OFFSET
0,2
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..200
EXAMPLE
G.f.: A(x) = 1 + 2*x + 10*x^2 + 68*x^3 + 550*x^4 + 5100*x^5 + 53668*x^6 + 644328*x^7 + 9018182*x^8 + 153030092*x^9 + 3321466604*x^10 + ...
where A(x) = 1/(x*F(oo,x)) where F(oo,x) is the limit of the following process.
Start with F(0,x) = 1/x, and continue F(n,x) = (F(n-1,x)^n - 2^(n+1)*x^n)^(1/2^n) for n > 0, like so:
F(0,x) = 1
F(1,x) = (F(0,x)^2 - 4*x)^(1/2)
F(2,x) = (F(1,x)^4 - 16*x^2)^(1/4)
F(3,x) = (F(2,x)^8 - 64*x^3)^(1/8)
F(4,x) = (F(3,x)^16 - 256*x^4)^(1/16)
F(5,x) = (F(4,x)^32 - 1024*x^5)^(1/32)
F(6,x) = (F(4,x)^64 - 4096*x^5)^(1/64)
...
then A(x) equals the limit of 1/F(n,x) as n approaches infinity:
1/F(1,x) = 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 + 924*x^6 + ...
1/F(2,x) = 1 + 2*x + 10*x^2 + 60*x^3 + 390*x^4 + 2652*x^5 + 18564*x^6 + ...
1/F(3,x) = 1 + 2*x + 10*x^2 + 68*x^3 + 534*x^4 + 4524*x^5 + 40068*x^6 + ...
1/F(4,x) = 1 + 2*x + 10*x^2 + 68*x^3 + 550*x^4 + 5068*x^5 + 51492*x^6 + ...
1/F(5,x) = 1 + 2*x + 10*x^2 + 68*x^3 + 550*x^4 + 5100*x^5 + 53604*x^6 + ...
1/F(6,x) = 1 + 2*x + 10*x^2 + 68*x^3 + 550*x^4 + 5100*x^5 + 53668*x^6 + ...
...
Related series.
Notice that A(x/2) is a power series in x with fractional coefficients, yet the logarithmic derivative appears to be an integer series:
A'(x/2)/A(x/2) = 1 + 4*x + 19*x^2 + 100*x^3 + 581*x^4 + 3742*x^5 + 27063*x^6 + 225608*x^7 + 2257705*x^8 + 28543914*x^9 + 478859723*x^10 + ...
Also, it appears that A(x/2)^2 is an integer series:
A(x/2)^2 = 1 + 2*x + 6*x^2 + 22*x^3 + 92*x^4 + 430*x^5 + 2240*x^6 + 13126*x^7 + 88606*x^8 + 718610*x^9 + 7429776*x^10 + 104210506*x^11 + 2065002874*x^12 + ...
PROG
(PARI) {a(n) = my(A=1, F); F=1; for(m=1, n, F = (F^(2^m) - 4^m*x^m +x*O(x^n) )^(1/2^m)); A = 1/F; polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))
(PARI) /* Faster */
{a(n) = my(A=1, F); F=1; for(m=1, n, F = (F^2 - 4^m*x^m +x*O(x^n) ); ); A = 1/F^(1/2^n); polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Apr 06 2023
STATUS
approved