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%I #30 Jun 15 2023 13:33:18
%S 0,5,62,363,1364,3905,9330,19607,37448,66429,111110,177155,271452,
%T 402233,579194,813615,1118480,1508597,2000718,2613659,3368420,4288305,
%U 5399042,6728903,8308824,10172525,12356630,14900787,17847788,21243689,25137930,29583455
%N a(n) = n^5 + n^4 + n^3 + n^2 + n.
%H Alois P. Heinz, <a href="/A152031/b152031.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6, -15, 20, -15, 6, -1).
%F a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6) n>5, a(0)=0, a(1)=5, a(2)=62, a(3)=363, a(4)=1364, a(5)=3905. [_Yosu Yurramendi_, Sep 03 2013]
%F From _Indranil Ghosh_, Apr 05 2017: (Start)
%F G.f.: x*(5 + 32x + 66x^2 + 16x^3 + x^4)/(x - 1)^6.
%F E.g.f.: exp(x)*x*(5 + 26x + 32x^2 + 11x^3 + x^4).
%F (End)
%F a(n) = n*A053699(n). - _Michel Marcus_, Apr 05 2017
%p a:= n-> `if`(n=1, 5, (n^6-n)/(n-1)):
%p seq(a(n), n=0..35); # _Alois P. Heinz_, Aug 20 2013
%t lst={};Do[AppendTo[lst,n^5+n^4+n^3+n^2+n],{n,0,5!}];lst
%t (* Other programs: *)
%t Table[Total[n^Range@ 5], {n, 0, 31}] (* or *)
%t CoefficientList[Series[x (5 + 32 x + 66 x^2 + 16 x^3 + x^4)/(x - 1)^6, {x, 0, 31}], x] (* _Michael De Vlieger_, Apr 05 2017 *)
%o (R)
%o a <- c(0, 5, 62, 363, 1364, 3905)
%o for(n in (length(a)+1):40) a[n] <- 6*a[n-1] -15*a[n-2] +20*a[n-3] -15*a[n-4] +6*a[n-5] -a[n-6]
%o a
%o [_Yosu Yurramendi_, Sep 03 2013]
%o (PARI) a(n) = n^5 + n^4 + n^3 + n^2 + n; \\ _Joerg Arndt_, Sep 03 2013
%o (Python) def a(n): return n**5 + n**4 + n**2 + n # _Indranil Ghosh_, Apr 05 2017
%Y Column k=5 of A228275.
%Y Cf. A053699.
%K nonn
%O 0,2
%A _Vladimir Joseph Stephan Orlovsky_, Nov 20 2008