A268355 Highest power of 8 dividing n.
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
Offset: 1
Examples
Since 16 = 8 * 2, a(16) = 8. Likewise, since 8 does not divide 15, a(15) = 1.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537
- 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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Maple
seq(8^floor(padic:-ordp(n,2)/3), n=1..100); # Robert Israel, Feb 03 2016
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Mathematica
8^Table[IntegerExponent[n, 8], {n, 150}] (* Vincenzo Librandi, Feb 03 2016 *) -
PARI
a(n) = 8^valuation(n, 8); \\ Michel Marcus, Feb 05 2016
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Python
def A268355(n): return (m:=(~n&n-1))+1>>(m.bit_length()%3) # Chai Wah Wu, Jul 09 2022
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Sage
[8^valuation(i, 8) for i in [1..100]]
Formula
a(n) = 8^valuation(n,8).
a(n) = 8^A244413(n).
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
From Amiram Eldar, Dec 31 2022: (Start)
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)
Extensions
Keyword:mult added by Andrew Howroyd, Jul 20 2018
Comments