|
%I #18 Aug 29 2022 06:12:07
%S 0,1,-3,-13,15,-37,23,-125,-467,-2269,-2269,-12147,-66877,-66877,
%T -66877,-66877,-370963,-2061725,-11464371,17899313,99555423,99555423,
%U -155471765,-864636987,1350136841,1350136841,-2108411107,-2108411107
%N Let f(n)=Floor[Mod[10^k*(7/(4*k + 1) - 6/(4*k + 3) - 1/(4*k + 5)), 3]]; M0 = {{0, 1}, {1, 1/2}}; M = {{0, 2}, {2, 1}}; as Mh=M0.M.(M0+I*f[n]); v[(n)=Mh.v(n-1), then a(n) is the first element of v.
%H G. C. Greubel, <a href="/A152270/b152270.txt">Table of n, a(n) for n = 0..1000</a>
%t f[k_] = Floor[Mod[10^k*(7/(4*k + 1) - 6/(4*k + 3) - 1/(4*k + 5)), 3]];
%t M0 = {{0, 1}, {1, 1/2}}; M = {{0, 2}, {2, 1}};
%t Mh[n_] := M0.(M.Inverse[f[n]*IdentityMatrix[2] + M0]);
%t v[0] = {0, 1};
%t v[n_] := v[n] = Mh[n].v[n - 1]
%t Table[ -v[n][[1]]/2, {n, 0, 30}]
%K sign,uned,obsc,base
%O 0,3
%A _Roger L. Bagula_ and _Alexander R. Povolotsky_, Dec 01 2008
|
