OFFSET
1,1
COMMENTS
For a prime p, denote by s(p,k) the odd part of the digital sum of p^k. Let, for the first time, s(p,k) be divisible by 5 for k=k_1, be divisible by 7 for k=k_2 and
be divisible by 11 for k=k_3.
Sequence lists primes p for which s(p,k_1)=5, s(p,k_2)=7 and s(p,k_3)=11.
Consider also sequence which lists primes p with s(p,k_1)=5, s(p,k_2)=7, s(p,k_3)=11 and s(p,k_4)=13; sequence which lists primes p with s(p,k_1)=5, s(p,k_2)=7, s(p,k_3)=11, s(p,k_4)=13 and s(p,k_5)=17; etc. Then it seems that we will be eventually left with 2 and 5.
For example, for s(p,k_1)=5, s(p,k_2)=7,
s(p,k_3)=11, s(p,k_4)=13, s(p,k_5)=17 and
s(p,k_6)=19, the known terms of the sequence are 2, 5, 2311, 4721, 43321.
A weaker conjecture: {2,5} is the intersection of all such sequences.
MATHEMATICA
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; Select[Range[2400], PrimeQ[#] && okQ[#, 5] && okQ[#, 7] && okQ[#, 11] &] (* Amiram Eldar, Dec 08 2018 *)
PROG
(PARI) s(p, k) = my(s=sumdigits(p^k)); s >> valuation(s, 2);
f5(p) = my(k=1); while(s(p, k) % 5, k++); k;
isok5(p) = s(p, f5(p)) == 5;
f7(p) = my(k=1); while(s(p, k) % 7, k++); k;
isok7(p) = s(p, f7(p)) == 7;
f11(p) = my(k=1); while(s(p, k) % 11, k++); k;
isok11(p) = s(p, f11(p)) == 11;
lista(nn) = forprime(p=2, nn, if (isok5(p) && isok7(p) && isok11(p), print1(p, ", "))); \\ Michel Marcus, Dec 08 2018
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Vladimir Shevelev and Peter J. C. Moses, Dec 16 2014
EXTENSIONS
More terms from Michel Marcus, Dec 08 2018
STATUS
approved