A360231 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n-1))^(n+1) for n >= 0.
1, 1, 1, 6, 53, 628, 9167, 156309, 3021720, 64960004, 1532234825, 39270176511, 1085601040372, 32185085432757, 1018593646880447, 34279111177431666, 1222648239226278333, 46084480032637208699, 1830881732391546532475, 76488074741796221197580, 3352854778050665597014436
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + x^2 + 6*x^3 + 53*x^4 + 628*x^5 + 9167*x^6 + 156309*x^7 + 3021720*x^8 + 64960004*x^9 + 1532234825*x^10 + ... RELATED SERIES. G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins: B(x) = 1 + x + 2*x^2 + 10*x^3 + 86*x^4 + 1004*x^5 + 14507*x^6 + 246218*x^7 + 4753205*x^8 + ... + b(n)*x^n + ... such that b(n) = [x^n] (1 + x*A(x)^(n-1))^(n+1) / (n+1), as well as b(n) = [x^n] A(x)^(n+1) / (n+1), so that b(n) begin: [1/1, 2/2, 6/3, 40/4, 430/5, 6024/6, 101549/7, 1969744/8, ...]. ILLUSTRATION OF DEFINITION. The table of coefficients of x^k in A(x)^(n+1) begins: n=0: [1, 1, 1, 6, 53, 628, 9167, 156309, ...]; n=1: [1, 2, 3, 14, 119, 1374, 19732, 332844, ...]; n=2: [1, 3, 6, 25, 201, 2259, 31891, 531933, ...]; n=3: [1, 4, 10, 40, 303, 3308, 45870, 756192, ...]; n=4: [1, 5, 15, 60, 430, 4551, 61930, 1008565, ...]; n=5: [1, 6, 21, 86, 588, 6024, 80373, 1292370, ...]; n=6: [1, 7, 28, 119, 784, 7770, 101549, 1611352, ...]; n=7: [1, 8, 36, 160, 1026, 9840, 125864, 1969744, ...]; ... Compare to the table of coefficients in (1 + x*A(x)^(n-1))^(n+1): n=0: [1, 1, -1, 0, -5, -42, -528, -7939, ...]; n=1: [1, 2, 1, 0, 0, 0, 0, 0, ...]; n=2: [1, 3, 6, 10, 30, 207, 2266, 31824, ...]; n=3: [1, 4, 14, 40, 141, 808, 7694, 101288, ...]; n=4: [1, 5, 25, 100, 430, 2376, 19680, 235165, ...]; n=5: [1, 6, 39, 200, 1035, 6024, 45879, 490524, ...]; n=6: [1, 7, 56, 350, 2135, 13601, 101549, 988338, ...]; n=7: [1, 8, 76, 560, 3950, 27888, 213952, 1969744, ...]; ... to see that the main diagonals of the tables are the same: [1, 2, 6, 40, 430, 6024, 101549, 1969744, ...].
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..400
Programs
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PARI
{a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(m-1))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]} for(n=0, 20, print1(a(n), ", "))
Formula
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n-1))^(n+1) for n>=0.
(2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(n-1))^(n+1) / (n+1).
a(n) ~ c * d^n * n! * n^alfa, where d = A360279, alfa = 0.311338934287018467072138011497837... and c = 0.1932932528309324180094... - Vaclav Kotesovec, Feb 03 2023