A352236 G.f. A(x) satisfies: A(x) = 1 + x*A(x)^2 / (A(x) - 2*x*A'(x)).

1, 1, 3, 19, 185, 2353, 36075, 638115, 12683761, 278485217, 6674259667, 173097575603, 4826128088489, 143896870347793, 4568544366818747, 153883892657000259, 5481761893234193889, 205939077652874352577, 8138639816942009694627, 337568614331296733526867

Offset: 0

Author

Paul D. Hanna, Mar 08 2022

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 19*x^3 + 185*x^4 + 2353*x^5 + 36075*x^6 + 638115*x^7 + 12683761*x^8 + ...
such that A(x) = 1 + x*A(x)^2/(A(x) - 2*x*A'(x)).
Related table.
The table of coefficients of x^k in A(x)^(2*n+1) begins:
n=0: [1,  1,   3,   19,   185,   2353,   36075, ...];
n=1: [1,  3,  12,   76,   705,   8595,  127680, ...];
n=2: [1,  5,  25,  165,  1490,  17506,  252050, ...];
n=3: [1,  7,  42,  294,  2632,  30016,  419454, ...];
n=4: [1,  9,  63,  471,  4239,  47295,  643017, ...];
n=5: [1, 11,  88,  704,  6435,  70785,  939312, ...];
n=6: [1, 13, 117, 1001,  9360, 102232, 1329016, ...]; ...
in which the following pattern holds:
[x^n] A(x)^(2*n+1) = [x^(n-1)] (2*n+1) * A(x)^(2*n+1), n >= 1,
as illustrated by
[x^1] A(x)^3 = 3 = [x^0] 3*A(x)^3 = 3*1;
[x^2] A(x)^5 = 25 = [x^1] 5*A(x)^5 = 5*5;
[x^3] A(x)^7 = 294 = [x^2] 7*A(x)^7 = 7*42;
[x^4] A(x)^9 = 4239 = [x^3] 9*A(x)^9 = 9*471;
[x^5] A(x)^11 = 70785 = [x^4] 11*A(x)^11 = 11*6435;
[x^6] A(x)^13 = 1329016 = [x^5] 13*A(x)^13 = 13*102232; ...
Also, compare the above terms along the diagonal to the series
B(x) = A(x*B(x)^2) = 1 + x + 5*x^2 + 42*x^3 + 471*x^4 + 6435*x^5 + 102232*x^6 + 1837630*x^7 + ... + A317352(n)*x^n + ...
where B(x)^2 = (1/x) * Series_Reversion( x/A(x)^2 ).
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..400

Crossrefs

Cf. A088715, A286797, A317352, A352235, A352237, A352238.

Programs

PARI

/* Using A(x) = 1 + x*A(x)^2/(A(x) - 2*x*A'(x)) */
{a(n) = my(A=1); for(i=1,n, A = 1 + x*A^2/(A - 2*x*A' + x*O(x^n)) );
polcoeff(A,n)}
for(n=0,30, print1(a(n),", "))

PARI

/* Using [x^n] A(x)^(2*n+1) = [x^(n-1)] (2*n+1)*A(x)^(2*n+1) */
{a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
A[#A] = polcoeff((x*Ser(A)^(2*(#A)-1) - Ser(A)^(2*(#A)-1)/(2*(#A)-1)),#A-1));A[n+1]}
for(n=0,30, print1(a(n),", "))

Formula

G.f. A(x) satisfies:

(1) [x^n] A(x)^(2*n+1) = [x^(n-1)] (2*n+1) * A(x)^(2*n+1) for n >= 1.

(2) A(x) = 1 + x*A(x)^2/(A(x) - 2*x*A'(x)).

(3) A'(x) = A(x) * (1 + x*A(x)/(1 - A(x))) / (2*x).

(4) A(x) = exp( Integral (1 + x*A(x)/(1 - A(x)))/(2*x) dx ).

a(n) ~ c * 2^n * n! * n^(3/2), where c = 0.06926688933886004638602492... - Vaclav Kotesovec, Nov 16 2023