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a(n) = n^4*(3*n+2).
3

%I #16 Jul 17 2024 09:00:40

%S 0,5,128,891,3584,10625,25920,55223,106496,190269,320000,512435,

%T 787968,1171001,1690304,2379375,3276800,4426613,5878656,7688939,

%U 9920000,12641265,15929408,19868711,24551424,30078125,36558080,44109603,52860416,62948009,74520000

%N a(n) = n^4*(3*n+2).

%C Number of ascending runs in {1,...,n}^5.

%H Alois P. Heinz, <a href="/A229147/b229147.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 G.f.: (x^4+58*x^3+198*x^2+98*x+5)*x/(x-1)^6.

%F a(0)=0, a(1)=5, a(2)=128, a(3)=891, a(4)=3584, a(5)=10625, 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). - _Harvey P. Dale_, Aug 14 2015

%F E.g.f.: exp(x)*x*(5 + 59*x + 87*x^2 + 32*x^3 + 3*x^4). - _Stefano Spezia_, Jul 17 2024

%p a:= n-> n^4*(3*n+2):

%p seq(a(n), n=0..40);

%t Table[n^4 (3n+2),{n,0,30}] (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{0,5,128,891,3584,10625},40] (* _Harvey P. Dale_, Aug 14 2015 *)

%Y Row n=5 of A229079.

%Y Cf. A000583, A016789.

%K nonn,easy

%O 0,2

%A _Alois P. Heinz_, Sep 15 2013