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%I #9 Dec 18 2022 06:09:36
%S 0,0,-1,0,1,0,-8,0,25,0,-97,0,367,0,-1398,0,5259,0,-19878,0,75319,0,
%T -285137,0,1078711,0,-4081933,0,15449144,0,-58467488,0,221260378,0,
%U -837337471,0,3168858565,0,-11992319160,0,45383925816,0,-171751869342,0,649981903584,0,-2459806349188,0
%N p[n_, k_]=Product[(E/(2-E))^i, {i, 1, n}]/Product[(E/(2-E))^i, {i, Floor[n/2^k], n}], a(n) = Sum[Floor[p[n, k]/(8*p[n-1, k])], {k, 1, 8}]
%t p[n_, k_]=Product[(E/(2-E))^i, {i, 1, n}]/Product[(E/(2-E))^i, {i, Floor[n/2^k], n}]
%t digits=80
%t f[n_]=Sum[Floor[p[n, k]/(8*p[n-1, k])], {k, 1, 8}]
%t at=Table[f[n], {n, 2, digits}]
%K sign,uned,less
%O 0,7
%A _Roger L. Bagula_, Nov 21 2003
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