A108210 - OEIS

A108210

Let M[n] be the 2 X 2 matrix {{0, -3}, {(n - 1), 5*(n - 1)}} and let v[1] = {0, 1}', v[n] = M[n]*v[n - 1]'. Then a[n] is the first entry of v[n].

%I #5 Mar 30 2012 17:34:19

%S 0,3,15,132,1845,35316,855225,25021062,857777445,33710592312,

%T 1493816663025,73679515381890,4003077396124125,237532181213699460,

%U 15283471760441624025,1059866671619938304430,78802244142275499751125

%N Let M[n] be the 2 X 2 matrix {{0, -3}, {(n - 1), 5*(n - 1)}} and let v[1] = {0, 1}', v[n] = M[n]*v[n - 1]'. Then a[n] is the first entry of v[n].

%C Derangement-type quadratic Markov chain.

%t M[n_] := {{0, -3}, {(n - 1), 5*(n - 1)}} v[1] = {0, 1} v[n_] := v[n] = M[n].v[n - 1] a = Table[Abs[v[n][[1]]], {n, 1, 25}]

%Y Cf. A000166.

%K nonn

%O 1,2

%A _Roger L. Bagula_, Jun 15 2005

%E Edited by _N. J. A. Sloane_, Mar 29 2007. The prime indicates transposition. Possible M should be transposed too, the Mathematica code is not clear to me.