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Number of compositions of 1 into parts 1/2^k with 0 <= k <= n.
5

%I #43 Mar 31 2021 12:06:41

%S 1,2,6,56,5272,47350056,3820809588459176,

%T 24878564279781563409541239097464,

%U 1054787931172699885204409659788147413348784265452313995416385160

%N Number of compositions of 1 into parts 1/2^k with 0 <= k <= n.

%C Equivalently, the number of compositions of 2^n into powers of 2.

%H Alois P. Heinz, <a href="/A248377/b248377.txt">Table of n, a(n) for n = 0..11</a>

%H B. Abdul-Baki, <a href="http://www.codeproject.com/Messages/4917146/Re-Coding-music-rhythms.aspx">Rhythmic Figures</a>

%F a(n) = A023359(2^n).

%F lim_{n->oo} a(n+1)/a(n)^2 = 1.704176310706592045608982.... - _Bassam Abdul-Baki_, Sep 03 2020

%e a(0) = 1: [1].

%e a(1) = 2: [1/2,1/2], [1].

%e a(2) = 6: [1/4,1/4,1/4,1/4], [1/2,1/4,1/4], [1/4,1/2,1/4], [1/4,1/4,1/2], [1/2,1/2], [1].

%p b:= proc(n) option remember; `if`(n=0, 1,

%p add(b(n-2^j), j=0..ilog2(n)))

%p end:

%p a:= n-> b(2^n):

%p seq(a(n), n=0..10); # _Alois P. Heinz_, Oct 20 2014

%t $RecursionLimit = 2000; Clear[b]; b[n_] := b[n] = If[n == 0, 1, Sum[b[n - 2^j], {j, 0, Log[2, n] // Floor}]]; a[n_] := b[2^n]; Table[a[n], {n, 0, 10}] (* _Jean-François Alcover_, Oct 30 2014, after _Alois P. Heinz_ *)

%Y Cf. A023359.

%Y Row sums of A323840.

%K nonn

%O 0,2

%A _Bassam Abdul-Baki_, Oct 05 2014