A079489 a(n) = (2*4^n*binomial(2*n, n) - binomial(4*n + 1, 2*n)) / (n + 1).
1, 3, 22, 211, 2306, 27230, 338444, 4362627, 57788170, 781825066, 10757497972, 150073096238, 2117778107732, 30176799215196, 433586825237912, 6274885068167651, 91383942213277530, 1338275570267001458, 19695358741104824036, 291137841642777382330, 4320734864185863437820
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..830
- Cyril Banderier and Michael Wallner, Young Tableaux with Periodic Walls: Counting with the Density Method, Séminaire Lotharingien de Combinatoire, 85B (2021), Art. 47, 12 pp.
- Feihu Liu and Guoce Xin, Simple Generating Functions for Certain Young Tableaux with Periodic Walls, arXiv:2401.14627 [math.CO], 2024.
- D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344 (Table A.1).
- Alon Regev, Enumerating triangulations by parallel diagonals, arXiv:1208.3915 [math.CO], 2012. - _N. J. A. Sloane_, Dec 25 2012
- Alon Regev, Enumerating Triangulations by Parallel Diagonals, Journal of Integer Sequences, Vol. 15 (2012), #12.8.5.
Crossrefs
Programs
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Maple
a := n -> (2*4^n*binomial(2*n, n) - binomial(4*n + 1, 2*n)) / (n + 1): seq(a(n), n = 0..20); # Peter Luschny, Aug 26 2024
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Mathematica
((Sqrt[2] Sqrt[1 + Sqrt[1 - 16 x]] - Sqrt[1 - 16 x] - 1)/(4 x) + O[x]^20)[[3]] (* Vladimir Reshetnikov, Sep 25 2016 *) CoefficientList[Series[-(1 - Sqrt[1 - 4*Sqrt[x]])*(1 - Sqrt[1 + 4*Sqrt[x]])/(4*x), {x,0,50}], x] (* G. C. Greubel, Apr 13 2017 *)
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PARI
a(n)=if(n<0,0,polcoeff(serreverse(x*(1-x^2)/(1+x^2)^2+O(x^(2*n+3))),2*n+1))
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PARI
{a(n)=polcoeff(exp(sum(m=1,n,binomial(4*m-1,2*m)*x^m/m)+x*O(x^n)),n)} \\ Paul D. Hanna, Dec 30 2010
Formula
Series reversion of x(1-x^2)/(1+x^2)^2 expanded in odd powers of x. [Previous name.]
If x = y*(1-y^2)/(1+y^2)^2 then y = x + 3*x^3 + 22*x^5 + 211*x^7 + 2306*x^9 + ...
G.f. A(x) satisfies x*A(x^2) = (C(x) - C(-x))/(C(x) + C(-x)) where C(x) is g.f. of the Catalan numbers A000108.
a(n) = ((2^(4n+2))/Gamma(1/2)) * ((Gamma(n+1/2)/(2*Gamma(n+2))) - Gamma(2n+3/2)/Gamma(2n+3)). [David Dickson (dcmd(AT)unimelb.edu.au), Nov 10 2009]
G.f.: exp( Sum_{n>=1} C(4n-1,2n)*x^n/n ). - Paul D. Hanna, Dec 30 2010
G.f.: C(sqrt(x))*C(-sqrt(x)) where C(x) is the g.f. for the Catalan numbers A000108. - David Callan, Apr 10 2012
D-finite with recurrence n*(n+1)*(2*n+1)*a(n) -2*n*(32*n^2-32*n+11)*a(n-1) +16*(4*n-5)*(4*n-3)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 29 2012
a(n) ~ (2-sqrt(2))*16^n/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 20 2013
a(n) = 2^(2*n+1)*Catalan(n) - Catalan(2*n+1) (see Regev). It follows that the 2-adic valuations of a(n) and Catalan(n) are equal. In particular, a(n) is odd iff n is of the form 2^m - 1. - Peter Bala, Aug 02 2016
G.f.: (sqrt(2) * sqrt(1 + sqrt(1-16*x)) - sqrt(1-16*x) - 1)/(4*x). - Vladimir Reshetnikov, Sep 25 2016
G.f. A(x) satisfies A(x^2) = C(x)^2*r(-x*C(x)^2), where C(x) is g.f. of the Catalan numbers A000108, and r(x) is g.f. of the large Schröder numbers A006318. - Alexander Burstein, Nov 21 2019
From Peter Bala, Sep 14 2021: (Start)
A(x) = exp( Sum_{n >= 1} (1/2)*binomial(4*n,2*n)*x^n/n ).
1 + x*A(x) is the o.g.f. of A066357.
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 3, 53, 1056, 22181, 480003, 10588508, 236720424, ...] and satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 3. See A333563. Cf. A060941. (End)
From Peter Bala, Oct 23 2024: (Start)
For integer r and positive integer s, define a sequence {u(n) : n >= 0} by setting u(n) = [x^(s*n)] A(x)^(r*n). We conjecture that the supercongruence u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and for all positive integers n and k.
Let B(x) = 1/x * series_reversion(x*A(x)). Define a sequence {v(n) : n >= 0} by setting v(n) = [x^(s*n)] B(x)^(r*n). We conjecture that the supercongruence v(n*p^k) == v(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and for all positive integers n and k. (End)
Extensions
New name by Peter Luschny, Aug 26 2024
Comments