A120970 G.f. A(x) satisfies A(x/A(x)^2) = 1 + x ; thus A(x) = 1 + Series_Reversion(x/A(x)^2).
1, 1, 2, 9, 60, 504, 4946, 54430, 655362, 8496454, 117311198, 1711459903, 26228829200, 420370445830, 7021029571856, 121859518887327, 2192820745899978, 40831103986939664, 785429260324068156, 15585831041632684997, 318649154587152781210, 6704504768568697046504
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 2*x^2 + 9*x^3 + 60*x^4 + 504*x^5 + 4946*x^6 + ... Related expansions. A(x)^2 = 1 + 2*x + 5*x^2 + 22*x^3 + 142*x^4 + 1164*x^5 + 11221*x^6 + ... A(A(x)-1) = 1 + x + 4*x^2 + 26*x^3 + 218*x^4 + 2151*x^5 + 23854*x^6 + ... A(A(x)-1)^2 = 1 + 2*x + 9*x^2 + 60*x^3 + 504*x^4 + 4946*x^5 + ... x/A(x)^2 = x - 2*x^2 - x^3 - 10*x^4 - 73*x^5 - 662*x^6 - 6842*x^7 - ... Series_Reversion(x/A(x)^2) = x + 2*x^2 + 9*x^3 + 60*x^4 + 504*x^5 + 4946*x^6 + ... To illustrate the formula a(n) = [x^(n-1)] 2*A(x)^(2*n)/(2*n), form a table of coefficients in A(x)^(2*n) as follows: A^2: [(1), 2, 5, 22, 142, 1164, 11221, 121848, ...]; A^4: [ 1, (4), 14, 64, 397, 3116, 29002, 306468, ...]; A^6: [ 1, 6, (27), 134, 825, 6270, 56492, 580902, ...]; A^8: [ 1, 8, 44, (240), 1502, 11200, 98144, 983016, ...]; A^10: [ 1, 10, 65, 390, (2520), 18672, 160115, 1565260, ...]; A^12: [ 1, 12, 90, 592, 3987, (29676), 250730, 2399388, ...]; A^14: [ 1, 14, 119, 854, 6027, 45458, (381010), 3582266, ...]; ... in which the main diagonal forms the initial terms of this sequence: [2/2*(1), 2/4*(4), 2/6*(27), 2/8*(240), 2/10*(2520), 2/12*(29676), ...].
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..350
Crossrefs
Programs
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Mathematica
terms = 21; A[] = 1; Do[A[x] = 1 + x*A[A[x] - 1]^2 + O[x]^j // Normal, {j, terms}]; CoefficientList[A[x], x] (* Jean-François Alcover, Jan 15 2018 *)
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PARI
{a(n)=local(A=[1,1]);for(i=2,n,A=concat(A,0); A[ #A]=-Vec(subst(Ser(A),x,x/Ser(A)^2))[ #A]);A[n+1]} for(n=0,30,print1(a(n),", ")) -
PARI
/* This sequence is generated when k=2, m=0: A(x/A(x)^k) = 1 + x*A(x)^m */ {a(n,k=2,m=0)=local(A=sum(i=0,n-1,a(i,k,m)*x^i));if(n==0,1,polcoeff((m+k)/(m+k*n)*A^(m+k*n),n-1))} \\ Paul D. Hanna, Nov 16 2008 for(n=0,30,print1(a(n),", ")) -
PARI
b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(2*n+k, j)/(2*n+k)*b(n-j, 2*j))); a(n) = if(n==0, 1, b(n-1, 2)); \\ Seiichi Manyama, Jun 04 2025
Formula
G.f. satisfies: A(x) = 1 + x*A(A(x) - 1)^2.
Let B(x) be the g.f. of A120971, then B(x) and g.f. A(x) are related by:
(a) B(x) = A(A(x)-1),
(b) B(x) = A(x*B(x)^2),
(c) A(x) = B(x/A(x)^2),
(d) A(x) = 1 + x*B(x)^2,
(e) B(x) = 1 + x*B(x)^2*B(A(x)-1)^2,
(f) A(B(x)-1) = B(A(x)-1) = B(x*B(x)^2).
a(n) = [x^(n-1)] (1/n)*A(x)^(2n) for n>=1 with a(0)=1; i.e., a(n) equals 1/n times the coefficient of x^(n-1) in A(x)^(2n) for n>=1. [Paul D. Hanna, Nov 16 2008]
From Seiichi Manyama, Jun 04 2025: (Start)
Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120971.
b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(2*n+k,j)/(2*n+k) * b(n-j,2*j).
a(n) = b(n-1,2) for n > 0. (End)
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