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%I #12 May 11 2019 18:32:57
%S 2,4,5,6,9,10,12,14,15,17,19,20,24,26,33,34,36,46,47,48,66,73,74,79,
%T 81,82,92,98,101,103,104,106,107,110,113,118,119,126,131,132,133,136,
%U 137,143,144,145,147,151,156,158,161,164,171,181,185,192,195,198,200,204
%N Numbers k such that the sum of the digits of 4^k is prime.
%C This is the 4th row of a table which begins as follows.
%C A(j,k) = numbers k such that the sum of the digits of j^k is prime.
%C j | A(j,k)
%C --+-------------------------------------------------------
%C 1 | none
%C 2 | A076203
%C 3 | none (3 | sum of digits)
%C 4 | 2, 4, 5, 6, 9, 10, 12, 14, 15, 17, ... (this sequence)
%C 5 | 1, 2, 4, 5, 6, 7, 19, ...
%F Numbers k such that A007953(A000302(k)) is in A000040.
%e a(1) = 2 because digit sum(4^2) = digit sum(16) = 1+6 = 7.
%e a(2) = 4 because digit sum(4^4) = digit sum(256) = 13.
%e a(3) = 5 because digit sum(4^5) = digit sum(1024) = 7.
%e a(4) = 6 because digit sum(4^6) = digit sum(4096) = 19.
%e a(5) = 9 because digit sum(4^9) = digit sum(262144) = 19.
%e a(6) = 10 because digit sum(4^10) = digit sum(1048576) = 31.
%e a(7) = 12 because digit sum(4^12) = digit sum(16777216) = 37.
%e a(8) = 14 because digit sum(4^14) = digit sum(268435456) = 43.
%e a(9) = 15 because digit sum(4^15) = digit sum(1073741824) = 37.
%e a(10) = 17 because digit sum(4^17) = digit sum(17179869184) = 61.
%p sd:=proc(n) options operator, arrow: add(convert(n, base, 10)[j], j=1..nops(convert(n, base, 10))) end proc: a:=proc(n) if isprime(sd(4^n)) = true then n else end if end proc: seq(a(n),n=1..150); # _Emeric Deutsch_, Nov 24 2007
%t Select[Range[500], PrimeQ[Plus @@ IntegerDigits[4^# ]] &] (* _Stefan Steinerberger_, Nov 20 2007 *)
%Y Cf. A000040, A000302, A007953, A076203.
%K base,easy,less,nonn
%O 1,1
%A _Jonathan Vos Post_, Nov 17 2007
%E More terms from _Stefan Steinerberger_ and _Emeric Deutsch_, Nov 20 2007
%E Edited by _Jon E. Schoenfield_, May 11 2019
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