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
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
Programs
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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 */
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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
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