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Expansion of 1/((1-x)*((1-5x)^2)*(1-8x)).
1

%I #13 Jun 11 2023 11:02:15

%S 1,19,238,2490,23631,211509,1823908,15348100,127057261,1040261799,

%T 8453319978,68343722910,550640774491,4426107030889,35521389816448,

%U 284771933350920,2281370275767321,18267889925254779,146232526369201318,1170331087647336130,9365122293936867751

%N Expansion of 1/((1-x)*((1-5x)^2)*(1-8x)).

%C This sequence was chosen to illustrate a method of solution.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (19, -123, 305, -200).

%F a(n) = (2*8^(n+4) - (84*n+287)*5^(n+2) - 9)/1008.

%F In general, for the expansion of 1/((1-t)*((1-s)^2)*(1-r)) with r > s > t, we have the formula: a(n) = (K*r^(n+3) + L*s^(n+3) + M*s^(n+2) + N*t^(n+3))/D, where K, L, M, N, D have the following values:

%F K = (s-t)^2;

%F L = (r-t)*(r-2*s+t);

%F M = -(r-s)*(r-t)*(s-t)*(n+3);

%F N = -(r-s)^2;

%F D = (r-t)*((s-t)^2)*((r-s)^2).

%F Directly using formula we get: a(n) = (16*8^(n+3) - 7*5^(n+3) -84*(n+3)*5^n+2) - 9)/1008. After transformation we obtain previous formula.

%K nonn

%O 0,2

%A _Yahia Kahloune_, Sep 18 2013