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A104184
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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.
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4
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1, 1, 3, 9, 32, 120, 473, 1925, 8034, 34188, 147787, 647141, 2864508, 12796238, 57615322, 261197436, 1191268350, 5462080688, 25162978925, 116414836445, 540648963645, 2519574506595, 11779011525030, 55225888341334, 259612579655392, 1223396051745310
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OFFSET
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0,3
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COMMENTS
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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.
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
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LINKS
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FORMULA
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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
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
Logarithmic derivative yields the central pentanomial coefficients (A005191).
G.f.: exp( Sum_{n>=1} A005191(n)*x^n/n ).
G.f.: (1/x)*Series_Reversion(x*(1-x^5)*(1-x^2)*(1-x)/(1-x^10)).
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)
a(n) ~ (3-sqrt(5)) * 5^(n+1) / (4 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Sep 09 2014
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
-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
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EXAMPLE
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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.
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MAPLE
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a:= proc(n) option remember; `if`(n<5, [1, 1, 3, 9, 32][n+1],
((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))/
(2*(n-1)*(n+1)*(n+2)*(2*n+1)))
end:
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MATHEMATICA
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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 *)
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PROG
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(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 */
(Maxima)
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 */
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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Nicholas Biller (billern(AT)gmail.com), Mar 11 2005
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STATUS
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approved
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