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A248377 Number of compositions of 1 into parts 1/2^k with 0 <= k <= n.

%I A248377 #43 Mar 31 2021 12:06:41
%S A248377 1,2,6,56,5272,47350056,3820809588459176,
%T A248377 24878564279781563409541239097464,
%U A248377 1054787931172699885204409659788147413348784265452313995416385160
%N A248377 Number of compositions of 1 into parts 1/2^k with 0 <= k <= n.
%C A248377 Equivalently, the number of compositions of 2^n into powers of 2.
%H A248377 Alois P. Heinz, <a href="/A248377/b248377.txt">Table of n, a(n) for n = 0..11</a>
%H A248377 B. Abdul-Baki, <a href="http://www.codeproject.com/Messages/4917146/Re-Coding-music-rhythms.aspx">Rhythmic Figures</a>
%F A248377 a(n) = A023359(2^n).
%F A248377 lim_{n->oo} a(n+1)/a(n)^2 = 1.704176310706592045608982.... - _Bassam Abdul-Baki_, Sep 03 2020
%e A248377 a(0) = 1: [1].
%e A248377 a(1) = 2: [1/2,1/2], [1].
%e A248377 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 A248377 b:= proc(n) option remember; `if`(n=0, 1,
%p A248377       add(b(n-2^j), j=0..ilog2(n)))
%p A248377     end:
%p A248377 a:= n-> b(2^n):
%p A248377 seq(a(n), n=0..10);  # _Alois P. Heinz_, Oct 20 2014
%t A248377 $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 A248377 Cf. A023359.
%Y A248377 Row sums of A323840.
%K A248377 nonn
%O A248377 0,2
%A A248377 _Bassam Abdul-Baki_, Oct 05 2014