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%I #17 Sep 06 2020 11:12:28
%S 16,32,80,224,720,2976,15120,92448,704592,6455520,69518032,897158048,
%T 12700875536,203746961696,3674124288080,74060581880032,
%U 1692860294726352,42020514867170208,1144626896435067664,33222330682700081440
%N a(n) = f(a(n-1)), with f(m) = Sum i*b(i)*2^(i-1), m = Sum b(i)*2^i, and starting value 16.
%C Previous name was : For N=Sum a(i).2^i, a(i)=0,1, set D(N)/D(2)=Sum i.a(i).2^(i-1); sequence gives n-th derivative of 16.
%C Actually, it appears that f(n) for n=0, 1, 2, ... gives 0, 0, 1, 1, 4, 4, 5, 5, 12, 12, 13, 13, 16, 16, 17, 17, 32, 32, 33, 33, 36, ... that is A136013. - _Michel Marcus_, Jul 18 2013 [This now follows from the comments in A136013. - _Pontus von Brömssen_, Sep 06 2020]
%F a(n) = 16*A007823(n). - _Pontus von Brömssen_, Sep 06 2020
%o (PARI) D(n) = {b = binary(n); lb = #b; rb = vector(lb, i, b[lb-i+1]); return(sum (i=0, lb-1, i*2^(i-1)*rb[i+1]));}
%o lista(nn) = {m = 16; for (n=0, nn, print1(m, ", "); m = D(m););} \\ _Michel Marcus_, Jul 18 2013
%Y Cf. A007823, A136013.
%K nonn
%O 0,1
%A Ralph Buchholz [ ralph(AT)defcen.gov.au ], Leisa Condie
%E Edited by _Michel Marcus_, Jul 18 2013
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