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A264981
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Highest power of 9 dividing n.
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2
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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
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OFFSET
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1,9
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COMMENTS
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The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 9. - Tom Edgar, Feb 02 2016
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LINKS
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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.
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FORMULA
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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
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)
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EXAMPLE
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Since 18 = 9 * 2, a(18) = 9. Likewise, since 9 does not divide 17, a(17) = 1. - Tom Edgar, Feb 02 2016
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MATHEMATICA
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Table[9^Length@ TakeWhile[Reverse@ IntegerDigits[n, 9], # == 0 &], {n, 99}] (* Michael De Vlieger, Dec 09 2015 *)
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PROG
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(Scheme)
(define (A264981 n) (let loop ((k 9)) (if (not (zero? (modulo n k))) (/ k 9) (loop (* 9 k)))))
(Sage) [9^valuation(i, 9) for i in [1..100]] # Tom Edgar, Feb 02 2016
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CROSSREFS
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KEYWORD
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nonn,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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