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%I #23 Oct 15 2019 11:47:18
%S 1,2,5,4,14,6,7,16,21,23,24,0,0,32,19,20,22,186,177,26,29,27,61,236,
%T 34,0,36,78,54,0,41,87,43,44,188,0,55,118,229,66,59,70,69,60,58,0,279,
%U 147,81,610,74,325,85,101,75,179,0,369,100,97,0,91,193,95,205
%N Let k be the smallest number such that s(k) = odd part of digital sum of 5^k is a multiple of prime(n); then a(n)=k, if s(k) = prime(n). Otherwise, or if there is no such k, a(n)=0.
%C We conjecture that k in the definition exists for every n>=3.
%C a(n)=0 for n=14,15,28,32,38,48,59,63,69,76,91,...
%C Note that we can continue the series of sequences A252666, A252668, ... by changing 2^k in the definition to 5^k, 7^k, 11^k, ..., prime(i)^k, ... .
%C Let the position of the first zero in the sequence corresponding to prime(i) be u(i). Then we call v(i)=u(i)-3 the exponential digital index (EDI) of prime(i). It is clear that in the case of i=2, prime(i)=3 and EDI(3)=0.
%C EDI(p) shows how many consecutive primes, beginning with 5, we obtain in the considered sequence corresponding to prime p.
%e If n=4, evidently, k=2, since 5^2=25, s(2)= 2+5 = 7 = prime(4). So a(4)=2.
%e If n=14, then k=57, but s(57)>prime(14)=43, so a(14)=0 (the equation s(x)=43 has the smallest solution x=107).
%o (PARI) s(k) = my(sd = sumdigits(5^k)); sd/2^valuation(sd, 2);
%o a(n) = {p = prime(n); k = 1; while ((sk=s(k)) % p, k++); if (sk == p, k, 0);} \\ _Michel Marcus_, Dec 29 2014
%Y Cf. A251964, A252280, A252281, A252282, A252283, A252666.
%K nonn,base
%O 3,2
%A _Vladimir Shevelev_, Dec 20 2014
%E More terms from _Peter J. C. Moses_, Dec 20 2014
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