OFFSET
0,4
COMMENTS
What is the limit (a(n+1)/a(n))/n ? (Value is near 1.44605... at n=300.)
Limit n->infinity (a(n+1)/a(n))/n = 1/log(2) = 1.4426950408889634... . - Vaclav Kotesovec, Aug 08 2014
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..300
FORMULA
G.f. satisfies:
(1) A(x)^2/(1+x) = A( x * A(x)^2/(1+x) ).
(2) A(x)^2/(1+x) = G(x) where G(x) = A(x*G(x)) and A(x) = G(x/A(x)) and G(x*G(x)) = G(x)^2/(1 + x*G(x)).
a(n) ~ c * n^(n + 1/2 + log(2)) / (exp(n) * (log(2))^n), where c = 0.2812864532720972025... . - Vaclav Kotesovec, Aug 08 2014
EXAMPLE
G.f.: A(x) = 1 + x + x^2 + 4*x^3 + 24*x^4 + 194*x^5 + 1910*x^6 + 21906*x^7 +...
Related series:
A(x)^2 = 1 + 2*x + 3*x^2 + 10*x^3 + 57*x^4 + 444*x^5 + 4272*x^6 + 48212*x^7 +...
A(x/A(x)) = 1 + x + 2*x^3 + 10*x^4 + 87*x^5 + 866*x^6 + 10067*x^7 +...
A(x/A(x))^2 = 1 + 2*x + x^2 + 4*x^3 + 24*x^4 + 194*x^5 + 1910*x^6 + 21906*x^7 +...
PROG
(PARI) {a(n)=local(A=[1, 1]); for(i=1, n, A=concat(A, 0); F=Ser(A); A[#A]=Vec(1+F-subst(F^2, x, x/F))[#A]); A[n+1]}
for(n=0, 25, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Aug 07 2014
STATUS
approved