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URL: https://oeis.org/A268357

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A268357
Highest power of 11 dividing n.
5
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 121, 1
OFFSET
1,11
COMMENTS
The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 11.
This first index where this differs from A109014 is 121; a(121) = 121 and A109014(121) = 11.
LINKS
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.
FORMULA
a(n) = 11^valuation(n,11).
Completely multiplicative with a(11) = 11, a(p) = 1 for prime p and p<>11. - Andrew Howroyd, Jul 20 2018
From Peter Bala, Feb 21 2019: (Start)
a(n) = gcd(n,11^n).
O.g.f.: x/(1 - x) + 10*Sum_{n >= 1} 11^(n-1)*x^(11^n)/ (1 - x^(11^n)). (End)
Sum_{k=1..n} a(k) ~ (10/(11*log(11)))*n*log(n) + (6/11 + 10*(gamma-1)/(11*log(11)))*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 15 2022
Dirichlet g.f.: zeta(s)*(11^s-1)/(11^s-11). - Amiram Eldar, Jan 03 2023
EXAMPLE
Since 22 = 11 * 2, a(22) = 11. Likewise, since 11 does not divide 21, a(21) = 1.
MATHEMATICA
Table[11^IntegerExponent[n, 11], {n, 130}] (* Bruno Berselli, Feb 03 2016 *)
PROG
(SageMath) [11^valuation(i, 11) for i in [1..130]]
(Magma) [11^Valuation(n, 11): n in [1..130]]; // Vincenzo Librandi, Feb 03 2016
KEYWORD
nonn,easy,mult
AUTHOR
Tom Edgar, Feb 02 2016
STATUS
approved