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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.

%I A243953 #46 Nov 02 2023 14:20:01
%S A243953 1,1,2,8,56,592,8512,155584,3456896,90501632,2728876544,93143809024,
%T A243953 3550380249088,149488545697792,6890674623094784,345131685337530368,
%U A243953 18664673706719019008,1083931601731053223936,67278418002152175960064,4444711314548967826259968
%N 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.
%H A243953 Vincenzo Librandi, <a href="/A243953/b243953.txt">Table of n, a(n) for n = 0..100</a>
%H A243953 Paul Barry, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Barry/barry84.html">A Catalan Transform and Related Transformations on Integer Sequences</a>, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
%F A243953 E.g.f. A(x) satisfies:
%F A243953 (1) A(x) = exp(1 - sqrt(1-4*x)) * (1 + sqrt(1-4*x))/2.
%F A243953 (2) A(x)^2 - A(x)*A'(x) + x*A'(x)^2 = 0 (differential equation).
%F A243953 (3) [x^n/n!] A(x)^(n+1) = (n+1)^(n-1)*2^n for n>=0.
%F A243953 (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!.
%F A243953 (5) A(x) = x / Series_Reversion(x*G(x)) where G(x) = Sum_{n>=0} (n+1)^(n-2)*2^n*x^n/n!.
%F A243953 (6) x = -LambertW(-2*x/A(x)) * (2 + LambertW(-2*x/A(x)))/4. [From a formula by _Vaclav Kotesovec_ in A127670]
%F A243953 a(n) ~ 2^(2*n-5/2) * n^(n-2) / exp(n-1). - _Vaclav Kotesovec_, Jun 22 2014
%F A243953 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
%F A243953 From _Peter Bala_, Apr 14 2017: (Start)
%F A243953 a(n+2) = 2^(n+1)*A001515(n).
%F A243953 a(n+1) = Sum_{k = 0..n} binomial(n+k-1,2*k)*2^(n-k)*(2*k)!/k!.
%F A243953 D-finite with recurrence a(n) = (4*n - 10)*a(n-1) + 4*a(n-2) with a(0) = a(1) = 1.
%F A243953 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)
%F A243953 E.g.f. A(x) satisfies (x/A(x))' = 1/A'(x). - _Alexander Burstein_, Oct 31 2023
%e A243953 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! +...
%e A243953 such that the logarithmic derivative of the e.g.f. equals the Catalan numbers:
%e A243953 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 +...
%e A243953 thus A'(x)/A(x) = C(x) where C(x) = 1 + x*C(x)^2.
%e A243953 Also, e.g.f. A(x) satisfies:
%e A243953 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! +...
%e A243953 If we form a table of coefficients of x^k/k! in A(x)^n, like so:
%e A243953 [1, 1,  2,    8,    56,    592,    8512,   155584,    3456896, ...];
%e A243953 [1, 2,  6,   28,   200,   2064,   28768,   511424,   11106432, ...];
%e A243953 [1, 3, 12,   66,   504,   5256,   72288,  1259712,   26822016, ...];
%e A243953 [1, 4, 20,  128,  1064,  11488,  158752,  2740480,   57517184, ...];
%e A243953 [1, 5, 30,  220,  2000,  22680,  319600,  5525600,  115094400, ...];
%e A243953 [1, 6, 42,  348,  3456,  41472,  602352, 10533024,  219321216, ...];
%e A243953 [1, 7, 56,  518,  5600,  71344, 1075648, 19176304,  401916032, ...];
%e A243953 [1, 8, 72,  736,  8624, 116736, 1835008, 33554432,  712166016, ...];
%e A243953 [1, 9, 90, 1008, 12744, 183168, 3009312, 56687040, 1224440064, ...]; ...
%e A243953 then the main diagonal equals (n+1)^(n-1) * 2^n for n>=0:
%e A243953 [1, 2, 12, 128, 2000, 41472, 1075648, 33554432, 1224440064, ...].
%e A243953 Note that Sum_{n>=0} (n+1)^(n-2) * 2^n * x^n/n! is an e.g.f. of A127670.
%t A243953 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 *)
%o A243953 (PARI) /* Explicit formula: */
%o A243953 {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)}
%o A243953 for(n=0,25,print1(a(n),", "))
%o A243953 (PARI) /* Logarithmic derivative of e.g.f. equals Catalan numbers: */
%o A243953 {A000108(n) = binomial(2*n,n)/(n+1)}
%o A243953 {a(n)=n!*polcoeff( exp(sum(m=1,n, A000108(m-1)*x^m/m)+x*O(x^n)),n)}
%o A243953 for(n=0,25,print1(a(n),", "))
%o A243953 (PARI) /* From [x^n/n!] A(x)^(n+1) = (n+1)^(n-1)*2^n */
%o A243953 {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)}
%o A243953 for(n=0,25,print1(a(n),", "))
%o A243953 (Maxima)
%o A243953 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 */
%Y A243953 Cf. A000108, A127670, A243954, A213507.
%Y A243953 Cf. A251663, A251664, A251665, A251666, A251667, A251668, A251669, A251670.
%Y A243953 Cf. A251573, A251574, A251575, A251576, A251577, A251578, A251579, A251580.
%Y A243953 Cf. A001515.
%K A243953 nonn,easy
%O A243953 0,3
%A A243953 _Paul D. Hanna_, Jun 21 2014