A055392 Number of bracketings of 0#0#0#...#0 giving result 0, where 0#0 = 1, 0#1 = 1#0 = 1#1 = 0.
1, 0, 2, 1, 12, 14, 100, 180, 990, 2310, 10920, 30030, 129612, 396576, 1620168, 5318841, 21029580, 72364578, 280735884, 997356360, 3828988020, 13905563100, 53108050320, 195875639310, 746569720572, 2784329809344, 10610782107800
Offset: 1
Keywords
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..1670
- Elżbieta Liszewska and Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
Programs
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Magma
[(1/n)*(&+[(-1)^j*Binomial(2*j,j)*Binomial(2*n-j-2,n-j-1): j in [0..n-1]]): n in [1..30]]; // G. C. Greubel, Jan 12 2022
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Mathematica
CoefficientList[ Series[1/2 + 1/2(3 - 2(1 - 4x)^(1/2))^(1/2), {x, 0, 27}], x] (* Robert G. Wilson v, May 04 2004 *) -
PARI
{a(n)=if(n<1,0,polcoeff(serreverse(x - 2*x^3 - x^4 +x*O(x^n)),n))} /* Paul D. Hanna, Apr 05 2012 */ -
Sage
[(1/n)*sum( (-1)^j*binomial(2*j,j)*binomial(2*n-j-2,n-j-1) for j in (0..n-1) ) for n in (1..30)] # G. C. Greubel, Jan 12 2022
Formula
G.f.: (1/2)*(1 + sqrt(3 - 2*sqrt(1 - 4*x))).
The g.f. Z is also given by Z(x) = C(x)U(xC(x)), where U(x) = C(-x) and C is the g.f. of the Catalan numbers. - D. G. Rogers, Oct 20 2005
a(n) = Sum_{j=0..n} (1/n)*(-1)^(j-1)*C(2*n-j-1, n-j)*C(2*(j-1), j-1). - Paul Barry, Sep 09 2005, corrected by Peter Bala, Aug 19 2014
G.f. A(x) satisfies: A(x) = x + 2*A(x)^3 + A(x)^4; thus, A(x - 2*x^3 - x^4) = x. - Paul D. Hanna, Apr 05 2012
G.f. A(x) satisfies: x = Sum_{n>=1} 1/(1+A(x))^(2*n-1) * Product_{k=1..n} (1 - 1/(1+A(x))^k). - Paul D. Hanna, Apr 05 2012
Conjecture: 500*n*(n-1)*a(n) +100*(n-1)*(5*n -12)*a(n-1) +20*(25*n^2 -463*n +846)*a(n-2) +(-140161*n^2 +966559*n -1637508)*a(n-3) +2*(250*n^2 -26509*n +105084)*a(n-4) +98036*(4*n -19)*(4*n -21)*a(n-5) = 0. - R. J. Mathar, Nov 26 2012
a(n) ~ 4^(n-1) / (sqrt(3*Pi) * n^(3/2)). - Vaclav Kotesovec, Sep 03 2019
Comments