cp's OEIS Frontend

A264981 Highest power of 9 dividing n.

%I A264981 #29 Dec 31 2022 03:42:18
%S A264981 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,
%T A264981 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,
%U A264981 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
%N A264981 Highest power of 9 dividing n.
%C A264981 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
%H A264981 Antti Karttunen, <a href="/A264981/b264981.txt">Table of n, a(n) for n = 1..6561</a>
%H A264981 Tyler Ball, Tom Edgar, and Daniel Juda, <a href="http://dx.doi.org/10.4169/math.mag.87.2.135">Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem</a>, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
%H A264981 Tom Edgar and Michael Z. Spivey, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Edgar/edgar3.html">Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers</a>, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.
%F A264981 a(n) = 9^valuation(n,9). - _Tom Edgar_, Feb 02 2016
%F A264981 G.f.: x/(1 - x) + 8 * Sum_{k>=1} 9^(k-1)*x^(9^k)/(1 - x^(9^k)). - _Ilya Gutkovskiy_, Jul 10 2019
%F A264981 From _Amiram Eldar_, Dec 31 2022: (Start)
%F A264981 Multiplicative with a(3^e) = 3^(2*floor(e/2)), and a(p^e) = 1 if p != 3.
%F A264981 Dirichlet g.f.: zeta(s)*(9^s-1)/(9^s-9).
%F A264981 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)
%e A264981 Since 18 = 9 * 2, a(18) = 9. Likewise, since 9 does not divide 17, a(17) = 1. - _Tom Edgar_, Feb 02 2016
%t A264981 Table[9^Length@ TakeWhile[Reverse@ IntegerDigits[n, 9], # == 0 &], {n, 99}] (* _Michael De Vlieger_, Dec 09 2015 *)
%t A264981 9^Table[IntegerExponent[n, 9], {n, 150}] (* _Vincenzo Librandi_, Feb 03 2016 *)
%o A264981 (Scheme)
%o A264981 (define (A264981 n) (let loop ((k 9)) (if (not (zero? (modulo n k))) (/ k 9) (loop (* 9 k)))))
%o A264981 (PARI) a(n) = 9^valuation(n, 9); \\ _Michel Marcus_, Dec 08 2015
%o A264981 (Sage) [9^valuation(i, 9) for i in [1..100]] # _Tom Edgar_, Feb 02 2016
%Y A264981 Cf. A001620, A264979.
%Y A264981 Similar sequences for other bases: A006519 (2), A038500 (3), A234957 (4), A060904 (5), A234959 (6).
%K A264981 nonn,mult
%O A264981 1,9
%A A264981 _Antti Karttunen_, Dec 07 2015
%E A264981 Keyword:mult added by _Andrew Howroyd_, Jul 20 2018