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A268355
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Highest power of 8 dividing n.
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1
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1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 64, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1
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OFFSET
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1,8
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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 8.
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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) = 8^valuation(n,8).
G.f.: Sum_{m>=0} 8^m * Sum_{j=1..7} x^(j*8^m)/(1-x^(8^(m+1))). - Robert Israel, Feb 03 2016
Multiplicative with a(2^e) = 2^(3*floor(e/3)), and a(p^e) = 1 if p >= 3.
Dirichlet g.f.: zeta(s)*(8^s-1)/(8^s-8).
Sum_{k=1..n} a(k) ~ (7/(24*log(2)))*n*log(n) + (9/16 + 7*(gamma-1)/(24*log(2)))*n, where gamma is Euler's constant (A001620). (End)
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EXAMPLE
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Since 16 = 8 * 2, a(16) = 8. Likewise, since 8 does not divide 15, a(15) = 1.
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MAPLE
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seq(8^floor(padic:-ordp(n, 2)/3), n=1..100); # Robert Israel, Feb 03 2016
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MATHEMATICA
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PROG
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(Sage) [8^valuation(i, 8) for i in [1..100]]
(Python)
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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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