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A268354
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Highest power of 7 dividing n.
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7
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1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 49, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 49, 1, 1, 1, 1, 1, 1, 7
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OFFSET
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1,7
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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 7.
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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) = 7^valuation(n,7).
Completely multiplicative with a(7) = 7, a(p) = 1 for prime p and p <> 7. - Andrew Howroyd, Jul 20 2018
a(n) = gcd(n,7^n).
O.g.f.: x/(1 - x) + 6*Sum_{n >= 1} 7^(n-1)*x^(7^n)/ (1 - x^(7^n)). (End)
Sum_{k=1..n} a(k) ~ (6/(7*log(7)))*n*log(n) + (4/7 + 6*(gamma-1)/(7*log(7)))*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 15 2022
Dirichlet g.f.: zeta(s)*(7^s-1)/(7^s-7). - Amiram Eldar, Jan 03 2023
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EXAMPLE
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Since 14 = 7 * 2, a(14) = 7. Likewise, since 7 does not divide 13, a(13) = 1.
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MATHEMATICA
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PROG
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(Sage) [7^valuation(i, 7) for i in [1..100]]
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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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EXTENSIONS
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STATUS
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approved
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