A029930 If 2n = Sum 2^e_i, a(n) = Product 2^e_i.
1, 2, 4, 8, 8, 16, 32, 64, 16, 32, 64, 128, 128, 256, 512, 1024, 32, 64, 128, 256, 256, 512, 1024, 2048, 512, 1024, 2048, 4096, 4096, 8192, 16384, 32768, 64, 128, 256, 512, 512, 1024, 2048, 4096, 1024, 2048, 4096, 8192, 8192, 16384, 32768, 65536, 2048
Offset: 0
Examples
14 = 8+4+2 so a(7) = 8*4*2 = 64.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..8191
- Arvind Ayyer, A. Prasad and S. Spallone, Representations of symmetric groups with non-trivial determinant, arXiv preprint arXiv:1604.08837 [math.RT], 2016. See Eq. (14).
Programs
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Maple
HammingWeight := n -> add(i, i = convert(n, base, 2)): a := proc(n) option remember; `if`(n = 0, 1, ifelse(n::even, 2^HammingWeight(n/2)*a(n/2), 2*a(n-1))) end: seq(a(n), n = 0..48); # Peter Luschny, Oct 30 2021
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Mathematica
e1[n_] := Total[IntegerDigits[n, 2]]; a[0] = 1; a[n_] := a[n] = If[EvenQ[ n], 2^e1[n/2] a[n/2], 2 a[n-1]]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Mar 07 2016 *) -
PARI
a(n) = {my(bd = Vecrev(binary(n))); prod(k=1, #bd, if (bd[k], 2^k, 1));} \\ Michel Marcus, Mar 07 2016
Formula
From Ralf Stephan, Jun 19 2003: (Start)
G.f.: Prod_{k>=0} 1+2^(k+1)x^2^k.
a(0) = 1, a(2n) = 2^e1(n)*a(n), a(2n+1) = 2a(2n), where e1(n) = A000120(n).
a(n) = 2^A029931(n). (End)