A243953 E.g.f.: exp( Sum_{n>=1} A000108(n-1)*x^n/n ), where A000108(n) = binomial(2*n,n)/(n+1) forms the Catalan numbers.

1, 1, 2, 8, 56, 592, 8512, 155584, 3456896, 90501632, 2728876544, 93143809024, 3550380249088, 149488545697792, 6890674623094784, 345131685337530368, 18664673706719019008, 1083931601731053223936, 67278418002152175960064, 4444711314548967826259968

Offset: 0

Author

Paul D. Hanna, Jun 21 2014

Keywords

• nonn
• easy

Examples

			G.f.: A(x) = 1 + x + 2*x^2/2! + 8*x^3/3! + 56*x^4/4! + 592*x^5/5! + 8512*x^6/6! +...
such that the logarithmic derivative of the e.g.f. equals the Catalan numbers:
log(A(x)) = x + x^2/2 + 2*x^3/3 + 5*x^4/4 + 14*x^5/5 + 42*x^6/6 + 132*x^7/7 + 429*x^8/8 +...+ A000108(n-1)*x^n/n +...
thus A'(x)/A(x) = C(x) where C(x) = 1 + x*C(x)^2.
Also, e.g.f. A(x) satisfies:
A(x) = 1 + x/A(x) + 4*(x/A(x))^2/2! + 32*(x/A(x))^3/3! + 400*(x/A(x))^4/4! + 6912*(x/A(x))^5/5! +...+ (n+1)^(n-2)*2^n*(x/A(x))^n/n! +...
If we form a table of coefficients of x^k/k! in A(x)^n, like so:
[1, 1,  2,    8,    56,    592,    8512,   155584,    3456896, ...];
[1, 2,  6,   28,   200,   2064,   28768,   511424,   11106432, ...];
[1, 3, 12,   66,   504,   5256,   72288,  1259712,   26822016, ...];
[1, 4, 20,  128,  1064,  11488,  158752,  2740480,   57517184, ...];
[1, 5, 30,  220,  2000,  22680,  319600,  5525600,  115094400, ...];
[1, 6, 42,  348,  3456,  41472,  602352, 10533024,  219321216, ...];
[1, 7, 56,  518,  5600,  71344, 1075648, 19176304,  401916032, ...];
[1, 8, 72,  736,  8624, 116736, 1835008, 33554432,  712166016, ...];
[1, 9, 90, 1008, 12744, 183168, 3009312, 56687040, 1224440064, ...]; ...
then the main diagonal equals (n+1)^(n-1) * 2^n for n>=0:
[1, 2, 12, 128, 2000, 41472, 1075648, 33554432, 1224440064, ...].
Note that Sum_{n>=0} (n+1)^(n-2) * 2^n * x^n/n! is an e.g.f. of A127670.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..100
• Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.

Crossrefs

Cf. A000108, A127670, A243954, A213507.

Cf. A251663, A251664, A251665, A251666, A251667, A251668, A251669, A251670.

Cf. A251573, A251574, A251575, A251576, A251577, A251578, A251579, A251580.

Cf. A001515.

Programs

Mathematica

CoefficientList[Series[E^(1 - Sqrt[1-4*x]) * (1 + Sqrt[1-4*x])/2, {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Jun 22 2014 *)

Maxima

a(n):=if n=0 then 1 else sum((n-1)!/(n-i-1)!*binomial(2*i,i)/(i+1)*a(n-i-1),i,0,n-1); /* Vladimir Kruchinin, Feb 22 2015 */

PARI

/* Explicit formula: */
{a(n)=n!*polcoeff( exp(1-sqrt(1-4*x +x*O(x^n))) * (1 + sqrt(1-4*x +x*O(x^n)))/2,n)}
for(n=0,25,print1(a(n),", "))

PARI

/* Logarithmic derivative of e.g.f. equals Catalan numbers: */
{A000108(n) = binomial(2*n,n)/(n+1)}
{a(n)=n!*polcoeff( exp(sum(m=1,n, A000108(m-1)*x^m/m)+x*O(x^n)),n)}
for(n=0,25,print1(a(n),", "))

PARI

/* From [x^n/n!] A(x)^(n+1) = (n+1)^(n-1)*2^n */
{a(n)=n!*polcoeff(x/serreverse(x*sum(m=0, n+1, (m+1)^(m-2)*(2*x)^m/m!)+x^2*O(x^n)), n)}
for(n=0,25,print1(a(n),", "))

Formula

E.g.f. A(x) satisfies:

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

(2) A(x)^2 - A(x)*A'(x) + x*A'(x)^2 = 0 (differential equation).

(3) [x^n/n!] A(x)^(n+1) = (n+1)^(n-1)*2^n for n>=0.

(4) A(x) = G(x/A(x)) such that A(x*G(x)) = G(x) = Sum_{n>=0} (n+1)^(n-2)*2^n*x^n/n!.

(5) A(x) = x / Series_Reversion(x*G(x)) where G(x) = Sum_{n>=0} (n+1)^(n-2)*2^n*x^n/n!.

(6) x = -LambertW(-2*x/A(x)) * (2 + LambertW(-2*x/A(x)))/4. [From a formula by Vaclav Kotesovec in A127670]

a(n) ~ 2^(2*n-5/2) * n^(n-2) / exp(n-1). - Vaclav Kotesovec, Jun 22 2014

a(n) = sum(i=0..n-1, (n-1)!/(n-i-1)!*binomial(2*i,i)/(i+1)*a(n-i-1)), a(0)=1. - Vladimir Kruchinin, Feb 22 2015

From Peter Bala, Apr 14 2017: (Start)

a(n+2) = 2^(n+1)*A001515(n).

a(n+1) = Sum_{k = 0..n} binomial(n+k-1,2*k)*2^(n-k)*(2*k)!/k!.

D-finite with recurrence a(n) = (4*n - 10)*a(n-1) + 4*a(n-2) with a(0) = a(1) = 1.

The derivative A'(x) of the e.g.f. is equal to exp(2*x*c(x)), that is, A'(x) is the Catalan transform of exp(2*x) as defined in Barry, Section 3. (End)

E.g.f. A(x) satisfies (x/A(x))' = 1/A'(x). - Alexander Burstein, Oct 31 2023