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Antidiagonal sums of table A088925, which lists coefficients T(n,k) of x^n*y^k in f(x,y) that satisfies f(x,y) = 1/(1-x-y) + xy*f(x,y)^3.
+20
3
1, 2, 5, 14, 43, 142, 496, 1808, 6807, 26270, 103357, 412942, 1670572, 6828824, 28159880, 116997296, 489271039, 2057800158, 8698624303, 36936288650, 157474552403, 673830974654, 2892864930292, 12457038200008, 53789813903620
FORMULA
a(n) = sum(k=0, n, sum(i=0, k, C(n, 2i)*C(n-2i, k-i)* A001764(i) )), where A001764(i)=(3i)!/[i!(2i+1)! ] (from Michael Somos). G.f. satisfies A(x) = 1/(1-2x) + x^2*A(x)^3.
a(n) ~ (2 + 3*sqrt(3)/2)^(n + 3/2) / (3^(7/4) * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Oct 10 2020
EXAMPLE
A(x) = 1/(1-2x) + x^2*A(x)^3 since 1/(1-2x) = 1 + 2x + 4x^2 + 8x^3 +... and x^2*A(x)^3 = 1x^2 + 6x^3 + 27x^4 + 110x^5 +...
MATHEMATICA
Table[Sum[Sum[Binomial[n, 2*i] * Binomial[n - 2*i, k - i] * (3*i)! / (i! * (2*i + 1)!), {i, 0, k}], {k, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Oct 10 2020 *)
Main diagonal of table A088925, which lists coefficients T(n,k) of x^n*y^k in f(x,y) that satisfies f(x,y) = 1/(1-x-y) + xy*f(x,y)^3.
+20
2
1, 3, 21, 212, 2617, 36345, 544080, 8577378, 140456625, 2368062095, 40859183247, 718386164556, 12829418522056, 232153200359592, 4248457201595622, 78508329463480160, 1463164022514939392, 27474112707608092672
COMMENTS
The g.f. for A001764 satisfies: g(x) = 1 + x*g(x)^3.
FORMULA
a(n) = sum(i=0, n, C(2n, 2i)*C(2n-2i, n-i)* A001764(i) ), where A001764(i)=(3i)!/[i!(2i+1)! ] (from Michael Somos). a(n) ~ (4 + 3*sqrt(3))^(2*n + 2) / (Pi * 3^(7/4) * n^2 * 2^(2*n + 4)). - Vaclav Kotesovec, Oct 10 2020
MATHEMATICA
Table[Sum[Binomial[2*n, 2*i] * Binomial[2*n - 2*i, n - i]*(3*i)!/(i!*(2*i + 1)!), {i, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Oct 10 2020 *)
Symmetric square table of coefficients, read by antidiagonals, where T(n,k) is the coefficient of x^n*y^k in f(x,y) that satisfies: f(x,y) = g(x,y) + xy*f(x,y)^4 and where g(x,y) satisfies: 1 + (x+y-1)*g(x,y) + xy*g(x,y)^2 = 0.
+10
3
1, 1, 1, 1, 4, 1, 1, 10, 10, 1, 1, 20, 48, 20, 1, 1, 35, 162, 162, 35, 1, 1, 56, 441, 841, 441, 56, 1, 1, 84, 1036, 3314, 3314, 1036, 84, 1, 1, 120, 2184, 10786, 18004, 10786, 2184, 120, 1, 1, 165, 4236, 30460, 77952, 77952, 30460, 4236, 165, 1, 1, 220, 7689, 77044
COMMENTS
Explicitly, g(x,y) = ((1-x-y)+sqrt((1-x-y)^2-4xy))/(2xy) = sum(n>=0, sum(k>=0, N(n,k)*x^n*y^k), where N(n,k) are the Narayana numbers: N(n,k) = C(n+k,k)*C(n+k+2,k+1)/(n+k+2). This array is directly related to sequence A002293, which has a g.f. h(x) that satisfies h(x) = 1 + x*h(x)^4. The inverse binomial transform of the rows grows by three terms per row.
EXAMPLE
Rows begin:
[1, 1, 1, 1, 1, 1, 1, 1, ...];
[1, 4, 10, 20, 35, 56, 84, 120, ...];
[1, 10, 48, 162, 441, 1036, 2184, 4236, ...];
[1, 20, 162, 841, 3314, 10786, 30460, 77044, ...];
[1, 35, 441, 3314, 18004, 77952, 284880, 912042, ...];
[1, 56, 1036, 10786, 77952, 435654, 2007456, 7951674, ...];
[1, 84, 2184, 30460, 284880, 2007456, 11427992, 55009548, ...];
[1, 120, 4236, 77044, 912042, 7951674, 55009548, 317112363, ...];
[1, 165, 7689, 178387, 2624453, 27870393, 231114465, 1576219474, ...]; ...
PROG
(PARI) {L=10; T=matrix(L, L, n, k, 1); for(n=1, L-1, for(k=1, L-1, T[n+1, k+1]=binomial(n+k, k)*binomial(n+k+2, k+1)/(n+k+2)+ sum(j3=1, k, sum(i3=1, n, T[n-i3+1, k-j3+1]* sum(j2=1, j3, sum(i2=1, i3, T[i3-i2+1, j3-j2+1]* sum(j1=1, j2, sum(i1=1, i2, T[i2-i1+1, j2-j1+1]*T[i1, j1])); )); )); )); T}
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