A214411 The maximum exponent k of 7 such that 7^k divides n.

0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0

Offset

1,49

Comments

7-adic valuation of n.

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Dario T. de Castro, P-adic Order of Positive Integers via Binomial Coefficients, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 22, Paper A61, 2022.

Formula

G.f.: Sum_{k>=1} x^(7^k)/(1-x^(7^k)). See A112765. - Wolfdieter Lang, Jun 18 2014

If n == 0 (mod 7) then a(n) = 1 + a(n/7), otherwise a(n) = 0. - M. F. Hasler, Mar 05 2020

Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/6. - Amiram Eldar, Jan 17 2022

a(n) = 7*Sum_{j=1..floor(log(n)/log(7))} frac(binomial(n, 7^j)*7^(j-1)/n). - Dario T. de Castro, Jul 12 2022

Example

n=147 = 3*7*7 is divisible by 7^2, so a(147)=2.

Maple

seq(padic:-ordp(n, 7), n=1..100); # Robert Israel, Mar 05 2020

Mathematica

mek[n_]:=Module[{k=Ceiling[Log[7, n]]}, While[!Divisible[n, 7^k], k--]; k]; Array[ mek, 140] (* Harvey P. Dale, Mar 27 2017 *)
IntegerExponent[Range[150], 7] (* Suggested by Amiram Eldar *) (* Harvey P. Dale, Mar 07 2020 *)

Prog

(PARI) a(n)=valuation(n, 7) \\ Charles R Greathouse IV, Jul 17 2012
(PARI) A=vector(1000); for(i=1, log(#A+.5)\log(7), forstep(j=7^i, #A, 7^i, A[j]++)); A \\ Charles R Greathouse IV, Jul 17 2012

Crossrefs

Cf. A007814 (2-adic), A007949 (3-adic), A112765 (5-adic), A082784.

Sequence in context: A347714 A089807 A089810 * A324179 A372357 A216577

Adjacent sequences: A214408 A214409 A214410 * A214412 A214413 A214414

Keyword

nonn,easy

Author

Redjan Shabani, Jul 16 2012

Status

approved