A365516 Expansion of g.f. A(x) satisfying A(x) = (1 + 2*x*A(x))^2 / (1 + 2*x*A(x) - 2*x^3*A(x)^3).

1, 2, 4, 10, 32, 112, 400, 1464, 5520, 21296, 83456, 331136, 1328320, 5379200, 21959936, 90271904, 373347840, 1552438016, 6486311680, 27217331456, 114649525760, 484640538112, 2055185596416, 8740711936000, 37273693649920, 159340373710848, 682708771254272, 2931290431277056

Offset

0,2

Comments

Related identities for the Catalan function C(x) = 1 + x*C(x)^2 (A000108):

(1) [x^(n-1)] (1 + n*x*C(x))^n / C(x)^n = n^(n-1) for n >= 1.

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

Related identity: [x^(n-1)] (1 + (n+1)*x*B(x))^n / B(x)^n = n*(n-1)^(n-1) for n >= 1 when B(x) = 1/(1 - 2*x).

Related identity: F(x) = (2/x) * Sum{n>=1} n*(n+1)^(n-2) * x^n * F(x)^n / (1 + (n+1)*x*F(x))^(n+1), which holds formally for all Maclaurin series F(x). - Paul D. Hanna, Oct 03 2023

Links

Paul D. Hanna, Table of n, a(n) for n = 0..400

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

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

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

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

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

(5) [x^(n-1)] (1 + (n-1)*x*A(x))^n / A(x)^n = -n*(n-3)^(n-2) for n > 1.

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

(7) A(x) = (2/x) * Sum{n>=1} n*(n+1)^(n-2) * x^n * A(x)^n / (1 + (n+1)*x*A(x))^(n+1). - Paul D. Hanna, Oct 03 2023

Example

G.f.: A(x) = 1 + 2*x + 4*x^2 + 10*x^3 + 32*x^4 + 112*x^5 + 400*x^6 + 1464*x^7 + 5520*x^8 + 21296*x^9 + 83456*x^10 + ...
where A(x) = (1 + 2*x*A(x))^2/(1 + 2*x*A(x) - 2*x^3*A(x)^3).
RELATED SERIES.
B(x) = A(x/B(x)) where B(x) = (1 + 2*x)^2/(1 + 2*x - 2*x^3) begins
B(x) = 1 + 2*x + 2*x^3 + 4*x^6 - 8*x^7 + 16*x^8 - 24*x^9 + 32*x^10 - 32*x^11 + 16*x^12 + 32*x^13 - 128*x^14 + ...
RELATED TABLE.
The table of coefficients of x^k in (1 + (n+1)*x*A(x))^n/A(x)^n begins:
n=1: [1,  0,    0,    -2,     -8,    -24,     -76,    -272, ...];
n=2: [1,  2,    1,    -4,    -20,    -64,    -196,    -664, ...];
n=3: [1,  6,   12,     2,    -48,   -192,    -600,   -1896, ...];
n=4: [1, 12,   54,   100,    -23,   -600,   -2224,   -6944, ...];
n=5: [1, 20,  160,   630,   1080,   -696,   -8660,  -30960, ...];
n=6: [1, 30,  375,  2488,   9027,  14406,  -15371, -144852, ...];
n=7: [1, 42,  756,  7546,  44800, 153552,  229376, -342728, ...];
n=8: [1, 56, 1372, 19192, 167222, 921400, 3026332, 4251528, ...]; ...
in which the main diagonal equals n*(n+1)^(n-2) for n > 1.

Mathematica

a[n_] := Module[{A = {1}, m, x}, Do[AppendTo[A, 0]; m = Length[A]; A[[-1]] = Coefficient[(1 + (m + 1)*x*SeriesData[x, 0, A, 0, m + 1, 1])^m/SeriesData[x, 0, A, 0, m + 1, 1]^m, x, m - 1]/m - (m + 1)^(m - 2), {i, 1, n}]; A[[n + 1]]]; Table[a[n], {n, 0, 27}] (* Robert P. P. McKone, Sep 07 2023 *)

Prog

(PARI) /* Using series reversion */
{a(n) = my(A = (1/x)*serreverse(x*(1 + 2*x - 2*x^3)/(1 + 2*x +x*O(x^n))^2));
polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) /* Formula [x^(n-1)] (1 + (n+1)*x*A(x))^n/A(x)^n = n*(n+1)^(n-2) */
{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A;
A[#A] = polcoeff( (1 + (m+1)*x*Ser(A))^m / Ser(A)^m , m-1)/m - (m+1)^(m-2) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))

Crossrefs

Cf. A365574, A303063, A365095.

Sequence in context: A151400 A363138 A367113 * A336614 A071954 A352279

Adjacent sequences: A365513 A365514 A365515 * A365517 A365518 A365519

Keyword

nonn

Author

Paul D. Hanna, Sep 07 2023

Status

approved