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%I #59 Mar 29 2024 08:46:10
%S 1,1,3,9,32,120,473,1925,8034,34188,147787,647141,2864508,12796238,
%T 57615322,261197436,1191268350,5462080688,25162978925,116414836445,
%U 540648963645,2519574506595,11779011525030,55225888341334,259612579655392,1223396051745310
%N a(n) is the number of paths from (0,0) to (n,0) using steps of the form (1,2),(1,1),(1,0),(1,-1) or (1,-2) and staying above the x-axis. Also, a(n) is the number of possible combinations of balls on the lawn after n turns, using a Motzkin variation of the (4,2)-case of the tennis ball problem considered by D. Merlini, R. Sprugnoli and M. C. Verri.
%C The (4,2)-case of the Motzkin Tennis Ball Problem is a variation of the Tennis Ball Problem that generates a(n). On each turn, i, four balls labeled i are placed in the bucket and then any two are removed and placed on the lawn. We consider all possible combinations of balls on the lawn after n turns.
%C The number of ways to choose n numbers, ranging from 0 to 4, so that their sum is 2n and so that when you take k numbers from the left, the sum of these numbers is <= 2k (e.g. the combination of {141} is impossible, for 1+4 > 2k). Thus a(1) = {2}; a(2) = {04}, {13} and {22}; a(3) = {024}, {033}, {042}, {114}, {123}, {132}, {204}, {213} and {222}. - Joost Vermeij (joost_vermeij(AT)hotmail.com), Jun 12 2005
%H Alois P. Heinz, <a href="/A104184/b104184.txt">Table of n, a(n) for n = 0..1437</a>
%H C. Banderier, C. Krattenthaler, A. Krinik, D. Kruchinin, V. Kruchinin, D. Nguyen, and M. Wallner, <a href="https://arxiv.org/abs/1609.06473">Explicit formulas for enumeration of lattice paths: basketball and the kernel method</a>, arXiv preprint arXiv:1609.06473 [math.CO], 2016.
%H AJ Bu and Doron Zeilberger, <a href="https://arxiv.org/abs/2305.09030">Using Symbolic Computation to Explore Generalized Dyck Paths and Their Areas</a>, arXiv:2305.09030 [math.CO], 2023.
%H P. Flajolet and R. Sedgewick, <a href="http://algo.inria.fr/flajolet/Publications/books.html">Analytic Combinatorics</a>, 2009; see page 512
%H D. Merlini, R. Sprugnoli and M. C. Verri, <a href="http://dx.doi.org/10.1006/jcta.2002.3273">The tennis ball problem</a>, J. Combin. Theory, A 99 (2002), pp. 307-344.
%F G.f. (for offset 1): (1/(4*x))*(1+x+sqrt((1-6*x+5*x^2)) - sqrt(2)*sqrt(1+sqrt((1-6*x+5*x^2)) + x*(-2-5*x+sqrt((1-6*x+5*x^2))))). - _N-E. Fahssi_, Jan 10 2008
%F Let M be the infinite pentadiagonal matrix with all 1's in the 1st and 2nd subdiagonals, the main diagonal, and the 1st and 2nd superdiagonals, and with the rest 0's. V = vector [1,0,0,0,...]. The sequence starting with offset 1 = iterates of M*V, leftmost column. - _Gary W. Adamson_, Jun 06 2011
%F From _Paul D. Hanna_, Oct 19 2011: (Start)
%F Logarithmic derivative yields the central pentanomial coefficients (A005191).
%F G.f.: exp( Sum_{n>=1} A005191(n)*x^n/n ).
%F G.f.: (1/x)*Series_Reversion(x*(1-x^5)*(1-x^2)*(1-x)/(1-x^10)).
%F G.f. satisfies: A(x) = (1-x^10*A(x)^10)/((1-x^5*A(x)^5)*(1-x^2*A(x)^2)*(1-x*A(x))). (See formula from _Michael Somos_ in A005191.) (End)
%F a(n) ~ (3-sqrt(5)) * 5^(n+1) / (4 * sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Sep 09 2014
%F a(n) = ((Sum_{l=1..n+1} (C(n+1,l)*Sum_{i=0..n-1} (C(i,2*l-1) * Sum_{j=0..n-l+1} (C(j,n-j-i-1)*C(n-l+1,j))))) + Sum_{j=0..n+1} (C(j,n-j) * C(n+1,j)))/(n+1). - _Vladimir Kruchinin_, Jun 26 2015
%F -2*(n-1)*(2*n+1)*(n+2)*(n+1)*a(n) +(n+1)*(43*n^3-48*n^2-7*n+2)*a(n-1) +(-124*n^4+370*n^3-255*n^2-15*n+14)*a(n-2) +5*(n-2)*(2*n^3-52*n^2+65*n-1)*a(n-3) +25*(n-2)*(n-3)*(8*n^2-8*n-1)*a(n-4) -125*n*(n-2)*(n-3)*(n-4)*a(n-5)=0. - _R. J. Mathar_, Jul 23 2017
%e a(3) = 9, since the possible combinations of balls on the lawn after 3 turns is 111122, 111123, 111133, 111222, 111223, 111233, 112222, 112223, 112233, if on each turn there are 4 identically labeled balls received and 2 selected.
%p a:= proc(n) option remember; `if`(n<5, [1, 1, 3, 9, 32][n+1],
%p ((n+1)*(43*n^3-48*n^2-7*n+2)*a(n-1)
%p -(124*n^4-370*n^3+255*n^2+15*n-14)*a(n-2)
%p +5*(n-2)*(2*n^3-52*n^2+65*n-1)*a(n-3)
%p +25*(n-2)*(n-3)*(8*n^2-8*n-1)*a(n-4)
%p -125*n*(n-2)*(n-3)*(n-4)*a(n-5))/
%p (2*(n-1)*(n+1)*(n+2)*(2*n+1)))
%p end:
%p seq(a(n), n=0..30); # _Alois P. Heinz_, Oct 11 2013
%t CoefficientList[Series[(1 + x + Sqrt[1 - 6*x + 5*x^2] - Sqrt[2]*Sqrt[1 + Sqrt[1 - 6*x + 5*x^2] + x*(-2 - 5*x + Sqrt[1 - 6*x + 5*x^2])])/(4*x^2), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Sep 09 2014 *)
%o (PARI) {a(n)=local(A=1);A=exp(sum(m=1,n+1,polcoeff(((1-x^5)/(1-x) +O(x^(2*m+1)))^m, 2*m)*x^m/m)+x*O(x^n));polcoeff(A,n)} /* _Paul D. Hanna_ */
%o (Maxima)
%o a(n):=((sum(binomial(n+1,l)*sum(binomial(i,2*l-1)*sum(binomial(j,n-j-i-1) *binomial(n-l+1,j),j,0,n-l+1),i,0,n-1),l,1,n+1))+sum(binomial(j,n-j) *binomial(n+1,j),j,0,n+1))/(n+1); /* _Vladimir Kruchinin_, Jun 26 2015 */
%Y Cf. A066357, A001006, A005191.
%K nonn
%O 0,3
%A Nicholas Biller (billern(AT)gmail.com), Mar 11 2005
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