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.

Showing 1-4 of 4 results.

A206950 Number of nonisomorphic graded posets with 0 and non-uniform Hasse graph of rank n, with no 3-element antichain.

Original entry on oeis.org

0, 0, 0, 3, 33, 259, 1762, 11093, 66592, 387264, 2202053, 12314587, 67995221, 371697914, 2015659707, 10859379024, 58190011080, 310409500291, 1649579166385, 8738000970251, 46158910515154, 243260704208613, 1279386591175904, 6716811592446952, 35209193397256085
Offset: 0

Views

Author

David Nacin, Feb 13 2012

Keywords

Comments

We do not assume all maximal elements have maximal rank and thus use graded poset to mean: For every element x, all maximal chains among those with x as greatest element have the same finite length. Here, the term uniform used in the sense of Retakh, Serconek and Wilson.

Crossrefs

Cf. A206949 (unique maximal element added.)
Cf. A206947, A206948 (requiring exactly two elements in each rank level above 0 with and without maximal element.)

Programs

  • Mathematica
    Join[{0},LinearRecurrence[{13, -59, 115, -109, 51, -9}, {0, 0, 3, 33, 259, 1762}, 40]]
  • Python
    def a(n,adict={0:0,1:0,2:0,3:3,4:33,5:259,6:1762}):
        if n in adict:
            return adict[n]
        adict[n]=13*a(n-1)-59*a(n-2)+115*a(n-3)-109*a(n-4)+51*a(n-5)-9*a(n-6)
        return adict[n]

Formula

a(n) = 13*a(n-1) - 59*a(n-2) + 115*a(n-3) - 109*a(n-4) + 51*a(n-5) - 9*a(n-6), a(1)=0, a(2)=0, a(3)=3, a(4)=33, a(5)=259, a(6)=1762.
G.f.: (3*x^3-6*x^4+7*x^5-3*x^6)/((-1+7*x-10*x^2+3*x^3)*(-1+6*x-7*x^2+3*x^3)).

A206902 Number of nonisomorphic graded posets with 0 and uniform Hasse diagram of rank n with no 3-element antichain.

Original entry on oeis.org

1, 2, 8, 36, 166, 768, 3554, 16446, 76102, 352152, 1629536, 7540458, 34892452, 161460114, 747134894, 3457265922, 15998031616, 74028732924, 342557973998, 1585140808368, 7335025230994, 33941839649382, 157061283704438, 726779900373936, 3363075935260696
Offset: 0

Views

Author

David Nacin, Feb 13 2012

Keywords

Comments

We do not assume all maximal elements have maximal rank and thus use graded poset to mean: For every element x, all maximal chains among those with x as greatest element have the same finite length.
Uniform (in the definition) used in the sense of Retakh, Serconek and Wilson (see paper in Links lines). - David Nacin, Mar 01 2012

Crossrefs

Cf. A025192 (adding a unique maximal element).
Cf. A124292, A206901 (dropping uniformity with and without maximal element).

Programs

  • GAP
    a:=[2,8,36];; for n in [4..30] do a[n]:=6*a[n-1]-7*a[n-2]+3*a[n-3]; od; Concatenation([1], a); # G. C. Greubel, May 21 2019
  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1-4*x +3*x^2-x^3)/(1-6*x+7*x^2-3*x^3) )); // G. C. Greubel, May 21 2019
    
  • Mathematica
    LinearRecurrence[{6,-7,3}, {1,2,8,36}, 30] (* Vincenzo Librandi, Feb 27 2012 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-4*x+3*x^2-x^3)/(1-6*x+7*x^2-3*x^3)) \\ G. C. Greubel, May 21 2019
    
  • Python
    def a(n, adict={1:2,2:8,3:36}):
        if n in adict:
            return adict[n]
        adict[n]=6*a(n-1)-7*a(n-2)+3*a(n-3)
        return adict[n]
    
  • Sage
    ((1-4*x+3*x^2-x^3)/(1-6*x+7*x^2-3*x^3)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 21 2019
    

Formula

a(n) = 6*a(n-1) - 7*a(n-2) + 3*a(n-3), a(1)=2, a(2)=8, a(3)=36.
G.f.: (1 -4*x +3*x^2 -x^3)/(1 -6*x +7*x^2 -3*x^3).

A208736 Number of nonisomorphic graded posets with 0 and 1 and non-uniform Hasse graph of rank n, with exactly 2 elements of each rank level between 0 and 1.

Original entry on oeis.org

0, 0, 0, 1, 5, 22, 91, 361, 1392, 5265, 19653, 72694, 267179, 977593, 3565600, 12975457, 47142021, 171075606, 620303547, 2247803785, 8141857808, 29481675889, 106728951109, 386314552438, 1398132674955, 5059626441177, 18308871648576, 66249898660801
Offset: 0

Views

Author

David Nacin, Mar 01 2012

Keywords

Comments

Uniform used in the sense of Retakh, Serconek and Wilson. We use Stanley's definition of graded poset: all maximal chains have the same length n (which also implies all maximal elements have maximal rank.)

References

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

Crossrefs

Programs

  • Mathematica
    Join[{0, 0}, LinearRecurrence[{8, -21, 20, -5}, {0, 1, 5, 22}, 40]]
  • Python
    def a(n, d={0:0,1:0,2:0,3:1,4:5,5:22}):
        if n in d:
            return d[n]
        d[n]=8*a(n-1) - 21*a(n-2) + 20*a(n-3) - 5*a(n-4)
        return d[n]

Formula

a(n) = 8*a(n-1) - 21*a(n-2) + 20*a(n-3) - 5*a(n-4), a(2) = 0, a(3) = 1, a(4) = 5, a(5) = 22.
G.f.: (x^3 - 3*x^4 + 3*x^5)/(1 - 8*x + 21*x^2 - 20*x^3 + 5*x^4); (x^3 * (1 - 3*x + 3*x^2))/((1 - 3*x + x^2)*(1 - 5*x + 5*x^2)) .
a(n) = A081567(n-2) - A001519(n-1).

A208737 Number of nonisomorphic graded posets with 0 and 1 and non-uniform Hasse graph of rank n, with no 3-element antichain.

Original entry on oeis.org

0, 0, 0, 1, 7, 37, 175, 778, 3325, 13837, 56524, 227866, 909832, 3607294, 14227447, 55894252, 218937532, 855650749, 3338323915, 13007422705, 50631143323, 196928737582, 765495534433, 2974251390529, 11552064922624, 44856304154086
Offset: 0

Views

Author

David Nacin, Mar 01 2012

Keywords

Comments

Uniform used in the sense of Retakh, Serconek and Wilson. We use Stanley's definition of graded poset: all maximal chains have the same length n (which also implies all maximal elements have maximal rank.)

References

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

Crossrefs

Programs

  • Mathematica
    Join[{0}, LinearRecurrence[{10, -36, 57, -39, 9}, {0, 0, 1, 7, 37}, 40]]
  • Python
    def a(n, d={0:0,1:0,2:0,3:1,4:7,5:37}):
        if n in d:
            return d[n]
        d[n]=10*a(n-1) - 36*a(n-2) + 57*a(n-3) - 39*a(n-4) + 9*a(n-5)
        return d[n]

Formula

a(n) = 10*a(n-1) - 36*a(n-2) + 57*a(n-3) - 39*a(n-4) + 9*a(n-5), a(1) = 0, a(2) = 0, a(3) = 1, a(4) = 7, a(5) = 37.
G.f: (x^3 - 3*x^4 + 3*x^5)/(1 - 10*x + 36*x^2 - 57*x^3 + 39*x^4 - 9*x^5); (x^3*(1 - 3*x + 3*x^2)) / ((1 - x) (1 - 3*x) (1 - 6*x + 9*x^2 - 3*x^3)).
a(n) = A124292(n) - A124302(n).
Showing 1-4 of 4 results.