A246653 G.f.: Sum_{n>=0} x^n / (1-x^2)^(2*n+1) * [Sum_{k=0..n} C(n,k)^2 * x^k] * [Sum_{k=0..n} C(n,k)^2 * x^(2*k)].
1, 1, 3, 9, 25, 87, 287, 993, 3519, 12525, 45369, 165519, 608569, 2253307, 8386575, 31370297, 117834837, 444258387, 1680516731, 6375706325, 24253227159, 92481509389, 353417696625, 1353280137975, 5191349266275, 19948136148837, 76771074483837, 295880290515411, 1141860205058433, 4412129801125275
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Keywords
Examples
G.f.: A(x) = 1 + x + 3*x^2 + 9*x^3 + 25*x^4 + 87*x^5 + 287*x^6 + 993*x^7 +... where the g.f. is given by the binomial series: A(x) = 1/(1-x^2) + x/(1-x^2)^3 * (1 + x) * (1 + x^2) + x^2/(1-x^2)^5 * (1 + 2^2*x + x^2) * (1 + 2^2*x^2 + x^4) + x^3/(1-x^2)^7 * (1 + 3^2*x + 3^2*x^2 + x^3) * (1 + 3^2*x^2 + 3^2*x^4 + x^6) + x^4/(1-x^2)^9 * (1 + 4^2*x + 6^2*x^2 + 4^2*x^3 + x^4) * (1 + 4^2*x^2 + 6^2*x^4 + 4^2*x^6 + x^8) + x^5/(1-x^2)^11 * (1 + 5^2*x + 10^2*x^2 + 10^2*x^3 + 5^2*x^4 + x^5) * (1 + 5^2*x^2 + 10^2*x^4 + 10^2*x^6 + 5^2*x^8 + x^10) +...
Crossrefs
Cf. A246563.
Programs
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Mathematica
Table[Sum[Binomial[2*k, k] * Binomial[n-k, k]^2,{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Sep 02 2014 *) -
PARI
/* By definition: */ {a(n)=local(A=1); A=sum(m=0, n, x^m/(1-x^2)^(2*m+1) * sum(k=0, m, binomial(m, k)^2 * x^k) * sum(k=0, m, binomial(m, k)^2 * x^(2*k)) +x*O(x^n)); polcoeff(A, n)} for(n=0, 30, print1(a(n), ", ")) -
PARI
/* By a binomial identity: */ {a(n)=polcoeff(sum(m=0, n, x^m*sum(k=0, m, binomial(m, k)^2 * x^(m-k) * sum(j=0, k, binomial(k, j)^2 * x^j )+x*O(x^n))), n)} for(n=0, 30, print1(a(n), ", ")) -
PARI
/* From a formula for a(n): */ {a(n)=sum(k=0, n\2, binomial(2*k, k)*binomial(n-k, k)^2)} for(n=0, 30, print1(a(n), ", "))
Formula
G.f.: Sum_{n>=0} x^n * Sum_{k=0..n} C(n,k)^2 * x^(n-k) * Sum_{j=0..k} C(k,j)^2 * x^j.
a(n) = Sum_{k=0..[n/2]} binomial(2*k, k) * binomial(n-k, k)^2.
Recurrence: n^2*(4*n-9)*a(n) = (8*n^3 - 26*n^2 + 18*n - 5)*a(n-1) + (28*n^3 - 119*n^2 + 162*n - 75)*a(n-2) + 2*(16*n^3 - 84*n^2 + 142*n - 75)*a(n-3) - 4*(2*n-5)^2*(4*n-5)*a(n-4). - Vaclav Kotesovec, Sep 02 2014
a(n) ~ 2^(2*n+1/2) / (Pi*n). - Vaclav Kotesovec, Sep 02 2014