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a(n) = 4^n * sum_{i=1,n} i^4/4^i.
4

%I #16 Jul 31 2015 11:49:03

%S 1,20,161,900,4225,18196,75185,304836,1225905,4913620,19669121,

%T 78697220,314817441,1259308180,5037283345,20149198916,80596879185,

%U 322387621716,1289550617185,5158202628740,20632810709441

%N a(n) = 4^n * sum_{i=1,n} i^4/4^i.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (9, -30, 50, -45, 21, -4).

%F 1/81 * [380*4^n - 27n^4 - 144n^3 - 360n^2 - 528n - 380]. - _Ralf Stephan_, May 08 2004

%F a(1)=1, a(2)=20, a(3)=161, a(4)=900, a(5)=4225, a(6)=18196, a(n)= 9*a(n-1)- 30*a(n-2)+50*a(n-3)-45*a(n-4)+21*a(n-5)-4*a(n-6). - _Harvey P. Dale_, Jul 15 2012

%F From _Peter Bala_, Nov 29 2012, (Start)

%F Recurrence equation: a(n) = 4*a(n-1) + n^4. See A047520 and A066999.

%F O.g.f.: (x + 11*x^2 + 11*x^3 + x^4)/((1 - 4*x)*(1 - x)^5) = x + 20*x^2 + 161*x^3 + .... (end)

%t Table[4^n*Sum[i^4/4^i,{i,n}], {n,30}] (* or *) LinearRecurrence[ {9,-30,50,-45,21,-4}, {1,20,161,900,4225,18196}, 30] (* _Harvey P. Dale_, Jul 15 2012 *)

%Y Cf. A047520, A066999.

%K nonn,easy

%O 1,2

%A _Benoit Cloitre_, Jan 27 2002