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A252668 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. 3
1, 2, 5, 4, 14, 6, 7, 16, 21, 23, 24, 0, 0, 32, 19, 20, 22, 186, 177, 26, 29, 27, 61, 236, 34, 0, 36, 78, 54, 0, 41, 87, 43, 44, 188, 0, 55, 118, 229, 66, 59, 70, 69, 60, 58, 0, 279, 147, 81, 610, 74, 325, 85, 101, 75, 179, 0, 369, 100, 97, 0, 91, 193, 95, 205 (list; graph; refs; listen; history; text; internal format)
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.

LINKS

Table of n, a(n) for n=3..67.

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

Cf. A251964, A252280, A252281, A252282, A252283, A252666.

Sequence in context: A212188 A298585 A102468 * A225046 A079053 A224272

Adjacent sequences:  A252665 A252666 A252667 * A252669 A252670 A252671

KEYWORD

nonn

AUTHOR

Vladimir Shevelev, Dec 20 2014

EXTENSIONS

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

STATUS

approved

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Last modified November 20 06:06 EST 2018. Contains 317385 sequences. (Running on oeis4.)