A264981 Highest power of 9 dividing n.
1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 81, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9
Offset: 1
Examples
Since 18 = 9 * 2, a(18) = 9. Likewise, since 9 does not divide 17, a(17) = 1. - _Tom Edgar_, Feb 02 2016
Links
- Antti Karttunen, Table of n, a(n) for n = 1..6561
- Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
- Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.
Crossrefs
Programs
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Mathematica
Table[9^Length@ TakeWhile[Reverse@ IntegerDigits[n, 9], # == 0 &], {n, 99}] (* Michael De Vlieger, Dec 09 2015 *) 9^Table[IntegerExponent[n, 9], {n, 150}] (* Vincenzo Librandi, Feb 03 2016 *) -
PARI
a(n) = 9^valuation(n, 9); \\ Michel Marcus, Dec 08 2015
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Sage
[9^valuation(i, 9) for i in [1..100]] # Tom Edgar, Feb 02 2016
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Scheme
(define (A264981 n) (let loop ((k 9)) (if (not (zero? (modulo n k))) (/ k 9) (loop (* 9 k)))))
Formula
a(n) = 9^valuation(n,9). - Tom Edgar, Feb 02 2016
G.f.: x/(1 - x) + 8 * Sum_{k>=1} 9^(k-1)*x^(9^k)/(1 - x^(9^k)). - Ilya Gutkovskiy, Jul 10 2019
From Amiram Eldar, Dec 31 2022: (Start)
Multiplicative with a(3^e) = 3^(2*floor(e/2)), and a(p^e) = 1 if p != 3.
Dirichlet g.f.: zeta(s)*(9^s-1)/(9^s-9).
Sum_{k=1..n} a(k) ~ (4/(9*log(3)))*n*log(n) + (5/9 + 4*(gamma-1)/(9*log(3)))*n, where gamma is Euler's constant (A001620). (End)
Extensions
Keyword:mult added by Andrew Howroyd, Jul 20 2018
Comments