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.

A003121 Strict sense ballot numbers: n candidates, k-th candidate gets k votes.

Original entry on oeis.org

1, 1, 1, 2, 12, 286, 33592, 23178480, 108995910720, 3973186258569120, 1257987096462161167200, 3830793890438041335187545600, 123051391839834932169117010215648000, 45367448380314462649742951646437285434233600, 207515126854334868747300581954534054343817468395494400
Offset: 0

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Author

Keywords

Comments

Also, number of even minus number of odd extensions of truncated (n-1) X n grid diagram.
Also, a(n) is the degree of the spinor variety, the complex projective variety SO(2n+1)/U(n). See Hiller's paper. - Burt Totaro (b.totaro(AT)dpmms.cam.ac.uk), Oct 29 2002
Also, number of ways of placing 1, ..., n*(n+1)/2 in a triangular array such that both rows and columns are increasing, i.e., the number of shifted standard Young tableaux of shape (n, n-1, ..., 1).
E.g., a(3) = 2 as we can write:
1 1
2 3 or 2 4
4 5 6 3 5 6
(or transpose these to have shifted tableaux). - Jon Perry, Jun 13 2003, edited by Joel B. Lewis, Aug 27 2011
Also, the number of symbolic sequences on the n symbols 0,1, ..., n-1. A symbolic sequence is a sequence that has n occurrences of 0, n-1 occurrences of 1, ..., one occurrence of n-1 and satisfies the condition that between any two consecutive occurrences of the symbol i it has exactly one occurrence of the symbol i+1. For example, the two symbolic sequences on 3 symbols are 012010 and 010210. The Shapiro-Shapiro paper shows how such sequences arise in the study of the arrangement of the real roots of a polynomial and its derivatives. There is a natural bijection between symbolic sequences and the triangular arrays described above. - Peter Bala, Jul 18 2007
a(n) also appears to be the number of chains from w_0 down to 1 in a certain suborder of the strong Bruhat order on S_n: we let v cover (ij)*v only if i,j straddle the leftmost descent in v. For n=3 and drawing that descent with a |, this order is 3|21 > 23|1 > 13|2 & 2|13 > 123, with two maximal chains. - Allen Knutson (allenk(AT)math.cornell.edu), Oct 13 2008
Number of ways to arrange the numbers 1,2,...,n(n+1)/2 in a triangle so that the rows interlace; e.g., one of the 12 triangles counted by a(4) is
6
4 8
2 5 9
1 3 7 10
- Clark Kimberling, Mar 25 2012
Also, the number of maps from n X n pipe dreams (rc-graphs) to words of adjacent transpositions in S_n that send a crossing of pipes x and y in square (i,j) to the transposition (i+j-1,i+j) swapping x and y. Taking the pictorial image of a permutation as a wiring diagram, these are maps from pipe dreams to wiring diagrams that send crossings of pipes to crossings of similarly labeled wires. - Cameron Marcott, Nov 26 2012
Number of words of length T(n)=n*(n+1)/2 with n 1's, (n-1) 2's, ..., and 1 n such that counting the numbers from left to right we always have |1| > |2| > |3| > ... > |n|. The 12 words for n=4 are 1111222334, 1111223234, 1112122334, 1112123234, 1112212334, 1112213234, 1112231234, 1121122334, 1121123234, 1121212334, 1121213234 and 1121231234. - Jon Perry, Jan 27 2013
Regarding the comment dated Mar 25 2012, it is assumed that each row is in increasing order, as in the example dated Jul 12 2012. How many row-interlacing triangles are there without that restriction? - Clark Kimberling, Dec 02 2014
Number of maximal chains of length binomial(n+1,2) in the Tamari lattice T_{n+1}. For n=2 there is 1 maximal chain of length 3 in the Tamari lattice T_3. - Alois P. Heinz, Dec 04 2015
The normalized volume of the subpolytope of the Birkhoff polytope obtained by taking the convex hull of all n X n permutation matrices corresponding to permutations that avoid the patterns 132 and 312. - Robert Davis, Dec 04 2016
From Emily Gunawan, Feb 26 2022: (Start)
The number of maximal chains in the lattice of permutations which avoid both the patterns 132 and 312, as a sublattice of the weak order on the symmetric group. For example, there are exactly 12 maximal chains in the sublattice for the weak order on the symmetric group on 5 elements.
The number of words in the commutation class of the c-sorting word of the longest permutation w_0 in the symmetric group for the Coxeter element c = s_1 s_2 s_3 s_4 s_5 ... . For example, the c-sorting word of w_0 for s_1 s_2 s_3 s_4 is the reduced word s_1 s_2 s_3 s_4 | s_1 s_2 s_3 | s_1 s_2 | s_1, and there are exactly 12 words in its commutation class.
The number of maximal chains in the lattice of c-singletons for the symmetric group, for the Coxeter element c = s_1 s_2 s_3 s_4 s_5 ... . For example, there are exactly 12 maximal chains in the lattice of c-singletons for c = s_1 s_2 s_3 s_4. (End)

Examples

			From _R. H. Hardin_, Jul 06 2012: (Start)
The a(4) = 12 ways to fill a triangle with the numbers 0 through 9:
.
     5         6         6         5
    3 7       3 7       2 7       2 7
   1 4 8     1 4 8     1 4 8     1 4 8
  0 2 6 9   0 2 5 9   0 3 5 9   0 3 6 9
.
     5         3         3         4
    3 6       2 6       2 7       3 7
   1 4 8     1 5 8     1 5 8     1 5 8
  0 2 7 9   0 4 7 9   0 4 6 9   0 2 6 9
.
     4         4         5         4
    2 6       2 7       2 6       3 6
   1 5 8     1 5 8     1 4 8     1 5 8
  0 3 7 9   0 3 6 9   0 3 7 9   0 2 7 9
.
(End)
		

References

  • G. Kreweras, Sur un problème de scrutin à plus de deux candidats, Publications de l'Institut de Statistique de l'Université de Paris, 26 (1981), 69-87.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

A213457 is also closely related.

Programs

  • Maple
    f:= n-> ((n*n+n)/2)!*mul((i-1)!/(2*i-1)!, i=1..n); seq(f(n), n=0..20);
  • Mathematica
    f[n_] := (n (n + 1)/2)! Product[(i - 1)!/(2 i - 1)!, {i, n}]; Array[f, 14, 0] (* Robert G. Wilson v, May 14 2013 *)
  • PARI
    a(n)=((n*n+n)/2)!*prod(i=1,n,(i-1)!/(2*i-1)!)

Formula

a(n) = binomial(n+1, 2)!*(1!*2!*...*(n-1)!)/(1!*3!*...*(2n-1)!).
a(n) ~ sqrt(Pi) * exp(n^2/4 + n/2 + 7/24) * n^(n^2/2 + n/2 + 23/24) / (A^(1/2) * 2^(3*n^2/2 + n + 5/24)), where A = 1.2824271291... is the Glaisher-Kinkelin constant (see A074962). - Vaclav Kotesovec, Nov 13 2014

Extensions

More terms from Michael Somos
Additional references from Frank Ruskey
Formula corrected by Eric Rowland, Jun 18 2010

A217800 Number of alternating permutations on 2n+1 letters that avoid a certain pattern of length 4 (see Lewis, 2012, Appendix, for precise definition).

Original entry on oeis.org

1, 2, 12, 110, 1274, 17136, 255816, 4124406, 70549050, 1264752060, 23555382240, 452806924752, 8939481277552, 180551099694400, 3719061442253520, 77933728043586630, 1658001861319441050, 35749633305661575300, 780123576993991461000, 17208112644166765652100
Offset: 0

Views

Author

N. J. A. Sloane, Oct 12 2012

Keywords

Comments

1 together with A007724. - Omar E. Pol, Aug 22 2016

Crossrefs

Programs

  • Magma
    [Factorial(3*n+3)/((4*(n+1)^2-1)*Factorial((n+1))^2*Factorial(n+ 2)): n in [0..20]]; // Vincenzo Librandi, Aug 30 2014
  • Maple
    a := n -> (-1)^n*hypergeom([-2-2*n, -2*n, -2*n-1], [2, 3], 1):
    seq(round(evalf(a(n), 32)), n=0..20); # Peter Luschny, Aug 29 2014
  • Mathematica
    Table[(3 n + 3)!/((4 (n + 1)^2 - 1) ((n + 1)!)^2 (n + 2)!), {n, 0, 20}] (* Vincenzo Librandi, Aug 30 2014 *)
    Table[(-1)^n HypergeometricPFQ[{-2 - 2 n, -2 n, -2 n - 1}, {2, 3}, 1], {n, 0, 20}] (* Michael De Vlieger, Aug 22 2016 *)
  • PARI
    a(n) = (3*n+3)!/((4*(n+1)^2-1)*((n+1)!)^2*(n+2)!); \\ Michel Marcus, Aug 10 2014
    

Formula

From Karol A. Penson, Aug 10 2014: (Start)
O.g.f.(in Maple notation): hypergeom([1/2, 1, 4/3, 5/3], [2, 5/2, 3], 27*z);a(n) ~ (1/93312)*sqrt(3)*27^n*(314928*n^4-1644624*n^3+5545260*n^2 -15387660*n+38310503)/(Pi*n^8), for n -> infinity.
Representation of a(n) as the n-th power moment of a positive function on the segment [0,27]:
a(n) = int(x^n*W(x),x=0..27),n=0,1,2..., where
W(x) = 1/(Pi*sqrt(x))+sqrt(x)/Pi-(9/20)*sqrt(3)*2^(1/3)* hypergeom([-2/3, -1/6, 1/3], [2/3, 11/6], (1/27)*x)*x^(1/3)/ (sqrt(Pi)*Gamma(5/6)*Gamma(2/3))-(27/56)*2^(2/3)*Gamma(5/6) *Gamma(2/3)*hypergeom([-1/3, 1/6, 2/3], [4/3, 13/6], (1/27)*x)* x^(2/3)/Pi^(5/2).
W(x) for x->0 has the singularity 1/sqrt(x), W(27)=0.
This is the solution of the Hausdorff moment problem and is unique.
a(n) = (1/2)*(n+3)!/((4*(n+1)^2-1)*(n+1)!)*A005789(n), where A005789(n) are the three-dimensional Catalan numbers (see the Gorska and Penson link).(End)
a(n) = A006480(n+1)/((2+n)*(1+2*n)*(3+2*n)). - Peter Luschny, Aug 15 2014
a(n) = (-1)^n*hypergeom([-2-2*n,-2*n,-2*n-1],[2,3],1). - Peter Luschny, Aug 29 2014
(2*n+3)*(n+2)*(n+1)*a(n) -3*(3*n+2)*(2*n-1)*(3*n+1)*a(n-1)=0. - R. J. Mathar, Jun 14 2016
a(n) ~ 3^(3*n + 7/2) / (8*Pi*n^4). - Vaclav Kotesovec, Jun 09 2019

Extensions

More terms from Alois P. Heinz, Aug 22 2016
Merged with A241958 by R. J. Mathar, Jul 07 2023

A368708 a(n) = hypergeom([-1 - n, -n, 1 - n], [2, 3], -2).

Original entry on oeis.org

1, 1, 3, 13, 69, 417, 2763, 19609, 146793, 1146833, 9278595, 77292261, 659973933, 5756169681, 51137399979, 461691066417, 4228199347281, 39216540096993, 367890444302787, 3486697883136957, 33353178454762389, 321754445379041601, 3127955713554766923, 30624486778208481993, 301790556354721667769, 2991957347531210976817
Offset: 0

Views

Author

Joerg Arndt, Jan 04 2024

Keywords

Crossrefs

Programs

  • Maple
    seq(simplify( hypergeom([-1 - n, -n, 1 - n], [2, 3], -2) ), n = 0..25); # Peter Bala, Sep 09 2024
  • Mathematica
    Table[HypergeometricPFQ[{-1-n, -n, 1-n}, {2, 3}, -2], {n, 0, 30}] (* Vaclav Kotesovec, Jan 04 2024 *)
    a[0] := 1; a[n_] := 2^(n + 1)/(n*(n + 1)^2)*Sum[(1/2)^k*Binomial[n + 1, k - 1]*Binomial[n + 1, k]*Binomial[n + 1, k + 1], {k, 1, n}]; Table[a[n], {n, 0, 25}] (* Detlef Meya, May 28 2024 *)
  • Python
    def A368708(n):
        if n == 0: return 1
        return sum(2**k * v for k, v in enumerate(A359363Row(n))) // 2
    print([A368708(n) for n in range(26)]) # Peter Luschny, Jan 04 2024
  • SageMath
    def A368708(n): return PolyA359363(n, 2) // 2 if n > 0 else 1
    print([A368708(n) for n in range(23)])  # Peter Luschny, Jan 04 2024
    

Formula

a(n) = (1/2)*B(n, 2) where B(n, x) are the Baxter polynomials with coefficients A359363, for n > 0. - Peter Luschny, Jan 04 2024
a(n) ~ 3^(n + 7/6) * (2^(2/3) + 2^(1/3) + 1)^(n + 5/3) / (2^(4/3) * Pi * n^4). - Vaclav Kotesovec, Jan 04 2024
a(0) = 1, a(n) = 2^(n + 1)/(n*(n + 1)^2)*Sum_{k=1..n} (1/2)^k*binomial(n + 1, k - 1)*binomial(n + 1, k)*binomial(n + 1, k + 1). - Detlef Meya, May 29 2024
From Peter Bala, Sep 09 2024: (Start)
a(n+1) = Sum_{k = 0..n} A056939(n, k)*2^k.
P-recursive: (n+1)*(n+3)*(n+2)*(3*n-2)*a(n) = 3*(9*n^3+3*n^2-4*n+4)*(n+1)*a(n-1) + 3*(n-2)*(3*n-1)*(9*n^2-3*n-10)*a(n-2) + 27*(3*n+1)*(n-3)*(n-2)^2*a(n-3) = 0 with a(0) = 1, a(1) = 1 and a(2) = 3. (End)

A241958 Duplicate of A217800.

Original entry on oeis.org

1, 2, 12, 110, 1274, 17136, 255816, 4124406, 70549050, 1264752060, 23555382240, 452806924752, 8939481277552, 180551099694400, 3719061442253520, 77933728043586630, 1658001861319441050, 35749633305661575300, 780123576993991461000, 17208112644166765652100, 383292388823513983713900
Offset: 0

Views

Author

Karol A. Penson, Aug 10 2014

Keywords

Comments

This is a duplicate of A217800 or of A007724. - Alois P. Heinz, Aug 22 2016

Programs

  • Magma
    [Factorial(3*n+3)/((4*(n+1)^2-1)*Factorial((n+1))^2*Factorial(n+ 2)): n in [0..20]]; // Vincenzo Librandi, Aug 30 2014
  • Maple
    a := n -> (-1)^n*hypergeom([-2-2*n,-2*n,-2*n-1],[2, 3],1):
    seq(round(evalf(a(n),32)),n=0..20); # Peter Luschny, Aug 29 2014
  • Mathematica
    Table[(3 n + 3)!/((4 (n + 1)^2 - 1) ((n + 1)!)^2 (n + 2)!), {n, 0, 20}] (* Vincenzo Librandi, Aug 30 2014 *)
    Table[(-1)^n HypergeometricPFQ[{-2 - 2 n, -2 n, -2 n - 1}, {2, 3},
    1], {n, 0, 20}] (* Michael De Vlieger, Aug 22 2016 *)
  • PARI
    a(n) = (3*n+3)!/((4*(n+1)^2-1)*((n+1)!)^2*(n+2)!); \\ Michel Marcus, Aug 10 2014
    

Formula

O.g.f.(in Maple notation): hypergeom([1/2, 1, 4/3, 5/3], [2, 5/2, 3], 27*z);
a(n) ~ (1/93312)*sqrt(3)*27^n*(314928*n^4-1644624*n^3+5545260*n^2 -15387660*n+38310503)/(Pi*n^8), for n -> infinity.
Representation of a(n) as the n-th power moment of a positive function on the segment [0,27]:
a(n) = int(x^n*W(x),x=0..27),n=0,1,2..., where
W(x) = 1/(Pi*sqrt(x))+sqrt(x)/Pi-(9/20)*sqrt(3)*2^(1/3)* hypergeom([-2/3, -1/6, 1/3], [2/3, 11/6], (1/27)*x)*x^(1/3)/ (sqrt(Pi)*Gamma(5/6)*Gamma(2/3))-(27/56)*2^(2/3)*Gamma(5/6) *Gamma(2/3)*hypergeom([-1/3, 1/6, 2/3], [4/3, 13/6], (1/27)*x)* x^(2/3)/Pi^(5/2).
W(x) for x->0 has the singularity 1/sqrt(x), W(27)=0.
This is the solution of the Hausdorff moment problem and is unique.
a(n) = (1/2)*(n+3)!/((4*(n+1)^2-1)*(n+1)!)*A005789(n), where A005789(n) are the three-dimensional Catalan numbers (see the Gorska and Penson link).
a(n) = A006480(n+1)/((2+n)*(1+2*n)*(3+2*n)). - Peter Luschny, Aug 15 2014
a(n) = (-1)^n*hypergeom([-2-2*n,-2*n,-2*n-1],[2,3],1). - Peter Luschny, Aug 29 2014
(2*n+3)*(n+2)*(n+1)*a(n) -3*(3*n+2)*(2*n-1)*(3*n+1)*a(n-1)=0. - R. J. Mathar, Jun 14 2016

A368709 a(n) = hypergeom([-1 - n, -n, 1 - n], [2, 3], +2).

Original entry on oeis.org

1, 1, -1, -3, 13, 17, -241, 121, 5081, -13327, -106705, 609589, 1850661, -23392159, -6796193, 811545073, -1688514383, -25224774367, 123764707231, 650087614573, -6385330335427, -9591188592399, 279171512779759, -318526766092183, -10665705513959287, 40625771132796817
Offset: 0

Views

Author

Joerg Arndt, Jan 04 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Table[HypergeometricPFQ[{-1-n, -n, 1-n}, {2, 3}, 2], {n, 0, 30}] (* Vaclav Kotesovec, Jan 04 2024 *)
    a[0] := 1; a[n_] := (-1)^n*2^(n + 1)/(n*(n + 1)^2)*Sum[(-1/2)^k*Binomial[n + 1, k - 1]*Binomial[n + 1, k]*Binomial[n + 1, k + 1], {k , 1, n}]; Table[a[n], {n, 0, 25}] (* Detlef Meya, May 29 2024 *)
  • Python
    def A368709(n):
        if n == 0: return 1
        return sum((-2)**k * v for k, v in enumerate(A359363Row(n))) // (-2)
    print([A368709(n) for n in range(26)]) # Peter Luschny, Jan 04 2024
  • SageMath
    def A368709(n): return PolyA359363(n, -2) // (-2) if n > 0 else 1
    print([A368709(n) for n in range(0, 26)])  # Peter Luschny, Jan 04 2024
    

Formula

a(n) = (-1/2)*B(n, -2) where B(n, x) are the Baxter polynomials with coefficients A359363, for n > 0. - Peter Luschny, Jan 04 2024
a(0) = 1, a(n) = (-1)^n*2^(n + 1)/(n*(n + 1)^2)*Sum_{k=1..n} (-1/2)^k*binomial(n + 1, k - 1)*binomial(n + 1, k)*binomial(n + 1, k + 1). - Detlef Meya, May 29 2024
Showing 1-5 of 5 results.