A100320 A Catalan transform of (1 + 2*x)/(1 - 2*x).
1, 4, 12, 40, 140, 504, 1848, 6864, 25740, 97240, 369512, 1410864, 5408312, 20801200, 80233200, 310235040, 1202160780, 4667212440, 18150270600, 70690527600, 275693057640, 1076515748880, 4208197927440, 16466861455200, 64495207366200, 252821212875504, 991837065896208
Offset: 0
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
- Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
- Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
- Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Programs
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Haskell
a100320 n = a124927 (2 * n) n -- Reinhard Zumkeller, Mar 04 2012
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Magma
[4*Binomial(2*n-1, n) - 3*0^n: n in [0..40]]; // G. C. Greubel, Feb 01 2023
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Mathematica
a[0]= 1; a[n_]:= 2 Binomial[2 n, n]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jul 31 2018 *) -
SageMath
def A100320(n): return 4*binomial(2*n-1, n) - 3*0^n [A100320(n) for n in range(41)] # G. C. Greubel, Feb 01 2023
Formula
G.f.: (1 + 2*x*c(x))/(1 - 2*x*c(x)), where c(x) is the g.f. of A000108.
a(n) = 4*binomial(2*n-1, n) - 3*0^n.
a(n) = binomial(2*n, n)*(4*2^(n-1) - 0^n)/2^n.
a(n) = Sum_{j=0..n} Sum_{k=0..n} C(2*n, n-k)*((2*k + 1)/(n + k + 1))*C(k, j)*(-1)^(j-k)*(4*2^(j-1) - 0^j).
a(n) = A028329(n), n > 0. - R. J. Mathar, Sep 02 2008
a(n) = T(2*n,n), where T(n,k) = A132046(n,k). - Paul Barry, Sep 19 2008
a(n) = A095660(2*n,n) for n > 0. - Reinhard Zumkeller, Apr 08 2012
G.f.: G(0) - 1, where G(k) = 1 + 1/(1 - 2*x*(2*k + 1)/(2*x*(2*k + 1) + (k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
a(n) = [x^n] (1 + 2*x)/(1 - x)^(n+1). - Ilya Gutkovskiy, Oct 12 2017
a(n) = 2*(2*n-1)*a(n-1)/n. - G. C. Greubel, Feb 01 2023
E.g.f.: 2*exp(2*x)*BesselI(0, 2*x) - 1. - Stefano Spezia, May 11 2024
Extensions
Incorrect connection with A046055 deleted by N. J. A. Sloane, Jul 08 2009
Comments