cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A124292 Number of free generators of degree n of symmetric polynomials in 4 noncommuting variables.

Original entry on oeis.org

1, 1, 2, 6, 21, 78, 297, 1143, 4419, 17118, 66366, 257391, 998406, 3873015, 15024609, 58285737, 226111986, 877174110, 3402893997, 13201132950, 51212274057, 198672129783, 770725711035, 2989941920334, 11599136512038, 44997518922327, 174562710686622
Offset: 1

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Author

Mike Zabrocki, Oct 24 2006

Keywords

Comments

Also the number of non-splitable set partitions (see Bergeron et al. reference) of length <= 4.
Also the number of nonisomorphic graded posets with 0 and 1 of rank n with no 3-element antichain. - Richard Stanley, Nov 30 2011
Also the number of nonisomorphic graded posets with 0 of rank n+1 with no 3-element antichain. (Using Stanley's definition of graded, that all maximal chains have length n.) - David Nacin, Feb 26 2012

References

  • R. Stanley, Enumerative combinatorics. Vol. 1, Cambridge University Press, Cambridge, 1997, pages 96-100.

Crossrefs

Programs

  • Maple
    a:= n-> (Matrix([[2,1,1]]). Matrix(3, (i,j)-> if i=j-1 then 1 elif j=1 then [6,-9,3][i] else 0 fi)^(n-1))[1,3]: seq(a(n), n=1..26); # Alois P. Heinz, Sep 05 2008
  • Mathematica
    m = {{2, 1, 1}, {1, 3, 0}, {1, 1, 1}}; Table[MatrixPower[m, n][[1,1]], {n, 0, 40}] (* David Nacin, Feb 11 2012 *)
    LinearRecurrence[{6, -9, 3}, {1, 1, 2}, 70] (* Vladimir Joseph Stephan Orlovsky, Feb 26 2012 *)
  • Python
    def a(n, adict={1:1, 2:1, 3:2}):
        if n in adict:
            return adict[n]
        adict[n]=6*a(n-1)-9*a(n-2)+3*a(n-3)
        return adict[n] # David Nacin, Mar 04 2012

Formula

O.g.f.: (1 - 5*q + 5*q^2)/(1 - 6*q + 9*q^2 - 3*q^3) = 1 - 1/(Sum_{k=0..4} q^k/(Product_{i=1..k} (1-i*q))).
a(n) = 6*a(n-1) - 9*a(n-2) + 3*a(n-3). - David Nacin, Feb 11 2012
a(n) = A055105(n,1) + A055105(n,2) + A055105(n,3) + A055105(n,4) = A055106(n,1) + A055106(n,2) + A055106(n,3).
Given matrix A = [[2,1,1],[1,3,0],[1,1,1]], a(n+1) = top left entry in A^n. - David Nacin, Feb 11 2012
a(n) = (1/3)*(x^(n-2) + y^(n-2) + z^(n-2)) for x = (2*cos(Pi/18))^2, y = (2*cos(5*Pi/18))^2, and z = (2*cos(7*Pi/18))^2. - Greg Dresden, Jan 28 2023