A268357 Highest power of 11 dividing n.
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
Examples
Since 22 = 11 * 2, a(22) = 11. Likewise, since 11 does not divide 21, a(21) = 1.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- 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.
Programs
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Magma
[11^Valuation(n,11): n in [1..130]]; // Vincenzo Librandi, Feb 03 2016
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Mathematica
Table[11^IntegerExponent[n, 11], {n, 130}] (* Bruno Berselli, Feb 03 2016 *)
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Sage
[11^valuation(i, 11) for i in [1..130]]
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
Comments