A065916 Denominator of sigma(8*n^2)/sigma(4*n^2).
7, 31, 7, 127, 7, 31, 7, 511, 7, 31, 7, 127, 7, 31, 7, 2047, 7, 31, 7, 127, 7, 31, 7, 511, 7, 31, 7, 127, 7, 31, 7, 8191, 7, 31, 7, 127, 7, 31, 7, 511, 7, 31, 7, 127, 7, 31, 7, 2047, 7, 31, 7, 127, 7, 31, 7, 511, 7, 31, 7, 127, 7, 31, 7, 32767, 7, 31, 7, 127, 7, 31, 7, 511, 7
Offset: 1
Examples
sigma(72)/sigma(36) = 15/7, so a(3) = 7.
Links
- Harry J. Smith, Table of n, a(n) for n = 1..1000
- Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple ordinary generating functions, 2004.
- Ralf Stephan, Table of generating functions.
- Index entries for sequences related to binary expansion of n.
Programs
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Maple
nmax:=73: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := 2*4^(p+1) - 1 od: od: seq(a(n), n=1..nmax); # Johannes W. Meijer, Feb 12 2013
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Mathematica
a[n_] := 2^(2*IntegerExponent[n, 2] + 3) - 1; Array[a, 100] (* Amiram Eldar, Jun 21 2024 *)
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PARI
a(n) = denominator(sigma(8*n^2)/sigma(4*n^2)) \\ Harry J. Smith, Nov 04 2009
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PARI
a(n)=2^(2*valuation(n,2)+3)-1 \\ Charles R Greathouse IV, Nov 18 2015
Formula
From Johannes W. Meijer, Feb 12 2013: (Start)
a((2*n-1)*2^p) = 2*4^(p+1) - 1 for p >= 0 and n >= 1. Observe that a(2^p) = A083420(p+1).
a(2^(p+3)*n + 2^(p+2) - 1) = a(2^(p+2)*n + 2^(p+1) - 1) for p >= 0. (End)
a(n) = 2^s-1, with s = 2*A007814(n) + 3. Recurrence: a(2n) = 4a(n)+3, a(2n+1) = 7. - Ralf Stephan, Aug 22 2013
Comments