A174158 -id:A174158 - cp's OEIS frontend

A319743 Row sums of A174158.

1, 2, 11, 74, 602, 5452, 53559, 558602, 6106034, 69298580, 811086718, 9740402476, 119550632788, 1495039156600, 19002275811887, 244983878813514, 3198363309664658, 42225545561470084, 563083734161627910, 7576864105285884420, 102790882283750139060, 1404908982711268821720

Offset: 1

Stefano Spezia, Dec 23 2018

nonn

G. C. Greubel, Table of n, a(n) for n = 1..830
Abderrahim Arabi, Hacène Belbachir, and Jean-Philippe Dubernard, Enumeration of parallelogram polycubes, arXiv:2105.00971 [cs.DM], 2021.

Cf. A000290, A007318, A174158, A174696.

List([1..20], n->Sum([1..n], m->(Binomial(n-1, m-1)*Binomial(n, m-1)/m)^2));

[(&+[(Binomial(n-1,j-1)*Binomial(n,j-1)/j)^2 : j in [1..n]]): n in [1..25]]; // G. C. Greubel, Feb 15 2021

a := n -> add(binomial(n-1, m-1)^2*binomial(n, m-1)^2/m^2, m = 1 .. n): seq(a(n), n = 1 .. 20)

Table[HypergeometricPFQ[{1-n,1-n,-n,-n},{1,2,2},1],{n,1,20}]

a(n) = sum(m=1, n, (binomial(n-1, m-1)*binomial(n, m-1)/m)^2);

[hypergeometric([-n, -n, -n+1, -n+1], [1, 2, 2], 1).simplify_hypergeometric() for n in (1..25)] # G. C. Greubel, Feb 15 2021

a(n) = Sum_{m=1..n} (binomial(n - 1, m - 1)*binomial(n, m - 1)/m)^2.
a(n) = Sum_{m=1..n} A000290(A007318(n - 1, m - 1)*A007318(n, m - 1)/m).
a(n) = 4F3([1 - n, 1 - n, - n, - n], [1, 2, 2], 1), where F is the generalized hypergeometric function.
From Vaclav Kotesovec, Dec 24 2018: (Start)
Recurrence: n*(n+1)^3*(5*n^2 - 10*n + 4)*a(n) = 2*n*(2*n - 1)*(15*n^4 - 30*n^3 + 7*n^2 + 8*n - 8)*a(n-1) + 4*(n-2)^2*(4*n - 5)*(4*n - 3)*(5*n^2 - 1)*a(n-2).
a(n) ~ 2^(4*n + 1/2) / (Pi^(3/2) * n^(7/2)).
(End)

A111910 Square array read by antidiagonals: S(p,q) = (p+q+1)!(2p+2q+1)!/((p+1)!(2p+1)!(q+1)!(2q+1)!) (p,q>=0).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 14, 14, 1, 1, 30, 84, 30, 1, 1, 55, 330, 330, 55, 1, 1, 91, 1001, 2145, 1001, 91, 1, 1, 140, 2548, 10010, 10010, 2548, 140, 1, 1, 204, 5712, 37128, 68068, 37128, 5712, 204, 1, 1, 285, 11628, 116280, 352716, 352716, 116280, 11628, 285, 1

Offset: 0

Emeric Deutsch, Aug 19 2005

			Array S(n,k) in rectangular form (n, k >= 0):
  1,  1,    1,     1,    1,       1,       1,       1,        1, ...
  1,  5,   14,    30,   55,      91,     140,     204,      285, ...
  1, 14,   84,   330,  1001,   2548,    5712,   11628,    21945, ...
  1, 30,  330,  2145, 10010,  37128,  116280,  319770,   793155, ...
  1, 55, 1001, 10010, 68068, 352716, 1492260, 5393454, 17185025, ...
  ...
Array T(n,k) in triangular form (n >= 0 and 0 <= k <= n):
  1,
  1,  1,
  1,  5,    1,
  1, 14,   14,   1,
  1, 30,   84,  30,     1,
  1, 55,  330, 330,    55,  1,
  1, 91, 1001, 2145, 1001, 91, 1,
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened).
Germain Kreweras and Heinrich Niederhausen, Solution of an enumerative problem connected with lattice paths, European J. Combin. 2 (1981), 55-60.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 9.
Anthony J. Wood, Richard A. Blythe, and Martin R. Evans, Combinatorial mappings of exclusion processes, arXiv:1908.00942 [cond-mat.stat-mech], 2019.

Cf. A091044, A111911 (main diagonal), A196148 (row sums of triangle).

Cf. A001263, A174158.

T:= func< n,k | Binomial(n+1, k)*Binomial(2*n+1, 2*k)/((k+1)*(2*k+1)) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 12 2021

a:=(p,q)->(p+q+1)!*(2*p+2*q+1)!/(p+1)!/(2*p+1)!/(q+1)!/(2*q+1)!: for n from 0 to 10 do seq(a(j,n-j),j=0..n) od; # yields sequence in triangular form

Table[(# + q + 1)! (2 # + 2 q + 1)!/((# + 1)! (2 # + 1)! (q + 1)! (2 q + 1)!) &[r - q], {r, 0, 9}, {q, 0, r}] // Flatten (* Michael De Vlieger, Oct 21 2019 *)
Table[Binomial[n+1, k]*Binomial[2*n+1, 2*k]/((k+1)*(2*k+1)), {n, 0, 12}, {k, 0,
n}]//Flatten (* G. C. Greubel, Feb 12 2021 *)

def A111910(n,k): return binomial(n+1, k)*binomial(2*n+1, 2*k)/((k+1)*(2*k+1))
flatten([[A111910(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 12 2021

S(n,n) = A111911(n).
From Peter Bala, Oct 13 2011: (Start)
Define a(n) = n!*(n+1/2)!*(n+1)!/(1/2)!.
S(n,k) = a(n+k)/(a(n)*a(k)) gives the sequence as a square array while T(n,k) = a(n)/(a(n-k)*a(k)) gives the sequence as a triangle.
S(n-1,k)*S(n,k+1)*S(n+1,k-1) = S(n-1,k+1)*S(n,k-1)*S(n+1,k). Cf. A091044.
(End)
From G. C. Greubel, Feb 12 2021: (Start)
As a number triangle:
T(n, k) = binomial(n+1, k)*binomial(2*n+1, 2*k)/((k+1)*(2*k+1)).
T(n, k) = binomial(2*n+1, 2*k)/((2*k+1)*binomial(n, k)) * A001263(n+1, k+1). (End)
From Peter Bala, Sep 19 2021: (Start)
As a triangle: T(n,k) = a(n)/(a(n-k)*a(k)), where a(n) = Product_{j = 1..n} s(p,j) with s(p,j) = Sum_{j = 1..n} j^p and p = 2. Note, p = 0 gives Pascal's triangle A007318, p = 1 gives the triangle of Narayana numbers A001263 and p = 3 gives the triangle A174158 whose entries are the squares of the Narayana numbers.
Let E(y) = Sum_{n >= 0} y^n/((n+1)!*(2*n+1)!). Then as a triangle this is the generalized Riordan array (E(y), y) as defined in Wang and Wang with respect to the sequence c_n = (n+1)!*(2*n+1)!.  Cf. A001263.
Generating function: E(y)*E(x*y) = 1 + (1 + x)*y/(2!*3!) + (1 + 5*x + x^2)*y^2/(3!*5!) + (1 + 14*x + 14*x^2 + x^3)*y^3/(4!*7!) + ....
The n-th power of this array has a generating function E(y)^n*E(x*y). In particular, the matrix inverse has a generating function E(x*y)/E(y).
exp(y)*E(y) = 1 + 13*y/(2!*3!) + 421*y^2/(3!*5!) + 25368*y^3/(4!*7!) + ... is essentially a generating function for A081442. (End)

Example section edited by Petros Hadjicostas, Sep 03 2019

A174696 Triangle T(n, k) = n!(1/k)^2(binomial(n-1, k-1)*binomial(n, k-1))^2 - n! + 1, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 49, 1, 1, 841, 841, 1, 1, 11881, 47881, 11881, 1, 1, 161281, 1799281, 1799281, 161281, 1, 1, 2217601, 55560961, 154344961, 55560961, 2217601, 1, 1, 31570561, 1548892801, 9680791681, 9680791681, 1548892801, 31570561, 1, 1, 469929601, 40967337601, 501853968001, 1129171881601, 501853968001, 40967337601, 469929601, 1

Offset: 1

Roger L. Bagula, Mar 27 2010

			Triangle begins as:
  1;
  1,        1;
  1,       49,          1;
  1,      841,        841,          1;
  1,    11881,      47881,      11881,          1;
  1,   161281,    1799281,    1799281,     161281,          1;
  1,  2217601,   55560961,  154344961,   55560961,    2217601,        1;
  1, 31570561, 1548892801, 9680791681, 9680791681, 1548892801, 31570561, 1;

G. C. Greubel, Rows n = 1..100 of the triangle, flattened

Cf. A174158, A174694, A176013, A319743.

A174696:= func< n, k | (Factorial(n)/k^2)*(Binomial(n-1, k-1)*Binomial(n, k-1))^2 - Factorial(n) + 1 >;
[A174696(n, k): k in [1..n], n in [1..12]]; // G. C. Greubel, Feb 09 2021

T[n_, m_]:= n!*(1/k)^2*(Binomial[n-1, k-1]*Binomial[n, k-1])^2 - n! + 1;
Table[T[n, k], {n,12}, {k,n}]//Flatten

def A174696(n, k): return (factorial(n)/k^2)*(binomial(n-1, k-1)*binomial(n, k-1))^2 - factorial(n) + 1
flatten([[A174696(n, k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Feb 09 2021

T(n, k) = n!*(1/k)^2*(binomial(n-1, k-1)*binomial(n, k-1))^2 - n! + 1.
From G. C. Greubel, Feb 09 2021: (Start)
T(n, k) = n! * A174158(n, k) - n! + 1.
Sum_{k=1..n} T(n,k) = n! * Hypergeometric4F3([-n, -n, 1-n, 1-n], [1, 2, 2], 1) - n*(n! - 1) = n! * A319743(n) - n*(n! - 1). (End)

Edited by G. C. Greubel, Feb 09 2021

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A319743 Row sums of A174158.

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A111910 Square array read by antidiagonals: S(p,q) = (p+q+1)!(2p+2q+1)!/((p+1)!(2p+1)!(q+1)!(2q+1)!) (p,q>=0).

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A174696 Triangle T(n, k) = n!*(1/k)^2*(binomial(n-1, k-1)*binomial(n, k-1))^2 - n! + 1, read by rows.

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Crossrefs

Programs

Magma

Mathematica

Sage

Formula

Extensions

A174696 Triangle T(n, k) = n!(1/k)^2(binomial(n-1, k-1)*binomial(n, k-1))^2 - n! + 1, read by rows.