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A339254 a(n) = 6*a(n - 1) - 12*a(n - 2) + 8*a(n - 3) for n >= 5, a(0) = 1, a(1) = 7, a(2) = 24, a(3) = 70, a(4) = 193.

%I A339254 #17 Nov 29 2020 08:32:11
%S A339254 1,7,24,70,193,510,1304,3248,7920,18976,44800,104448,240896,550400,
%T A339254 1247232,2805760,6270976,13934592,30801920,67764224,148439040,
%U A339254 323878912,704118784,1525678080,3295674368,7098859520,15250489344,32682016768,69877104640,149082341376
%N A339254 a(n) = 6*a(n - 1) - 12*a(n - 2) + 8*a(n - 3) for n >= 5, a(0) = 1, a(1) = 7, a(2) = 24, a(3) = 70, a(4) = 193.
%H A339254 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-12,8).
%F A339254 a(n) = [x^n] (5*x^4 + 2*x^3 - 6*x^2 + x + 1) / (1 - 2*x)^3.
%F A339254 a(n) = n! [x^n] (exp(2*x)*(18*x^2 + 52*x + 35) - 10*x - 19)/16.
%F A339254 a(n) = 2^(n-5)*(70 + 43*n + 9*n^2) for n >= 2. - _Stefano Spezia_, Nov 29 2020
%p A339254 a := proc(n) option remember; if n < 5 then return [1, 7, 24, 70, 193][n + 1] fi;
%p A339254 6*a(n - 1) - 12*a(n - 2) + 8*a(n - 3) end: seq(a(n), n = 0..29);
%t A339254 CoefficientList[Series[(5 x^4 + 2 x^3 - 6 x^2 + x + 1)/(1 - 2 x)^3, {x,0,29}], x]
%K A339254 nonn,easy
%O A339254 0,2
%A A339254 _Peter Luschny_, Nov 29 2020