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a(n) = a(n-4) + a(n-3) + a(n-2) + a(n-1) + (n-5).
1

%I #29 Mar 28 2024 13:19:10

%S 1,1,1,1,4,8,16,32,64,125,243,471,911,1759,3394,6546,12622,24334,

%T 46910,90427,174309,335997,647661,1248413,2406400,4638492,8940988,

%U 17234316,33220220,64034041,123429591,237918195,458602075,883983931,1703933822,3284438054

%N a(n) = a(n-4) + a(n-3) + a(n-2) + a(n-1) + (n-5).

%H Alois P. Heinz, <a href="/A196875/b196875.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3,-2,0,0,-1,1).

%F G.f.: (x^5-3*x^4+2*x-1)*x / ((x^4+x^3+x^2+x-1)*(x-1)^2 ).

%F a(n) = +3*a(n-1) -2*a(n-2) -a(n-5) +a(n-6).

%F a(n) = 5/9-n/3 +(10*A000078(n) +17*A000078(n+1) +21*A000078(n+2) -14*A000078(n+3))/9. - _R. J. Mathar_, Oct 16 2011

%p a:= n-> (Matrix(6, (i, j)-> `if`(i=j-1, 1, `if`(i=6, [1, -1, 0, 0, -2, 3][j], 0)))^n. <<-1, 1, 1, 1, 1, 4>>)[1, 1]: seq(a(n), n=1..50); # _Alois P. Heinz_, Oct 15 2011

%t nn = 40; a[1] = a[2] = a[3] = a[4] = 1; Do[a[n] = a[n - 1] + a[n - 2] + a[n - 3] + a[n - 4] + (n - 5), {n, 5, nn}]; Table[a[n], {n, nn}] (* _T. D. Noe_, Oct 07 2011 *)

%t RecurrenceTable[{a[1]==a[2]==a[3]==a[4]==1,a[n]==a[n-1]+a[n-2]+a[n-3]+a[n-4]+(n-5)},a,{n,40}] (* or *) LinearRecurrence[{3,-2,0,0,-1,1},{1,1,1,1,4,8},40] (* _Harvey P. Dale_, Aug 25 2014 *)

%Y Cf. A000126, A196787.

%K nonn,easy

%O 1,5

%A _Aditya Subramanian_, Oct 07 2011