A295763 G.f. satisfies: A(x) = Sum_{n>=0} binomial(n*(n+1),n)/(n+1) * x^n/A(x)^n.
1, 1, 4, 42, 744, 18570, 596929, 23457763, 1089601420, 58424516424, 3553205095552, 241765128267597, 18202737707568180, 1502857964050898494, 135033771405912550765, 13119213786776385900734, 1370572549521961522812200, 153224265349198540163190599, 18253426026439076436840194131, 2308479498698233016622014842489
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 4*x^2 + 42*x^3 + 744*x^4 + 18570*x^5 + 596929*x^6 + 23457763*x^7 + 1089601420*x^8 + 58424516424*x^9 + 3553205095552*x^10 +... such that A(x) = 1 + x/A(x) + 5*(x/A(x))^2 + 55*(x/A(x))^3 + 969*(x/A(x))^4 + 23751*(x/A(x))^5 + 749398*(x/A(x))^6 +...+ binomial(n*(n+1),n)/(n+1)*(x/A(x))^n +... The table of coefficients of x^k in A(x)^(n+1) begins: [1, 1, 4, 42, 744, 18570, 596929, 23457763, 1089601420, ...]; [1, 2, 9, 92, 1588, 38964, 1238714, 48320440, 2233007214, ...]; [1, 3, 15, 151, 2544, 61356, 1928659, 74668905, 3432698217, ...]; [1, 4, 22, 220, 3625, 85936, 2670332, 102589280, 4691284160, ...]; [1, 5, 30, 300, 4845, 112911, 3467585, 132173305, 6011511390, ...]; [1, 6, 39, 392, 6219, 142506, 4324575, 163518732, 7396271082, ...]; [1, 7, 49, 497, 7763, 174965, 5245786, 196729744, 8848607971, ...]; [1, 8, 60, 616, 9494, 210552, 6236052, 231917400, 10371729633, ...]; [1, 9, 72, 750, 11430, 249552, 7300581, 269200107, 11969016345, ...]; ... in which the main diagonal begins: [1, 2, 15, 220, 4845, 142506, 5245786, ..., binomial(n*(n+1),n), ...], thus [x^n] A(x)^(n+1) = [x^n] (1 + x)^(n*(n+1)) for n>=0.
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..300
Programs
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Mathematica
terms = 20; A[] = 1; Do[A[x] = Sum[Binomial[n*(n+1), n]/(n+1)*x^n/A[x]^n, {n, 0, terms}] + O[x]^terms // Normal, terms]; CoefficientList[A[x], x] (* Jean-François Alcover, Jan 14 2018 *)
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PARI
{a(n) = my(A=[1]); for(m=1,n, A = concat(A,0); V = Vec( Ser(A)^(m+1) ); A[m+1] = (binomial(m*(m+1),m) - V[m+1])/(m+1);); A[n+1]} for(n=0,20,print1(a(n),", "))
Formula
G.f. A(x) satisfies: [x^n] A(x)^(n+1) = binomial(n*(n+1),n) for n>=0.
a(n) ~ c * exp(n) * n^(n - 3/2), where c = exp(1/2 - exp(-2)) / sqrt(2*Pi) = 0.5744892944370457395619... - Vaclav Kotesovec, Oct 17 2020, updated Mar 18 2024