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%I #35 Jun 06 2020 11:14:12
%S 10,20,50,146,470,1610,5750,21146,79430,303050,1169750,4554746,
%T 17852390,70322090,278050550,1102537946,4381257350,17438542730,
%U 69495104150,277204002746,1106488342310,4418973508970,17654960746550
%N a(n) = 1*4^n + 2*3^n + 3*2^n + 4*1^n.
%C This is the sequence of fourth terms of "second partial sums of m-th powers".
%H Colin Barker, <a href="/A254030/b254030.txt">Table of n, a(n) for n = 0..1000</a>
%H Luciano Ancora, <a href="/A254030/a254030_3.pdf">Demonstration of formulas</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (10,-35,50,-24).
%F G.f.: -2*(77*x^3-100*x^2+40*x-5) / ((x-1)*(2*x-1)*(3*x-1)*(4*x-1)). - _Colin Barker_, Jan 26 2015
%F From _Peter Bala_, Jan 31 2016: (Start)
%F a(n) = (x + 1)*( Bernoulli(n + 1, x + 1) - Bernoulli(n + 1, 1) )/(n + 1) - ( Bernoulli(n + 2, x + 1) - Bernoulli(n + 2, 1) )/(n + 2) at x = 4.
%F a(n) = 1/3!*Sum_{k = 0..n} (-1)^(k+n)*(k + 5)!*Stirling2(n,k)/
%F ((k + 1)*(k + 2)). (End)
%p seq(add(i*(5 - i)^n, i = 1..4), n = 0..20); # _Peter Bala_, Jan 31 2017
%t Table[3 2^n + 2^(2 n) + 2 3^n + 4, {n, 0, 25}] (* _Bruno Berselli_, Jan 27 2015 *)
%t LinearRecurrence[{10,-35,50,-24},{10,20,50,146},30] (* _Harvey P. Dale_, Jun 06 2020 *)
%o (PARI) Vec(-2*(77*x^3-100*x^2+40*x-5)/((x-1)*(2*x-1)*(3*x-1)*(4*x-1)) + O(x^100)) \\ _Colin Barker_, Jan 26 2015
%Y Cf. A052548, A254028, A254031, A254144, A254145, A254146.
%K nonn,easy
%O 0,1
%A _Luciano Ancora_, Jan 26 2015
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