%I
%S 5,19,37,43,97,107,6091,6389,7121,21727,147107,148151,148279,148429,
%T 148469,172877,173209,173741,2621387,5642293,5642321,8932771,8981827,
%U 8981879,9094979,9095089,9997783,10010687,10010789,10037749,10144523,40179929,40365217,40379077,40379197,40386811,40612933
%N Primes p such that p = the cumulative sum of the digitsum in base 15 of the digitproduct in base 4 of each prime < p.
%F sum = sum + digitsum(digitmult(prime,base=4),base=15). The function digitmult(n) multiplies all digits d of n, where d > 0. For example, digitmult(1230) = 1 * 2 * 3 = 6. Therefore, the digitsum in base 15 of the digitmult(333) in base 4 = digitsum(3 * 3 * 3) = digitsum(1C) = 1 + C = 13. (1C in base 15 = 27 in base 10).
%e 5 = digitsum(digitmult(2,b=4),b=15) + sum(mult(3,b=4),b=15) = 2 + 3.
%e 19 = digitsum(digitmult(2,b=4),b=15) + sum(mult(3,b=4),b=15) + sum(mult(11,b=4),b=15) + sum(mult(13,b=4),b=15) + sum(mult(23,b=4),b=15) + sum(mult(31,b=4),b=15) + sum(mult(101,b=4),b=15) = 2 + 3 + 1 + 3 + 6 + 3 + 1.
%Y Cf. A240886 (similar sequence with digit sums in base 3).
%K nonn,base
%O 1,1
%A _Anthony Sand_, May 20 2014
