login
A361767
Expansion of e.g.f. A(x) = 1/F(oo,x) where F(oo,x) is the limit of the process F(n,x) = (F(n-1,x)^n - x^n)^(1/n) for n > 0, starting with F(0,x) = 1.
2
1, 1, 3, 17, 143, 1599, 22369, 376417, 7409793, 167120657, 4249941371, 120323916591, 3753781567183, 127950507522967, 4731189132093033, 188631199008696389, 8066710048305641729, 368331028066068082977, 17885422396274431047283, 920319838571287315515007, 50024628127300451995229871
OFFSET
0,3
COMMENTS
a(n+1)/a(n) tends to 2.9058... - Vaclav Kotesovec, Mar 31 2023
LINKS
EXAMPLE
E.g.f.: A(x) = 1 + x + 3*x^2/2! + 17*x^3/3! + 143*x^4/4! + 1599*x^5/5! + 22369*x^6/6! + 376417*x^7/7! + 7409793*x^8/8! + 167120657*x^9/9! + 4249941371*x^10/10! + ... + a(n)*x^n/n! + ...
where A(x) = 1/F(oo,x) where F(oo,x) is the limit of the following process.
Start with F(0,x) = 1, and continue F(n,x) = (F(n-1,x)^n - 1)^(1/n) for n > 0, like so:
F(0,x) = 1
F(1,x) = (F(0,x)^1 - x^1)^(1/1)
F(2,x) = (F(1,x)^2 - x^2)^(1/2)
F(3,x) = (F(2,x)^3 - x^3)^(1/3)
F(4,x) = (F(3,x)^4 - x^4)^(1/4)
F(5,x) = (F(4,x)^5 - x^5)^(1/5)
...
then A(x) is the limit of 1/F(n,x) as n approaches infinity:
1/F(1,x) = 1 + x + 2*x^2/2! + 6*x^3/3! + 24*x^4/4! + 120*x^5/5! + ...
1/F(2,x) = 1 + x + 3*x^2/2! + 15*x^3/3! + 105*x^4/4! + 945*x^5/5! + ...
1/F(3,x) = 1 + x + 3*x^2/2! + 17*x^3/3! + 137*x^4/4! + 1425*x^5/5! + ...
1/F(4,x) = 1 + x + 3*x^2/2! + 17*x^3/3! + 143*x^4/4! + 1575*x^5/5! + ...
1/F(5,x) = 1 + x + 3*x^2/2! + 17*x^3/3! + 143*x^4/4! + 1599*x^5/5! + ...
...
PROG
(PARI) {a(n) = my(A=1, F); F=1; for(m=1, n, F = (F^m - x^m +x*O(x^(n+1)))^(1/m)); A = 1/F; n!*polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
Cf. A361768.
Sequence in context: A189001 A087885 A178685 * A268254 A333331 A198860
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 29 2023
STATUS
approved