A217800 -id:A217800 - cp's OEIS frontend

A241958 Duplicate of A217800.

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

Offset: 0

Karol A. Penson, Aug 10 2014

dead

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

Vincenzo Librandi, Table of n, a(n) for n = 0..200
K. Gorska and K. A. Penson, Multidimensional Catalan and related numbers as Hausdorff moments, arXiv preprint arXiv:1304.6008, 2013

[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

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

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 *)

a(n) = (3*n+3)!/((4*(n+1)^2-1)*((n+1)!)^2*(n+2)!); \\ Michel Marcus, Aug 10 2014

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

A007724 Even minus odd extensions of truncated 3 X 2n grid diagram.

Original entry on oeis.org

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

Offset: 2

Frank Ruskey

Number of standard tableaux of shapes (n-1,n-1,k), k=0,1,...,n-1. Example: a(3)=12 because there are 2, 5 and 5 standard tableaux of shapes (2,2), (2,2,1) and (2,2,2), respectively. - Emeric Deutsch, May 25 2004
From Joel B. Lewis, Oct 05 2009: (Start)
Also the number of standard shifted Young tableaux of shape (n+1, n, n-1).
Also the number of 2143-avoiding up-down permutations of length 2n - 1. (End)

J. B. Lewis, Pattern Avoidance for Alternating Permutations and Reading Words of Tableaux, Ph. D. Dissertation, Department of Mathematics, MIT, 2012.
F. Ruskey, Generating linear extensions of posets by transpositions, J. Combin. Theory, B 54 (1992), 77-101.
Dennis White, Sign-balanced posets

Cf. A003121.
2143-avoiding up-down permutations of length 2n are given by A005789. - Joel B. Lewis, Oct 05 2009
After corrections, is very similar to A217800.

A007724 := proc(n)
    combinat[multinomial](3*n,n-1,n,n+1)/n/(2*n-1)/(2*n+1) ;
end proc:
seq(A007724(n),n=2..40) ; # R. J. Mathar, Jul 07 2023

Table[(3*n)!/((n-1)!*n!*(n+1)!)/(n*(2*n-1)*(2*n+1)),{n,2,10}] (* Vaclav Kotesovec, Nov 13 2014 *)
Table[(-1)^n HypergeometricPFQ[{-2 - 2 n, -2 n, -2 n - 1}, {2, 3}, 1], {n, 19}] (* Michael De Vlieger, Aug 22 2016 *)

{a(n) = if(n<2, 0, (3*n)!/((2*n+1) * (2*n-1) * (n+1)! * n!^2))}; /* Michael Somos, Jul 04 2020 */

a(n) = multinomial(3n; n-1, n, n+1)/(n(2n-1)(2n+1)).
a(n) ~ 3^(3*n+1/2) / (8*Pi*n^4). - Vaclav Kotesovec, Nov 13 2014
D-finite with recurrence n*(n+1)*(2*n+1)*a(n) -3*(3*n-1)*(2*n-3)*(3*n-2)*a(n-1)=0. - R. J. Mathar, Jul 07 2023

More terms from James Sellers, Dec 24 1999

a(16)-a(18) corrected and a(19)-a(20) added by Alois P. Heinz, Aug 22 2016

A359363 Triangle read by rows. The coefficients of the Baxter polynomials p(0, x) = 1 and p(n, x) = x*hypergeom([-1 - n, -n, 1 - n], [2, 3], -x) for n >= 1.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 4, 1, 0, 1, 10, 10, 1, 0, 1, 20, 50, 20, 1, 0, 1, 35, 175, 175, 35, 1, 0, 1, 56, 490, 980, 490, 56, 1, 0, 1, 84, 1176, 4116, 4116, 1176, 84, 1, 0, 1, 120, 2520, 14112, 24696, 14112, 2520, 120, 1, 0, 1, 165, 4950, 41580, 116424, 116424, 41580, 4950, 165, 1

Offset: 0

Peter Luschny, Dec 28 2022

This triangle is a member of a family of Pascal-like triangles. Let T(n, k, m) = sf(m)*F(n - 1) / (F(k - 1)*F(n - k)) if k > 0 and otherwise k^n, where F(n) = Product_{j=0..m} (n + j)! and sf(m) are the superfactorials A000178. The case m = 2 gives this triangle, some other cases are given in the crossreferences. See also A342889 for a related representation of generalized binomial coefficients.

			Triangle T(n, k) starts:
[0] 1
[1] 0, 1
[2] 0, 1,   1
[3] 0, 1,   4,    1
[4] 0, 1,  10,   10,     1
[5] 0, 1,  20,   50,    20,     1
[6] 0, 1,  35,  175,   175,    35,     1
[7] 0, 1,  56,  490,   980,   490,    56,    1
[8] 0, 1,  84, 1176,  4116,  4116,  1176,   84,   1
[9] 0, 1, 120, 2520, 14112, 24696, 14112, 2520, 120, 1
.
Let p = (p1, p2,..., pn) denote a permutation of {1, 2,..., n}. The pair (p(i), p(i+1)) is a 'rise' if p(i) < p(i+1). Additionally a conventional rise is counted at the beginning of p.
T(n, k) is the number of Baxter permutations of {1,2,...,n} with k rises. For example for n = 4, [T(n, k) for k = 0..n] = [0, 1, 10, 10, 1]. The permutations, with preceding number of rises, are:
.
1 [4, 3, 2, 1],  3 [2, 3, 4, 1],  2 [3, 4, 2, 1],  3 [2, 3, 1, 4],
2 [3, 2, 4, 1],  3 [2, 1, 3, 4],  2 [3, 2, 1, 4],  3 [1, 3, 4, 2],
2 [2, 4, 3, 1],  3 [1, 3, 2, 4],  2 [4, 2, 3, 1],  3 [3, 4, 1, 2],
2 [2, 1, 4, 3],  3 [3, 1, 2, 4],  2 [4, 2, 1, 3],  3 [1, 2, 4, 3],
2 [1, 4, 3, 2],  3 [1, 4, 2, 3],  2 [4, 1, 3, 2],  3 [4, 1, 2, 3],
2 [4, 3, 1, 2],  4 [1, 2, 3, 4].

Special cases of the general formula: A097805 (m = 0), (0,1)-Pascal triangle; A090181 (m = 1), triangle of Narayana; this triangle (m = 2); A056940 (m = 3), with 1,0,0...; A056941 (m = 4), with 1,0,0...; A142465 (m = 5), with 1,0,0....
Variant: A056939. Diagonals: A000292, A006542, A047819.
Cf. A000178, A001181, A046996, A217800, A342889.

p := (n, x) -> ifelse(n = 0, 1, x*hypergeom([-1-n, -n, 1-n], [2, 3], -x)):
seq(seq(coeff(simplify(p(n, x)), x, k), k = 0..n), n = 0..10);
# Alternative:
T := proc(n, k) local F; F := n -> n!*(n+1)!*(n+2)!;
ifelse(k = 0, k^n, 2*F(n-1)/(F(k-1)*F(n-k))) end:
for n from 0 to 9 do seq(T(n, k), k = 0..n) od;

C=binomial;
T(n, k) = if(n==0 && k==0, 1, ( C(n+1,k-1) * C(n+1,k) * C(n+1,k+1) ) / ( C(n+1,1) * C(n+1,2) ) );
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print());
\\ Joerg Arndt, Jan 04 2024

from functools import cache
from math import factorial
@cache
def A359363Row(n: int) -> list[int]:
    @cache
    def F(n: int): return factorial(n) ** 3 * ((n+1) * (n+1) * (n+2))
    if n == 0: return [1]
    return [0] + [(2*F(n-1))//(F(k-1) * F(n-k)) for k in range(1, n+1)]
for n in range(0, 10): print(A359363Row(n))
# Peter Luschny, Jan 04 2024

def A359363(n):
    if n == 0: return SR(1)
    h = x*hypergeometric([-1 - n, -n, 1 - n], [2, 3], -x)
    return h.series(x, n + 1).polynomial(SR)
for n in range(10): print(A359363(n).list())
def PolyA359363(n, t): return Integer(A359363(n)(x=t).n())
# Peter Luschny, Jan 04 2024

T(n, k) = [x^k] p(n, x).
T(n, k) = 2*F(n-1)/(F(k-1)*F(n-k)) for k > 0 where F(n) = n!*(n+1)!*(n+2)!.
p(n, 1) = A001181(n), i.e. the Baxter numbers are the values of the Baxter polynomials at x = 1.
(-1)^(n + 1)*p(2*n + 1, -1) = A217800(n) .

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

Joerg Arndt, Jan 04 2024

Cf. A001181, A007724, A056939, A217800, A359363, A368709.

seq(simplify( hypergeom([-1 - n, -n, 1 - n], [2, 3], -2) ), n = 0..25); # Peter Bala, Sep 09 2024

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 *)

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

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

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)

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

Joerg Arndt, Jan 04 2024

sign

Cf. A368708, A001181, A007724, A217800, A359363.

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 *)

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

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

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

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A241958 Duplicate of A217800.

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A007724 Even minus odd extensions of truncated 3 X 2n grid diagram.

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A359363 Triangle read by rows. The coefficients of the Baxter polynomials p(0, x) = 1 and p(n, x) = x*hypergeom([-1 - n, -n, 1 - n], [2, 3], -x) for n >= 1.

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

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

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