A352237 G.f. A(x) satisfies: A(x) = 1 + x*A(x)^2 / (A(x) - 3*x*A'(x)).

1, 1, 4, 37, 532, 9994, 226252, 5910445, 173581060, 5634589906, 199792389160, 7671942375898, 316936631324368, 14011781050744984, 660054967923455212, 33008607551445324157, 1746771084107236755604, 97536010045722766992778, 5731874036042145864368824

Offset: 0

Author

Paul D. Hanna, Mar 08 2022

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 37*x^3 + 532*x^4 + 9994*x^5 + 226252*x^6 + 5910445*x^7 + 173581060*x^8 + ...
such that A(x) = 1 + x*A(x)^2/(A(x) - 3*x*A'(x)).
Related table.
The table of coefficients of x^k in A(x)^(3*n+1) begins:
n=0: [1,  1,   4,   37,   532,   9994,   226252, ...];
n=1: [1,  4,  22,  200,  2717,  48788,  1069122, ...];
n=2: [1,  7,  49,  462,  6069, 104664,  2219784, ...];
n=3: [1, 10,  85,  850, 11020, 183832,  3777355, ...];
n=4: [1, 13, 130, 1391, 18083, 294203,  5869734, ...];
n=5: [1, 16, 184, 2112, 27852, 445632,  8659920, ...];
n=6: [1, 19, 247, 3040, 41002, 650161, 12353059, ...]; ...
in which the following pattern holds:
[x^n] A(x)^(3*n+1) = [x^(n-1)] (3*n+1) * A(x)^(3*n+1), n >= 1,
as illustrated by
[x^1] A(x)^4 = 4 = [x^0] 4*A(x)^4 = 4*1;
[x^2] A(x)^7 = 49 = [x^1] 7*A(x)^7 = 7*7;
[x^3] A(x)^10 = 850 = [x^2] 10*A(x)^10 = 10*85;
[x^4] A(x)^13 = 18083 = [x^3] 13*A(x)^13 = 13*1391;
[x^5] A(x)^16 = 445632 = [x^4] 16*A(x)^16 = 16*27852;
[x^6] A(x)^19 = 12353059 = [x^5] 19*A(x)^19 = 19*650161; ...
Also, compare the above terms along the diagonal to the series
B(x) = A(x*B(x)^3) = 1 + x + 7*x^2 + 85*x^3 + 1391*x^4 + 27852*x^5 + 650161*x^6 + 17204220*x^7 + ...
where B(x)^3 = (1/x) * Series_Reversion( x/A(x)^3 ).
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..378

Crossrefs

Cf. A088715, A286797, A317352, A352235, A352236, A352238.

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 - 3*x*A' + x*O(x^n)) );
polcoeff(A,n)}
for(n=0,20, print1(a(n),", "))

PARI

/* Using [x^n] A(x)^(3*n+1) = [x^(n-1)] (3*n+1)*A(x)^(3*n+1) */
{a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
A[#A] = polcoeff((x*Ser(A)^(3*(#A)-2) - Ser(A)^(3*(#A)-2)/(3*(#A)-2)),#A-1));A[n+1]}
for(n=0,20, print1(a(n),", "))

Formula

G.f. A(x) satisfies:

(1) [x^n] A(x)^(3*n+1) = [x^(n-1)] (3*n+1) * A(x)^(3*n+1) for n >= 1.

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

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

(4) A(x) = exp( Integral (1 + x*A(x)/(1 - A(x)))/(3*x) dx ).

a(n) ~ c * 3^n * n! * n^(4/3), where c = 0.0543186200722307001992331... - Vaclav Kotesovec, Nov 16 2023