OFFSET

3,2

COMMENTS

We conjecture that k in the definition exists for every n>=3.

a(n)=0 for n=14,15,28,32,38,48,59,63,69,76,91,...

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, ... .

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.

EDI(p) shows how many consecutive primes, beginning with 5, we obtain in the considered sequence corresponding to prime p.

EXAMPLE

If n=4, evidently, k=2, since 5^2=25, s(2)= 2+5 = 7 = prime(4). So a(4)=2.

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).

PROG

(PARI) s(k) = my(sd = sumdigits(5^k)); sd/2^valuation(sd, 2);

a(n) = {p = prime(n); k = 1; while ((sk=s(k)) % p, k++); if (sk == p, k, 0); } \\ Michel Marcus, Dec 29 2014

CROSSREFS

KEYWORD

nonn,base

AUTHOR

Vladimir Shevelev, Dec 20 2014

EXTENSIONS

More terms from Peter J. C. Moses, Dec 20 2014

STATUS

approved