A120589 Self-convolution of A120588, such that a(n) = 3*A120588(n) for n >= 2.
1, 2, 3, 6, 15, 42, 126, 396, 1287, 4290, 14586, 50388, 176358, 624036, 2228700, 8023320, 29084535, 106073010, 388934370, 1432916100, 5301789570, 19692361260, 73398801060, 274447690920, 1029178840950, 3869712441972, 14585839204356
Offset: 0
Examples
A(x) = 1 + 2*x + 3*x^2 + 6*x^3 + 15*x^4 + 42*x^5 + 126*x^6 + 396*x^7 + ... A(x)^(1/2) = 1 + x + x^2 + 2*x^3 + 5*x^4 + 14*x^5 + 42*x^6 + 132*x^7 + ...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
- Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
- S. Kitaev, J. Remmel and M. Tiefenbruck, Marked mesh patterns in 132-avoiding permutations I, arXiv:1201.6243 [math.CO], 2012. - From _N. J. A. Sloane_, May 09 2012
- Sergey Kitaev, Jeffrey Remmel, and Mark Tiefenbruck, Quadrant Marked Mesh Patterns in 132-Avoiding Permutations II, arXiv:1302.2274 [math.CO], 2013.
- Sergey Kitaev, Jeffrey Remmel, Mark Tiefenbruck, Quadrant Marked Mesh Patterns in 132-Avoiding Permutations II, Electronic Journal of Combinatorial Number Theory, 15 (2015), #A16.
Programs
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Magma
m:=30; R
:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (5-2*x-3*Sqrt(1-4*x))/2 )); // G. C. Greubel, Feb 18 2019 -
Maple
A120589List := proc(m) local A, P, n; A := [1,2,3]; P := [3]; for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), P[-1]]); A := [op(A), P[-1]] od; A end: A120589List(26); # Peter Luschny, Mar 26 2022
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Mathematica
Join[{1,2,3}, Table[3*(2*n)!/n!/(n+1)!, {n,2,40}]] CoefficientList[Series[(5-2x -3Sqrt[1-4x])/2, {x, 0, 30}], x] (* G. C. Greubel, Feb 18 2019 *)
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PARI
{a(n)=local(A=1+x+x^2+x*O(x^n));for(i=0,n,A=A-3*A+2+x+A^2);polcoeff(A^2,n)}
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PARI
my(x='x+O('x^30)); Vec((5-2*x-3*sqrt(1-4*x))/2) \\ G. C. Greubel, Feb 18 2019
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Sage
((5-2*x-3*sqrt(1-4*x))/2).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 18 2019
Formula
G.f.: (5 - 2*x - 3*sqrt(1-4*x))/2. - G. C. Greubel, Feb 18 2019
a(n) ~ 3 * 2^(2*n-2) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Aug 20 2025
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