OFFSET
0,4
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..120
FORMULA
E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies the following formulas.
(1) [x^n] A(x)^n = [x^n] B^n(x) for n > 0 where B(x) = x*A(x) and B^n(x) is the n-th iteration of B(x) with B^0(x) = x.
(2) [x^n] A(x)^n = n * [x^n] log(F(x)) for n >= 1 where F(x) = A(x*F(x)) and F(x) = (1/x)*Series_Reversion(x/A(x)) is the e.g.f. of A395119.
EXAMPLE
E.g.f.: A(x) = 1 + x + x^2/2! + 7*x^3/3! + 163*x^4/4! + 6851*x^5/5! + 470056*x^6/6! + 47608282*x^7/7! + 6710089289*x^8/8! + ...
RELATED TABLES.
The table of coefficients of x^k/k! in A(x)^n begins
n = 1: [1, 1, 1, 7, 163, 6851, 470056, 47608282, ...];
n = 2: [1, 2, 4, 20, 388, 15472, 1028194, 102164960, ...];
n = 3: [1, 3, 9, 45, 711, 26373, 1691694, 164660940, ...];
n = 4: [1, 4, 16, 88, 1192, 40304, 2482276, 236261848, ...];
n = 5: [1, 5, 25, 155, 1915, 58375, 3427900, 318350030, ...];
n = 6: [1, 6, 36, 252, 2988, 82176, 4565286, 412581672, ...];
n = 7: [1, 7, 49, 385, 4543, 113897, 5943154, 520964080, ...]; ...
Let B(x) = x*A(x) so that
B(x) = x + 2*x^2/2! + 3*x^3/3! + 28*x^4/4! + 815*x^5/5! + 41106*x^6/6! + 3290392*x^7/7! + ...
then the table of coefficients of x^k/k! in B^n(x), the n-th iteration of B(x), begins
n = 1: [0, 1, 2, 3, 28, 815, 41106, 3290392, ...];
n = 2: [0, 1, 4, 18, 140, 2860, 127722, 9545830, ...];
n = 3: [0, 1, 6, 45, 480, 8895, 327708, 21880824, ...];
n = 4: [0, 1, 8, 84, 1192, 24560, 826284, 48152524, ...];
n = 5: [0, 1, 10, 135, 2420, 58375, 2012430, 107401840, ...];
n = 6: [0, 1, 12, 198, 4308, 121740, 4565286, 239923362, ...];
n = 7: [0, 1, 14, 273, 7000, 228935, 9540552, 520964080, ...]; ...
Notice that the following diagonal is found in both tables
D = [1, 4, 45, 1192, 58375, 4565286, 520964080, ...],
illustrating that
D(n) = [x^n/n!] A(x)^n = [x^n/n!] B^n(x) for n >= 1.
RELATED SERIES.
Let F(x) = A(x*F(x)) = (1/x)*Series_Reversion(x/A(x)), then
F(x) = 1 + x + 3*x^2/2! + 22*x^3/3! + 383*x^4/4! + 13696*x^5/5! + 849022*x^6/6! + 80744770*x^7/7! + ... + A395119(n)*x^n/n! + ...
Compare the diagonal D given above with the coefficients in the series
log(F(x)) = x + 2*x^2/2! + 15*x^3/3! + 298*x^4/4! + 11675*x^5/5! + 760881*x^6/6! + 74423440*x^7/7! + ...
to see that
[x^n/n!] log(F(x)) = D(n)/n for n >= 1.
PROG
(PARI) \\ Routine to get the n-th iteration of function F
{IT(n, F) = my(G=x); for(i=1, n, G = subst(F, x, G)); G}
\\ Obtain e.g.f. A(x) using the definition:
{a(n) = my(A=[1], m); for(i=1, n, A = concat(A, 0); m = #A-1;
A[#A] = (Vec(IT(m, x*Ser(A)))[m] - Vec(Ser(A)^m)[m+1])/m ); EGF=Ser(A); n!*A[n+1]}
\\ Display the terms a(0) - a(n):
{upto(n) = a(n); Vec(serlaplace(EGF))}
upto(20)
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Apr 23 2026
STATUS
approved
