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URL: https://oeis.org/A316700

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A316700
E.g.f. A(x) satisfies: A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k/A(x).
4
1, 2, 5, 19, 87, 481, 3058, 22317, 183501, 1695937, 17383266, 196331895, 2413283755, 32071547509, 457005861978, 6958913121081, 112742453743929, 1940037369861185, 35336786759749378, 679714283742254627, 13755601059097927791, 292116789342048656525, 6489891770655364327818, 150589804371710317610221, 3642747130658567662759333, 91770842180615381158770081
OFFSET
0,2
COMMENTS
More generally, we have the following identity. Given the biexponential series
W(x,y) = Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + k*y,
then for fixed p and q,
Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k + p)*x + (k + q)*y = W(x,y)^(p+q+1) / ( (1 + x*W(x,y))^q * (1 + y*W(x,y))^p ).
Further, W(x,y) satisfies the biexponential functional equation
( W(x,y)/(1 + x*W(x,y)) )^x = ( W(x,y)/(1 + y*W(x,y)) )^y.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..447 (terms 0..70 from Paul D. Hanna)
FORMULA
E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:
(1) A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k/A(x).
(2) Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k + p) + (k + q)/A(x) = A(x)^(p+q+1) / ( (1 + x*A(x))^q * (1 + x)^p ), for fixed p and q.
(3) A(x)/(1 + x) = ( A(x)/(1 + x*A(x)) )^A(x).
EXAMPLE
E.g.f.: A(x) = 1 + 2*x + 5*x^2/2! + 19*x^3/3! + 87*x^4/4! + 481*x^5/5! + 3058*x^6/6! + 22317*x^7/7! + 183501*x^8/8! + 1695937*x^9/9! + ...
such that A = A(x) satisfies
A(x) = 1 + (1 + 1/A)*x + (2 + 1/A)*(1 + 2/A)*x^2/2! + (3 + 1/A)*(2 + 2/A)*(1 + 3/A)*x^3/3! + (4 + 1/A)*(3 + 2/A)*(2 + 3/A)*(1 + 4/A)*x^4/4! + (5 + 1/A)*(4 + 2/A)*(3 + 3/A)*(2 + 4/A)*(1 + 5/A)*x^5/5! + ...
Also,
A(x)^2/(1 + x*A(x)) = 1 + (1 + 2/A)*x + (2 + 2/A)*(1 + 3/A)*x^2/2! + (3 + 2/A)*(2 + 3/A)*(1 + 4/A)*x^3/3! + (4 + 2/A)*(3 + 3/A)*(2 + 4/A)*(1 + 5/A)*x^4/4! + (5 + 2/A)*(4 + 3/A)*(3 + 4/A)*(2 + 5/A)*(1 + 6/A)*x^5/5! + ...
And,
A(x)^3/((1 + x*A(x))*(1 + x)) = 1 + (2 + 2/A)*x + (3 + 2/A)*(2 + 3/A)*x^2/2! + (4 + 2/A)*(3 + 3/A)*(2 + 4/A)*x^3/3! + (5 + 2/A)*(4 + 3/A)*(3 + 4/A)*(2 + 5/A)*x^4/4! + (6 + 2/A)*(5 + 3/A)*(4 + 4/A)*(3 + 5/A)*(2 + 6/A)*x^5/5! + ...
RELATED SERIES.
A(x)/(1+x) = 1 + x + 3*x^2/2! + 10*x^3/3! + 47*x^4/4! + 246*x^5/5! + 1582*x^6/6! + 11243*x^7/7! + 93557*x^8/8! + 853924*x^9/9! + ...
A(x)/(1 + x*A(x)) = 1 + x - x^2/2! - 5*x^3/3! - 5*x^4/4! + 41*x^5/5! + 256*x^6/6! + 533*x^7/7! - 4451*x^8/8! - 57479*x^9/9! + ...
where ( A(x)/(1 + x*A(x)) )^A(x) = A(x)/(1 + x).
Let G(x) = A(x*G(x)) and A(x) = G(x/A(x)), where G(x) begins
G(x) = 1 + 2*x + 13*x^2/2! + 157*x^3/3! + 2819*x^4/4! + 67621*x^5/5! + 2036230*x^6/6! + 73907639*x^7/7! + 3142556933*x^8/8! + ... + A316701(n)*x^n/n! + ...
then G(x)/(1 + x*G(x)) = ( G(x)/(1 + x*G(x)^2) )^G(x)
and G(x) = (1/x)*Series_Reversion( x/A(x) ).
MATHEMATICA
nmax = 25; aa = ConstantArray[0, nmax]; aa[[1]] = 2; Do[y = 1 + 2*x + Sum[aa[[k]]*x^k, {k, 2, j - 1}] + koef*x^j; sol = Solve[SeriesCoefficient[(1 + x)*(y/(1 + x*y))^y - y, {x, 0, j + 1}] == 0, koef][[1]]; aa[[j]] = koef /. sol[[1]], {j, 2, nmax}]; Flatten[{1, aa}] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 16 2020 *)
PROG
(PARI) /* From Biexponential Series: */
{a(n) = my(A=1); for(i=1, n, A = sum(m=0, n, x^m/m! * prod(k=1, m, m+1-k + k/A +x*O(x^n)))); n!*polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jul 13 2018
STATUS
approved