A075045 Coefficients A_n for the s=3 tennis ball problem.
1, 9, 69, 502, 3564, 24960, 173325, 1196748, 8229849, 56427177, 386011116, 2635972920, 17974898872, 122430895956, 833108684637, 5664553564440, 38488954887171, 261369752763963, 1774016418598269, 12035694958994142, 81624256468292016, 553377268856455968
Offset: 0
Keywords
Links
- T. Amdeberhan, Integrality of a sum.
- Roland Bacher, On generating series of complementary plane trees, arXiv:math/0409050 [math.CO], 2004.
- Toufik Mansour and I. L. Ramirez, Enumerations of polyominoes determined by Fuss-Catalan words, Australas. J. Combin. 81 (3) (2021) 447-457, table 1.
- D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), pp. 307-344 (A_n for s=3).
Programs
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Maple
FussArea := proc(s,n) local a,i,j ; a := binomial((s+1)*n,n)*n/(s*n+1) ; ; add(j *(n-j) *binomial((s+1)*j,j) *binomial((s+1)*(n-j),n-j) /(s*j+1) /(s*(n-j)+1),j=0..n) ; a := a+binomial(s+1,2)*% ; for j from 0 to n-1 do for i from 0 to j do i*(j-i) /(s*i+1) /(s*(j-i)+1) /(n-j) *binomial((s+1)*i,i) *binomial((s+1)*(j-i),j-i) *binomial((s+1)*(n-j)-2,n-1-j) ; a := a-%*binomial(s+1,2) ; end do: end do: a ; end proc: seq(FussArea(2,n),n=1..30) ; # R. J. Mathar, Mar 31 2023 -
Mathematica
FussArea[s_, n_] := Module[{a, i, j, pc}, a = Binomial[(s + 1)*n, n]*n/(s*n + 1); pc = Sum[j*(n - j)*Binomial[(s + 1)*j, j]*Binomial[(s + 1)*(n - j), n - j]/(s*j + 1)/(s*(n - j) + 1), {j, 0, n}]; a = a + Binomial[s + 1, 2]*pc; For[j = 0, j <= n - 1 , j++, For[i = 0, i <= j, i++, pc = i*(j - i)/(s*i + 1)/(s*(j - i) + 1)/(n - j)*Binomial[(s + 1)*i, i]* Binomial[(s + 1)*(j - i), j - i]*Binomial[(s + 1)*(n - j) - 2, n - 1 - j]; a = a - pc*Binomial[s + 1, 2]; ]]; a]; Table[FussArea[2, n], {n, 1, 30}] (* Jean-François Alcover, Apr 02 2023, after R. J. Mathar *)
Formula
G.f.: seems to be (3*g-1)^(-2)*(1-g)^(-3) where g*(1-g)^2 = x. - Mark van Hoeij, Nov 10 2011
Conjecture: D-finite with recurrence 8*(2*n+3)*(7*n+1)*(n+1)*a(n) +6*(-252*n^3-477*n^2-220*n-11)*a(n-1) +81*(7*n+8)*(3*n-1)*(3*n+1)*a(n-2)=0. - Jean-François Alcover, Feb 07 2019
a(n) = (3n+2)*(n+1)*binomial(3n+3,n+1)/2/(2n+3) - A049235(n). [Merlini Theorem 2.5 for s=3] - R. J. Mathar, Oct 01 2021
From Seiichi Manyama, Jul 28 2025: (Start)
a(n) = Sum_{k=0..n} binomial(3*k+3+l,k) * binomial(3*n-3*k-l,n-k) for every real number l.
a(n) = Sum_{k=0..n} 2^(n-k) * binomial(3*n+4,k).
a(n) = Sum_{k=0..n} 3^(n-k) * binomial(2*n+k+3,k). (End)