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A366232
Expansion of e.g.f. A(x) satisfying A(x) = 1 + x*A(x) * exp(2*x*A(x)).
7
1, 1, 6, 54, 728, 13000, 290352, 7800016, 245115264, 8826560640, 358463525120, 16212238054144, 808215885708288, 44035925223746560, 2603618739995621376, 166031767704180111360, 11359670347331723952128, 830065763154102656204800, 64518486557995327748898816
OFFSET
0,3
COMMENTS
Related identities which hold formally for all Maclaurin series F(x):
(1) F(x) = (1/x) * Sum{n>=1} n^(n-1) * x^n/n! * F(x)^n * exp(-n*x*F(x)),
(2) F(x) = (2/x) * Sum{n>=1} n*(n+1)^(n-2) * x^n/n! * F(x)^n * exp(-(n+1)*x*F(x)),
(3) F(x) = (3/x) * Sum{n>=1} n*(n+2)^(n-2) * x^n/n! * F(x)^n * exp(-(n+2)*x*F(x)),
(4) F(x) = (4/x) * Sum{n>=1} n*(n+3)^(n-2) * x^n/n! * F(x)^n * exp(-(n+3)*x*F(x)),
(5) F(x) = (k/x) * Sum{n>=1} n*(n+k-1)^(n-2) * x^n/n! * F(x)^n * exp(-(n+k-1)*x*F(x)) for all fixed nonzero k.
FORMULA
a(n) = n! * Sum{k=0..n} binomial(n+1, n-k)/(n+1) * 2^k * (n-k)^k / k!.
Let A(x)^m = Sum_{n>=0} a(n,m) * x^n/n! then a(n,m) = n!*Sum_{k=0..n} binomial(n+m, n-k)*m/(n+m) * 2^k * (n-k)^k/k!.
E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies the following formulas.
(1) A(x) = 1 + x*A(x) * exp(2*x*A(x)).
(2) A(x) = (1/x) * Series_Reversion( x/(1 + x*exp(2*x)) ).
(3) A( x/(1 + x*exp(2*x)) ) = 1 + x*exp(2*x).
(4) A(x) = 1 + (m+1) * Sum{n>=1} n*(n+m)^(n-2) * x^n/n! * A(x)^n * exp(-(n+m-2)*x*A(x)) for all fixed nonnegative m.
(4.a) A(x) = 1 + Sum{n>=1} n^(n-1) * x^n/n! * A(x)^n * exp(-(n-2)*x*A(x)).
(4.b) A(x) = 1 + 2 * Sum{n>=1} n*(n+1)^(n-2) * x^n/n! * A(x)^n * exp(-(n-1)*x*A(x)).
(4.c) A(x) = 1 + 3 * Sum{n>=1} n*(n+2)^(n-2) * x^n/n! * A(x)^n * exp(-n*x*A(x)).
(4.d) A(x) = 1 + 4 * Sum{n>=1} n*(n+3)^(n-2) * x^n/n! * A(x)^n * exp(-(n+1)*x*A(x)).
a(n) ~ n^(n-1) * (1 + 2*LambertW(1/sqrt(2)))^(n + 3/2) / (sqrt(1 + LambertW(1/sqrt(2))) * 2^(n+2) * exp(n) * LambertW(1/sqrt(2))^(2*n + 3/2)). - Vaclav Kotesovec, Oct 06 2023
EXAMPLE
E.g.f.: A(x) = 1 + x + 6*x^2/2! + 54*x^3/3! + 728*x^4/4! + 13000*x^5/5! + 290352*x^6/6! + 7800016*x^7/7! + 245115264*x^8/8! + ...
where A(x) satisfies A(x) = 1 + x*A(x) * exp(2*x*A(x))
also
A(x) = 1 + 1^0*x*A(x)*exp(+1*x*A(x))/1! + 2^1*x^2*A(x)^2*exp(-0*x*A(x))/2! + 3^2*x^3*A(x)^3*exp(-1*x*A(x))/3! + 4^3*x^4*A(x)^4*exp(-2*x*A(x))/4! + 5^4*x^5*A(x)^5*exp(-3*x*A(x))/5! + 6^5*x^6*A(x)^6*exp(-4*x*A(x))/6! + ...
and
A(x) = 1 + 3*1*3^(-1)*x*A(x)*exp(-1*x*A(x))/1! + 3*2*4^0*x^2*A(x)^2*exp(-2*x*A(x))/2! + 3*3*5^1*x^3*A(x)^3*exp(-3*x*A(x))/3! + 3*4*6^2*x^4*A(x)^4*exp(-4*x*A(x))/4! + 3*5*7^3*x^5*A(x)^5*exp(-5*x*A(x))/5! + ...
RELATED SERIES.
exp(x*A(x)) = 1 + x + 3*x^2/2! + 25*x^3/3! + 313*x^4/4! + 5341*x^5/5! + ... + A212722(n)*x^n/n! + ...
MATHEMATICA
nmax = 20; A[_] = 0; Do[A[x_] = 1 + x*A[x] * E^(2*x*A[x]) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] * Range[0, nmax]! (* Vaclav Kotesovec, Oct 06 2023 *)
PROG
(PARI) /* a(n, m) = coefficient of x^n/n! in A(x)^m, here at m = 1 */
{a(n, m=1) = n!*sum(k=0, n, binomial(n+m, n-k)*m/(n+m) * 2^k * (n-k)^k/k!)}
for(n=0, 20, print1(a(n), ", "))
(PARI) {a(n) = my(A = (1/x) * serreverse( x/(1 + x*exp(2*x +O(x^(n+2)))) )); n!*polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
Cf. A365772 (dual), A212722 (exp(x*A(x)).
Sequence in context: A356927 A072034 A167571 * A138434 A360545 A217238
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 05 2023
STATUS
approved