OFFSET
0,3
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 22*x^3 + 294*x^4 + 5911*x^5 + 158293*x^6 + 5251690*x^7 + 206696194*x^8 + 9387611937*x^9 + 482745371458*x^10 + 27717788095397*x^11 + 1757818683339028*x^12 + 122058148921357056*x^13 + 9212494564360610855*x^14 + 751138761646263512978*x^15 + ...
such that
A(x) = 1 + C(2,1)/2*x/A(x)^2 + C(6,2)/3*x^2/A(x)^6 + C(12,3)/4*x^3/A(x)^12 + C(20,4)/5*x^4/A(x)^20 + C(30,5)/6*x^5/A(x)^30 + C(42,6)/7*x^6/A(x)^42 + C(56,7)/8*x^7/A(x)^56 + ...
more explicitly,
A(x) = 1 + x/A(x)^2 + 5*x^2/A(x)^6 + 55*x^3/A(x)^12 + 969*x^4/A(x)^20 + 23751*x^5/A(x)^30 + 749398*x^6/A(x)^42 + 28989675*x^7/A(x)^56 + ... + A135861(n)*x^n/A(x)^(n*(n+1)) + ...
MATHEMATICA
terms = 21; A[_] = 1; Do[A[x_] = 1 + Sum[Binomial[n*(n+1), n]/(n+1)*x^n/ A[x]^(n*(n+1)), {n, terms}] + O[x]^terms, {terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Feb 09 2018 *)
PROG
(PARI) {a(n) = my(A=[1]); for(i=1, n, A = Vec(sum(m=0, #A, binomial(m*(m+1), m)/(m+1) * x^m/Ser(A)^(m*(m+1)) ))); A[n+1]}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Feb 03 2018
STATUS
approved