A127804 a(2n) = 2^(2n), a(2n+1) = 2^(2n+1) + a(n).
1, 3, 4, 11, 16, 36, 64, 139, 256, 528, 1024, 2084, 4096, 8256, 16384, 32907, 65536, 131328, 262144, 524816, 1048576, 2098176, 4194304, 8390692, 16777216, 33558528, 67108864, 134225984, 268435456, 536887296, 1073741824, 2147516555, 4294967296, 8590000128
Offset: 0
Links
- Tilman Piesk, Table of n, a(n) for n = 0..1023 (terms 0..61 from R. J. Mathar)
- Tilman Piesk, Illustrated triangles with 32 rows:
- related to necklaces (similar to A385665),
- binary (A115361),
- signed (A127803)
- Tilman Piesk, Serration of Boolean functions (Wikiversity)
Programs
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Maple
A127804 := proc(n) add( A127803(n,k),k=0..n) ; end proc: # R. J. Mathar, Feb 12 2024 # second Maple program: a:= proc(n) option remember; 2^n+`if`(n::odd, a((n-1)/2), 0) end: seq(a(n), n=0..33); # Alois P. Heinz, Jul 01 2025
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Mathematica
rows = 30; A[n_, k_] := If[k <= n, If[n <= 2 k, 1/(2*2^n - 1), 0], 0]; T = Table[A[n, k], {n, 0, rows-1}, {k, 0, rows-1}] // Inverse; a[n_] := T[[n+1]] // Total; Table[a[n], {n, 0, rows-1}] (* Jean-François Alcover, Jul 03 2018 *)
Formula
Conjecture: a(n) = 1 + A187767(n+1). - Andrew Howroyd, Jun 02 2017
From Tilman Piesk, Jun 30 2025: (Start)
a(n) = Sum_{i=0..A001511(n+1)-1} 2^((n+1) / 2^i - 1)
= Sum_{i=0..A001511(n+1)-1} 2^floor(n / 2^i).
a(n) = A045654(n+1) / 2. (End)
Extensions
Name changed by Tilman Piesk, Jun 30 2025
Comments