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 1-5 of 5 results.

A206948 Number of nonisomorphic graded posets with 0 and non-uniform Hasse graph of rank n, with exactly 2 elements of each rank level above 0.

Original entry on oeis.org

0, 0, 0, 2, 19, 131, 791, 4446, 23913, 124892, 638878, 3218559, 16027375, 79093773, 387540260, 1887974063, 9154751912, 44221373872, 212931964415, 1022594028515, 4900116587043, 23437066655010, 111923110602497
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.

References

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

Crossrefs

a(n) = A086405(n) - A012781(n+1).
Cf. A206947 (unique maximal element added).
Cf. A206949, A206950 (allowing one or two elements in each rank level above 0 with and without maximal element).

Programs

  • Mathematica
    LinearRecurrence[{11, -40, 55, -30, 6}, {0, 0, 0, 2, 19, 131}, 23] (* David Nacin, Feb 29 2012; a(0) added by Georg Fischer, Apr 03 2019 *)
  • Python
    def a(n, adict={0:0, 1:0, 2:0, 3:2, 4:19, 5:131}):
      if n in adict:
        return adict[n]
      adict[n]=11*a(n-1)-40*a(n-2)+55*a(n-3)-30*a(n-4)+6*a(n-5)
      return adict[n]
    for n in range(0,40):
      print(a(n))

Formula

a(n) = 11*a(n-1) - 40*a(n-2) + 55*a(n-3) - 30*a(n-4) + 6*a(n-5), a(0)=0, a(1)=0, a(2)=0, a(3)=2, a(4)=19, a(5)=131.
G.f.: (x^3*(2 - 3*x + 2*x^2))/((1 - 6*x + 6*x^2)*(1 - 5*x + 4*x^2 - x^3)).

A012772 Take every 5th term of Padovan sequence A000931, beginning with the sixth term.

Original entry on oeis.org

1, 3, 12, 49, 200, 816, 3329, 13581, 55405, 226030, 922111, 3761840, 15346786, 62608681, 255418101, 1042002567, 4250949112, 17342153393, 70748973084, 288627200960, 1177482265857, 4803651498529, 19596955630177
Offset: 0

Views

Author

Keywords

Crossrefs

Cf. A012781 (partial sums).

Programs

  • Magma
    I:=[1, 3, 12]; [n le 3 select I[n] else 5*Self(n-1)-4*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Apr 18 2012
  • Mathematica
    CoefficientList[Series[(1-x)^2/(1-5*x+4*x^2-x^3),{x,0,33}],x] (* Vincenzo Librandi, Apr 18 2012 *)
    LinearRecurrence[{5,-4,1},{1,3,12},30] (* Harvey P. Dale, Aug 15 2024 *)

Formula

a(n+3) = 5*a(n+2) - 4*a(n+1) + a(n).
G.f.: (1-x)^2/(1-5*x+4*x^2-x^3). - Colin Barker, Feb 02 2012

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).

A084084 Length of lists created by n substitutions k -> Range[0,1+Mod[k+1,3]] starting with {0}.

Original entry on oeis.org

1, 3, 9, 28, 86, 265, 816, 2513, 7739, 23833, 73396, 226030, 696081, 2143648, 6601569, 20330163, 62608681, 192809420, 593775046, 1828587033, 5631308624, 17342153393, 53406819691, 164471408185, 506505428836, 1559831901918
Offset: 0

Views

Author

Wouter Meeussen, May 11 2003

Keywords

Comments

Transformation invert T109 gave a match with A078039; T100 binomial gave a match with A012781; equivalent to replacements 0 -> {0,1,2}; 1 -> {0,1,2,3}; 2 -> {0,1}, 3 -> {0,1,2} operating n times with {0}.

Examples

			{0}, {0,1,2}, {0,1,2,0,1,2,3,0,1}, {0,1,2,0,1,2,3,0,1,0,1,2,0,1,2,3,0,1,0,1,2,0,1,2,0,1,2,3} have lengths 1, 3, 9, 28.
G.f. = 1 + 3*x + 9*x^2 + 28*x^3 + 86*x^4 + 265*x^5 + 816*x^6 + ...
		

Crossrefs

Programs

  • Magma
    [n le 3 select 3^(n-1) else 2*Self(n-1) +3*Self(n-2) +Self(n-3): n in [1..41]]; // G. C. Greubel, Oct 16 2022
    
  • Mathematica
    Length/@Flatten/@NestList[ # /. k_Integer:>Range[0, 1+Mod[k+1, 3]]&, {0}, 8]
    LinearRecurrence[{2,3,1}, {1,3,9}, 41] (* G. C. Greubel, Oct 16 2022 *)
  • SageMath
    def A084084_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( (1+x)/(1-2*x-3*x^2-x^3) ).list()
    A084084_list(40) # G. C. Greubel, Oct 16 2022

Formula

G.f.: (1+x)/(1-2*x-3*x^2-x^3).
a(n) = A000931(4*n + 6). - Michael Somos, Sep 18 2012
Showing 1-5 of 5 results.