This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A099578 #54 Dec 26 2023 11:23:04
%S A099578 1,1,4,5,21,28,120,165,715,1001,4368,6188,27132,38760,170544,245157,
%T A099578 1081575,1562275,6906900,10015005,44352165,64512240,286097760,
%U A099578 417225900,1852482996,2707475148,12033222880,17620076360,78378960360,114955808528
%N A099578 a(n) = binomial(floor((3n+2)/2), floor(n/2)).
%C A099578 Main diagonal of triangle A099575.
%C A099578 With offset 2, this is the number of compositions of n-1 into floor(n/2) parts. - _T. D. Noe_, Jan 05 2007
%C A099578 From _Petros Hadjicostas_, Jul 19 2018: (Start)
%C A099578 We clarify the above comment by T. D. Noe. The number of compositions of N into K positive parts is C(N-1, K-1). This was proved by MacMahon in 1893 (and probably by others before him). The number of compositions of N into K nonnegative parts is C(N+K-1, K-1) because for every composition b_1 + ... + b_K = N with b_i >= 0 for all i, we may create another composition c_1 + ... + c_K = N+K with c_i = b_i + 1 >= 1.
%C A099578 The statement of T. D. Noe above means that, for n>=2, a(n-2) is the number of compositions of N = n-1 into K = floor(n/2) nonnegative parts. Thus, a(n-2) = C(N+K-1, K-1) = C(n-1+floor(n/2)-1, floor(n/2)-1) = C(floor((3(n-2)+2)/2), floor((n-2)/2)).
%C A099578 This interpretation is important for T. D. Noe's comments for sequence A030077, whose unknown general formula remains an unsolved problem (as of July 2018).
%C A099578 It should be noted, however, that for most authors "composition" means "composition into positive parts". The phrase "weak composition" is sometimes used for a "composition into nonnegative parts".
%C A099578 (End)
%H A099578 T. D. Noe, <a href="/A099578/b099578.txt">Table of n, a(n) for n=0..200</a>
%F A099578 a(n) = Sum_{k=0..floor(n/2)} binomial(n+k, k).
%F A099578 40*n*(n+1)*a(n) +36*n*(n-2)*a(n-1) -6*(45*n^2-23)*a(n-2) -27*(3*n-4)*(3*n-5)*a(n-3) = 0. - _R. J. Mathar_, Oct 30 2014
%F A099578 From _Benedict W. J. Irwin_, Aug 15 2016: (Start)
%F A099578 G.f.: -( (sqrt(4-27*x^2) - 2*(cos(arcsin(3*sqrt(3)*x/2)/3) + sqrt(3)*sin(2*arcsin(3*sqrt(3)*x/2)/3)) )/( 3*x*sqrt(4-27*x^2)) ).
%F A099578 E.g.f.: Hypergeometric2F3(2/3,4/3;1/2,1,3/2;27*x^2/16) + x*Hypergeometric2F3(4/3,5/3;3/2,3/2,2;27*x^2/16).
%F A099578 (End)
%F A099578 Recurrence: 4*n*(n+1)*(6*n-1)*a(n) = 18*n*a(n-1) + 3*(3*n-2)*(3*n-1)*(6*n+5)*a(n-2). - _Vaclav Kotesovec_, Aug 15 2016
%F A099578 a(n) = binomial(2*n+1, n)*hypergeom([-n, n+1], [-2*n-1], -1). - _Detlef Meya_, Dec 25 2023
%e A099578 From _Petros Hadjicostas_, Jul 19 2018: (Start)
%e A099578 With n=2 there are a(2-2) = a(0) = 1 compositions of 2-1 = 1 into floor(2/2) = 1 nonnegative parts, namely 1 (only).
%e A099578 With n=3 there are a(3-2) = a(1) = 1 compositions of 3-1 = 2 into floor(3/2) = 1 nonnegative parts, namely 2 (only).
%e A099578 With n=4 there are a(4-2) = a(2) = 4 compositions of 4-1 = 3 into floor(4/2) = 2 nonnegative parts, namely 0+3, 3+0, 1+2, and 2+1.
%e A099578 With n=5 there are a(5-2) = a(3) = 5 compositions of 5-1 = 4 into floor(5/2) = 2 nonnegative parts, namely 0+4, 4+0, 1+3, 3+1, and 2+2.
%e A099578 With n=6 there are a(6-2) = a(4) = 21 compositions of 6-1 = 5 into floor(6/2) = 3 nonnegative parts, namely the 3 permutations of 1+1+3, the 3 permutations of 1+2+2, the 3 permutations of 0+0+5, the 6 permutations of 0+1+4, and the 6 permutations of 0+2+3.
%e A099578 (End)
%p A099578 A099578:=n->binomial(floor((3*n+2)/2), floor(n/2)); seq(A099578(k), k=0..50); # _Wesley Ivan Hurt_, Nov 01 2013
%t A099578 Table[Binomial[Floor[(3n+2)/2], Floor[n/2]], {n, 0, 50}] (* _Wesley Ivan Hurt_, Nov 01 2013 *)
%t A099578 CoefficientList[Series[-((Sqrt[4 -27 x^2] -2(Cos[1/3 ArcSin[(3 Sqrt[3] x)/2]] + Sqrt[3] Sin[2/3 ArcSin[(3 Sqrt[3] x)/2]]))/(3 x Sqrt[4 -27 x^2])), {x, 0, 20}], x] (* _Benedict W. J. Irwin_, Aug 15 2016 *)
%t A099578 a[n_] := Binomial[2*n+1, n]*Hypergeometric2F1[-n, n+1, -2*n-1, -1]; Flatten[Table[a[n], {n, 0, 29}]] (* _Detlef Meya_, Dec 25 2023 *)
%o A099578 (PARI) a(n) = binomial((3*n+2)\2, n\2); \\ _Michel Marcus_, Nov 02 2013
%o A099578 (Magma) [(&+[Binomial(n+j, j): j in [0..Floor(n/2)]]): n in [0..40]]; // _G. C. Greubel_, Jul 24 2022
%o A099578 (SageMath) [binomial((3*n+2)//2, n//2) for n in (0..40)] # _G. C. Greubel_, Jul 24 2022
%Y A099578 Cf. A025174 (bisection), A030077, A045721 (bisection), A099575, A127040.
%K A099578 easy,nonn
%O A099578 0,3
%A A099578 _Paul Barry_, Oct 23 2004