A352238 G.f. A(x) satisfies: A(x) = 1 + x*A(x)^2 / (A(x) - 4*x*A'(x)).

1, 1, 5, 61, 1161, 28857, 864141, 29861749, 1160382737, 49854838897, 2340623599381, 119051103325613, 6516915195123097, 381912592990453545, 23856225840952434333, 1582482450123627473637, 111113139625779846025761, 8234335766045466358238433

Offset: 0

Author

Paul D. Hanna, Mar 08 2022

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 61*x^3 + 1161*x^4 + 28857*x^5 + 864141*x^6 + 29861749*x^7 + 1160382737*x^8 + ...
such that A(x) = 1 + x*A(x)^2/(A(x) - 4*x*A'(x)).
Related table.
The table of coefficients of x^k in A(x)^(4*n+1) begins:
n=0: [1,  1,   5,   61,   1161,   28857,   864141, ...];
n=1: [1,  5,  35,  415,   7430,  176286,  5107530, ...];
n=2: [1,  9,  81,  993,  17127,  389583, 10916559, ...];
n=3: [1, 13, 143, 1859,  31564,  693212, 18802212, ...];
n=4: [1, 17, 221, 3077,  52309, 1118549, 29427153, ...];
n=5: [1, 21, 315, 4711,  81186, 1704906, 43640030, ...];
n=6: [1, 25, 425, 6825, 120275, 2500555, 62513875, ...]; ...
in which the following pattern holds:
[x^n] A(x)^(4*n+1) = [x^(n-1)] (4*n+1) * A(x)^(4*n+1), n >= 1,
as illustrated by
[x^1] A(x)^5 = 5 = [x^0] 5*A(x)^5 = 5*1;
[x^2] A(x)^9 = 81 = [x^1] 9*A(x)^9 = 9*9;
[x^3] A(x)^13 = 1859 = [x^2] 13*A(x)^13 = 13*143;
[x^4] A(x)^17 = 52309 = [x^3] 17*A(x)^17 = 17*3077;
[x^5] A(x)^21 = 1704906 = [x^4] 21*A(x)^21 = 21*81186;
[x^6] A(x)^25 = 62513875 = [x^5] 25*A(x)^25 = 25*2500555; ...
Also, compare the above terms along the diagonal to the series
B(x) = A(x*B(x)^4) = 1 + x + 9*x^2 + 143*x^3 + 3077*x^4 + 81186*x^5 + 2500555*x^6 + 87388600*x^7 + ...
where B(x)^4 = (1/x) * Series_Reversion( x/A(x)^4 ).
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..363

Crossrefs

Cf. A088715, A286797, A317352, A352235, A352236, A352237.

Programs

PARI

/* Using A(x) = 1 + x*A(x)^2/(A(x) - 3*x*A'(x)) */
{a(n) = my(A=1); for(i=1,n, A = 1 + x*A^2/(A - 4*x*A' + x*O(x^n)) );
polcoeff(A,n)}
for(n=0,20, print1(a(n),", "))

PARI

/* Using [x^n] A(x)^(4*n+1) = [x^(n-1)] (4*n+1)*A(x)^(4*n+1) */
{a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
A[#A] = polcoeff((x*Ser(A)^(4*(#A)-3) - Ser(A)^(4*(#A)-3)/(4*(#A)-3)),#A-1));A[n+1]}
for(n=0,20, print1(a(n),", "))

Formula

G.f. A(x) satisfies:

(1) [x^n] A(x)^(4*n+1) = [x^(n-1)] (4*n+1) * A(x)^(4*n+1) for n >= 1.

(2) A(x) = 1 + x*A(x)^2/(A(x) - 4*x*A'(x)).

(3) A'(x) = A(x) * (1 + x*A(x)/(1 - A(x))) / (4*x).

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

a(n) ~ c * 4^n * n! * n^(5/4), where c = 0.0440035900116077498469559... - Vaclav Kotesovec, Nov 16 2023