[go: up one dir, main page]

login
A214411
The maximum exponent k of 7 such that 7^k divides n.
21
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
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
KEYWORD
nonn,easy
AUTHOR
Redjan Shabani, Jul 16 2012
STATUS
approved