A295764 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1) * x^n/A(x)^n.
1, 2, 8, 84, 1522, 38940, 1278880, 51136512, 2407921070, 130495143060, 8002598818976, 548083639109204, 41481408076217480, 3439242379130298624, 310088460754461374608, 30213233395705566500248, 3164041937175571224372118, 354458174755809132668847604, 42301583754989091491716706272, 5358139267082652649000120167456
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + 2*x + 8*x^2 + 84*x^3 + 1522*x^4 + 38940*x^5 + 1278880*x^6 + 51136512*x^7 + 2407921070*x^8 + 130495143060*x^9 + 8002598818976*x^10 +... such that A(x) = 1 + 2*(x/A(x)) + 12*(x/A(x))^2 + 140*(x/A(x))^3 + 2530*(x/A(x))^4 + 62832*(x/A(x))^5 + 1997688*(x/A(x))^6 +...+ binomial((n+1)^2,n)/(n+1)*(x/A(x))^n +... The table of coefficients of x^k in A(x)^(n+1) begins: [1, 2, 8, 84, 1522, 38940, 1278880, 51136512, 2407921070, ...]; [1, 4, 20, 200, 3444, 85312, 2744928, 108267280, 5049708672, ...]; [1, 6, 36, 356, 5862, 140508, 4424984, 172064160, 7946443350, ...]; [1, 8, 56, 560, 8888, 206176, 6350112, 243284064, 11121338640, ...]; [1, 10, 80, 820, 12650, 284252, 8556240, 322780800, 14599990830, ...]; [1, 12, 108, 1144, 17292, 376992, 11084864, 411518448, 18410660208, ...]; [1, 14, 140, 1540, 22974, 487004, 13983816, 510586400, 22584587382, ...]; [1, 16, 176, 2016, 29872, 617280, 17308096, 621216192, 27156348512, ...]; [1, 18, 216, 2580, 38178, 771228, 21120768, 744800256, 32164253550, ...]; ... in which the main diagonal begins: [1, 4, 36, 560, 12650, 376992, 13983816, ..., binomial((n+1)^2,n), ...], thus [x^n] A(x)^(n+1) = [x^n] (1 + x)^((n+1)^2) 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+1)^2, 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+1)^2,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+1)^2,n) for n>=0.
a(n) ~ c * exp(n) * n^(n - 3/2), where c = exp(3/2 - 2*exp(-2)) / sqrt(2*Pi) = 1.3639581816297942453935... - Vaclav Kotesovec, Aug 11 2021, updated Apr 20 2024