(n) is the third term of the intersection of sequences "5", ..., "prime(n)".
%C Is this sequence finite?
%C Up to n=13, the first two terms of the intersection of sequences "5", ..., "prime(n)" are 2 and 5 respectively.
%t s[p_, k_] := Module[{s = Total[IntegerDigits[p^k]]}, s/2^IntegerExponent[s, 2]]; f[p_,q_] := Module[{k = 1}, While[ ! Divisible[s[p, k], q], k++]; k]; okQ[p_,q_] := s[p, f[p,q]] == q; okpQ[p_,nbseq_] := Module[{ans=True}, Do[If[!okQ[p,Prime[k+2]], ans=False; Break[]],{k,1,nbseq}]; ans]; a[n_]:=Module[{c=0, p=2},While[c<3 , If[okpQ[p,n],c++];p=NextPrime[p]];NextPrime[p,-1]]; Array[a,6] (* _Amiram Eldar_, Dec 09 2018 *)
%o (PARI) s(p, k) = my(s=sumdigits(p^k)); s >> valuation(s, 2);
%o f(p, vp) = my(k=1); while(s(p,k) % vp, k++); k;
%o isok(p, vp) = s(p, f(p, vp)) == vp;
%o isokp(p, nbseq) = {for (k=1, nbseq, if (! isok(p, prime(k+2)), return (0));); return (1);}
%o a(n) = {my(nbpok = 0); forprime(p=2, oo, if (isokp(p, n), nbpok ++); if (nbpok == 3, return (p)););} \\ _Michel Marcus_, Dec 09 2018
%Y Cf. A251964, A252280, A252281, A252282, A252283, A252666, A252668, A252670.
%K nonn,base,more
%O 3,1
%A _Vladimir Shevelev_, Dec 21 2014
%E More terms from _Peter J. C. Moses_, Dec 21 2014
%E a(10)-a(11) from _Michel Marcus_, Dec 09 2018
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