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A268357
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Highest power of 11 dividing n.
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5
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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
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OFFSET
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1,11
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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 11.
This first index where this differs from A109014 is 121; a(121) = 121 and A109014(121) = 11.
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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) = 11^valuation(n,11).
Completely multiplicative with a(11) = 11, a(p) = 1 for prime p and p<>11. - Andrew Howroyd, Jul 20 2018
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
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EXAMPLE
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Since 22 = 11 * 2, a(22) = 11. Likewise, since 11 does not divide 21, a(21) = 1.
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MATHEMATICA
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Table[11^IntegerExponent[n, 11], {n, 130}] (* Bruno Berselli, Feb 03 2016 *)
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PROG
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(Sage) [11^valuation(i, 11) for i in [1..130]]
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CROSSREFS
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KEYWORD
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nonn,easy,mult
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AUTHOR
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STATUS
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approved
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