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A234957
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Highest power of 4 dividing n.
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15
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1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 16, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 16, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 16, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 64, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 16
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OFFSET
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1,4
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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 4.
In the binary representation of n, remove zeros from the right until the number of zeros is even, then remove all but the rightmost one bit. - Ralf Stephan, Jan 05 2014
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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) = 4^(valuation(n,4)).
a(n) = 4^(floor(valuation(n,2)/2)) = 4^A004526(A007814(n)). Recurrence: a(4n) = 4a(n), a(4n+k) = 1 for k=1,2,3. - Ralf Stephan, Jan 05 2014
G.f.: x/(1 - x) + 3 * Sum_{k>=1} 4^(k-1)*x^(4^k)/(1 - x^(4^k)). - Ilya Gutkovskiy, Jul 10 2019
Multiplicative with a(2^e) = 2^(2*floor(e/2)), and a(p^e) = 1 if p >= 3.
Dirichlet g.f.: zeta(s)*(4^s-1)/(4^s-4).
Sum_{k=1..n} a(k) ~ (3/(8*log(2)))*n*log(n) + (5/8 + 3*(gamma-1)/(8*log(2)))*n, where gamma is Euler's constant (A001620). (End)
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EXAMPLE
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Since 8=4*2, then a(8)=4. Likewise, since 4 does not divide 9, a(9)=1.
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MATHEMATICA
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Table[4^(IntegerExponent[n, 4]), {n, 1, 50}] (* G. C. Greubel, Apr 13 2017 *)
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PROG
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(Sage)
n=200 #change n for more terms
[4^(valuation(i, 4)) for i in [1..n]]
(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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