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A084636
Binomial transform of (1,0,1,0,1,0,2,0,2,0,2,0,...).
3
1, 1, 2, 4, 8, 16, 33, 71, 157, 349, 768, 1662, 3534, 7398, 15291, 31297, 63595, 128555, 258930, 520240, 1043540, 2090956, 4186757, 8379499, 16766313, 33541481, 67093588, 134199826, 268414602, 536846754, 1073713983, 2147451717, 4294930839, 8589893143
OFFSET
0,3
COMMENTS
Partial sums are A084637 (without leading 1).
The sequence starting 1,2,4,... is the binomial transform of (1,1,1,1,1,2,2,2,...) with b(n) = Sum_{k=0..4} C(n,k) + 2*Sum_{k=5..n} C(n,k) = 2^(n+1) - (n^4 -2*n^3 + 11*n^2 + 14*n + 24)/24. This gives the partial sums of A084635.
FORMULA
a(n) = Sum_{k=0..2} C(n, 2*k) + 2*Sum_{k=3..floor(n/2)} C(n, 2*k).
a(n) = (n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24 + 2*Sum_{k=3..floor(n/2)} C(n, 2*k).
O.g.f.: (1-2*x+2*x^2)*(1-4*x+5*x^2-2*x^3+x^4)/((1-x)^5*(1-2*x)). - R. J. Mathar, Apr 07 2008
a(n) = A000225(n) - (1/24)*n*(n-1)*(n^2 - 5*n + 18) + [n=0]. - G. C. Greubel, Mar 19 2023
MATHEMATICA
Table[Boole[n==0] +(2^n-1) -(1/24)*n*(n^3-6*n^2+23*n-18), {n, 0, 50}] (* G. C. Greubel, Mar 19 2023 *)
PROG
(Magma) [(2^n-1) -(1/24)*n*(n^3-6*n^2+23*n-18) +0^n: n in [0..50]]; // G. C. Greubel, Mar 19 2023
(SageMath) [(2^n-1) -(1/24)*n*(n^3-6*n^2+23*n-18) +0^n for n in range(51)] # G. C. Greubel, Mar 19 2023
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Jun 06 2003
STATUS
approved