A351918 G.f. A(x) satisfies: [x^n] (1 + n*x - A(x))^(2*n+1) = 0, for n > 0.
1, 2, 16, 262, 6688, 231938, 10130888, 532099310, 32582983264, 2275724862530, 178357712565096, 15489332978290006, 1475661504819976992, 152969444613431797250, 17137064243138924753480, 2062951616302994096028894
Offset: 1
Keywords
Examples
G.f.: A(x) = x + 2*x^2 + 16*x^3 + 262*x^4 + 6688*x^5 + 231938*x^6 + 10130888*x^7 + 532099310*x^8 + 32582983264*x^9 + 2275724862530*x^10 + ... Related table. Here we illustrate the formula [x^n] (1 + n*x - A(x))^(2*n+1) = 0 (n > 0). The table of coefficients of x^k in (1 + n*x - A(x))^(2*n+1) begins: n=0: [1, -1, -2, -16, -262, -6688, -231938, ...]; n=1: [1, 0, -6, -48, -774, -19872, -691910, ...]; n=2: [1, 5, 0, -110, -1645, -38439, -1286640, ...]; n=3: [1, 14, 70, 0, -3374, -74928, -2272326, ...]; n=4: [1, 27, 306, 1692, 0, -144342, -4262946, ...]; n=5: [1, 44, 858, 9504, 58938, 0, -7887286, ...]; n=6: [1, 65, 1924, 33982, 388401, 2694523, 0, ...]; ... in which the main diagonal is all zeros after the initial term, illustrating that [x^n] (1 + n*x - A(x))^(2*n+1) = 0, for n > 0.
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..330
Programs
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PARI
{a(n) = my(A=[1], m=1); for(i=1, n, A=concat(A, 0); m=#A; A[m] = polcoeff( (1 + m*x - x*Ser(A))^(2*m+1), m)/(2*m+1) ); A[n]} for(n=1, 20, print1(a(n), ", "))
Formula
a(n) ~ c * d^n * n! / n^2, where d = (1+r) / ((-1 + exp(r + LambertW(-1, -exp(-r)*r))) * LambertW(-exp(-1-r)*(1+r))) = 8.406107401279769476199925123910168..., r = 0.7545302104650497245839827141610818561001159135034... is the root of the equation r*(1 + r + LambertW(-exp(-1 - r)*(1 + r))) = -(1 + r)*(r + LambertW(-1, -exp(-r)*r)) and c = 0.053954066... - Vaclav Kotesovec, Feb 27 2022
Comments