A056939 -id:A056939 - cp's OEIS frontend

A342889 Triangle read by rows: T(n,k) = generalized binomial coefficients (n,k)_10 (n >= 0, 0 <= k <= n).

1, 1, 1, 1, 11, 1, 1, 66, 66, 1, 1, 286, 1716, 286, 1, 1, 1001, 26026, 26026, 1001, 1, 1, 3003, 273273, 1184183, 273273, 3003, 1, 1, 8008, 2186184, 33157124, 33157124, 2186184, 8008, 1, 1, 19448, 14158144, 644195552, 2254684432, 644195552, 14158144, 19448, 1

Offset: 0

N. J. A. Sloane, Apr 01 2021

			Triangle begins:
  [1],
  [1, 1],
  [1, 11, 1],
  [1, 66, 66, 1],
  [1, 286, 1716, 286, 1],
  [1, 1001, 26026, 26026, 1001, 1],
  [1, 3003, 273273, 1184183, 273273, 3003, 1],
  [1, 8008, 2186184, 33157124, 33157124, 2186184, 8008, 1],
...

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Seiichi Manyama, Rows n = 0..139, flattened
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993.
Richard L. Ollerton, Counting i-paths, Slides of talk presented at Thirteenth International Conference on Fibonacci Numbers and Their Applications, University of Patras (Greece), 2008.
Richard L. Ollerton, Counting i-paths, Background notes for slides of talk presented at Thirteenth International Conference on Fibonacci Numbers and Their Applications, University of Patras (Greece), 2008.
Richard L. Ollerton and Anthony G. Shannon, Extensions of generalized binomial coefficients, In: Howard F.T. (eds) Applications of Fibonacci Numbers. Springer, Dordrecht, pp. 187-199.
Richard L. Ollerton and Anthony G. Shannon, Further properties of generalized binomial coefficient k-extensions, Fibonacci Quarterly 43.2 (2005): 124-129.
Daniel B. Shapiro, Divisibility Properties of Integer Sequences, arXiv:2302.02243 [math.NT], 2023. See also Integers, (2023) Vol. 23, #A57.

Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

# Generalized binomial coefficient:
GBC := proc(n,k,m) local a,j;
a := mul((binomial(n+m-j,m)/binomial(j+m-1,m)),j=1..k);
end;
# Returns first M rows of triangle:
GBCT := proc(m,M) local a,b,n,k; global GBC;
a:=[];
for n from 0 to M do
  b:=[seq(GBC(n,k,m),k=0..n)];
  a:=[op(a),b];
od: a; end;
GBCT(10,12);

f[n_, k_, m_] := Product[Binomial[n + m - j, m]/Binomial[j + m - 1, m], {j, k}]; Table[f[n, k, 10], {n, 0, 8}, {k, 0, n}] // Flatten (* Michael De Vlieger, Sep 25 2023 *)

f(n, k, m) = prod(j=1, k, binomial(n-j+m, m)/binomial(j-1+m, m));
T(n, k) = f(n, k, 10); \\ Seiichi Manyama, Apr 02 2021

The generalized binomial coefficient (n,k)m = Product{j=1..k} binomial(n+m-j,m)/binomial(j+m-1,m).

A342890 Triangle read by rows: T(n,k) = generalized binomial coefficients (n,k)_11 (n >= 0, 0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 12, 1, 1, 78, 78, 1, 1, 364, 2366, 364, 1, 1, 1365, 41405, 41405, 1365, 1, 1, 4368, 496860, 2318680, 496860, 4368, 1, 1, 12376, 4504864, 78835120, 78835120, 4504864, 12376, 1, 1, 31824, 32821152, 1837984512, 6892441920, 1837984512, 32821152, 31824, 1

Offset: 0

N. J. A. Sloane, Apr 01 2021

For references, links, programs, etc., see earlier sequences in this series, especially A342889.

			Triangle begins:
  [1],
  [1, 1],
  [1, 12, 1],
  [1, 78, 78, 1],
  [1, 364, 2366, 364, 1],
  [1, 1365, 41405, 41405, 1365, 1],
  [1, 4368, 496860, 2318680, 496860, 4368, 1],
  [1, 12376, 4504864, 78835120, 78835120, 4504864, 12376, 1],
  [1, 31824, 32821152, 1837984512, 6892441920, 1837984512, 32821152, 31824, 1],
...

Seiichi Manyama, Rows n = 0..139, flattened

Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

f(n, k, m) = prod(j=1, k, binomial(n-j+m, m)/binomial(j-1+m, m));
T(n, k) = f(n, k, 11); \\ Seiichi Manyama, Apr 02 2021

The generalized binomial coefficient (n,k)m = Product{j=1..k} binomial(n+m-j,m)/binomial(j+m-1,m).

A342891 Triangle read by rows: T(n,k) = generalized binomial coefficients (n,k)_12 (n >= 0, 0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 13, 1, 1, 91, 91, 1, 1, 455, 3185, 455, 1, 1, 1820, 63700, 63700, 1820, 1, 1, 6188, 866320, 4331600, 866320, 6188, 1, 1, 18564, 8836464, 176729280, 176729280, 8836464, 18564, 1, 1, 50388, 71954064, 4892876352, 19571505408, 4892876352, 71954064, 50388, 1

Offset: 0

N. J. A. Sloane, Apr 01 2021

For references, links, programs, etc., see earlier sequences in this series, especially A342889.

			Triangle begins:
  [1],
  [1, 1],
  [1, 13, 1],
  [1, 91, 91, 1],
  [1, 455, 3185, 455, 1],
  [1, 1820, 63700, 63700, 1820, 1],
  [1, 6188, 866320, 4331600, 866320, 6188, 1],
  [1, 18564, 8836464, 176729280, 176729280, 8836464, 18564, 1],
...

Seiichi Manyama, Rows n = 0..139, flattened

Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

f(n, k, m) = prod(j=1, k, binomial(n-j+m, m)/binomial(j-1+m, m));
T(n, k) = f(n, k, 12); \\ Seiichi Manyama, Apr 02 2021

The generalized binomial coefficient (n,k)m = Product{j=1..k} binomial(n+m-j,m)/binomial(j+m-1,m).

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) .

A120258 Triangle of central coefficients of generalized Pascal-Narayana triangles.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 6, 3, 1, 1, 20, 20, 4, 1, 1, 70, 175, 50, 5, 1, 1, 252, 1764, 980, 105, 6, 1, 1, 924, 19404, 24696, 4116, 196, 7, 1, 1, 3432, 226512, 731808, 232848, 14112, 336, 8, 1, 1, 12870, 2760615, 24293412, 16818516, 1646568, 41580, 540, 9, 1

Offset: 0

Paul Barry, Jun 13 2006

Columns are the central coefficients of the triangles T(n, k;r) with T(n, k;r)=Product{j=0..r, C(n+j, k+j)/C(n-k+j, j)}*[k<=n]; (r=0,A007318), (r=1;A001263),(r=2,A056939),(r=3,A056940),(r=4,A056941). Essentially A103905 as a number triangle with an extra diagonal of 1's. Central coefficients T(2n, n) are A008793. Row sums are A120259. Diagonal sums are A120260.

			Triangle begins:
  1;
  1,   1;
  1,   2,     1;
  1,   6,     3,     1;
  1,  20,    20,     4,    1;
  1,  70,   175,    50,    5,   1;
  1, 252,  1764,   980,  105,   6, 1;
  1, 924, 19404, 24696, 4116, 196, 7, 1;
  ...

Seiichi Manyama, Rows n = 0..100, flattened

Row sums give A120259.

Cf. A000891, A000984, A103905, A120257.

T(n, k) = prod(j=0, k-1, binomial(2*n-2*k+j, n-k)/binomial(n-k+j, j)); \\ Seiichi Manyama, Apr 02 2021

Number triangle T(n, k)=[k<=n]*Product{j=0..k-1, C(2n-2k+j, n-k)/C(n-k+j, j)}

As a square array, this is T(n,m)=product{k=1..m, product{j=1..n, product{i=1..n, (i+j+k-1)/(i+j+k-2)}}}; - Paul Barry, May 13 2008

A197208 Triangular array: T(n,k) = sqrt(C(n-1,k-1)C(n-1,k)C(n,k+1)* C(n+1,k+1)C(n+1,k)C(n,k-1)), where C(n,k) = binomial(n,k).

Original entry on oeis.org

3, 12, 12, 30, 120, 30, 60, 600, 600, 60, 105, 2100, 5250, 2100, 105, 168, 5880, 29400, 29400, 5880, 168, 252, 14112, 123480, 246960, 123480, 14112, 252, 360, 30240, 423360, 1481760, 1481760, 423360, 30240, 360, 495, 59400, 1247400, 6985440, 12224520, 6985440, 1247400, 59400, 495

Offset: 2

Peter Bala, Oct 12 2011

In Pascal's triangle, the product of the six entries surrounding C(n,k) is a perfect square.
.............................................
..............C(n-1,k-1)____C(n-1,k).........
.............../.................\...........
............C(n,k-1)...C(n,k)....C(n,k+1)....
...............\................./...........
..............C(n+1,k)______C(n+1,k+1).......
.............................................
In fact, C(n-1,k-1)*C(n,k+1)*C(n+1,k) = C(n-1,k)*C(n+1,k+1)*C(n,k-1).

			.n\k.|....1......2......3......4......5......6
= = = = = = = = = = = = = = = = = = = = = = = =
..2..|....3...
..3..|...12.....12
..4..|...30....120.....30
..5..|...60....600....600.....60
..6..|..105...2100...5250...2100....105
..7..|..168...5880..29400..29400...5880....168
...
T(4,3) = sqrt(1*3*6*10*5*1) = sqrt(900) = 30
..............1..............
............1...1............
..........1...2...1..........
........1...3...3____1.......
.............../......\......
......1...4...6...4....1.....
...............\....../......
...1...5...10...10___5.....1.

Seiichi Manyama, Rows n = 2..141, flattened

Cf. A007318, A056939, A197209 (row sums).

T(n,k) = sqrt(C(n-1,k-1)*C(n-1,k)*C(n,k+1)*C(n+1,k+1)*C(n+1,k)* C(n,k-1)).
T(n,k) = C(n-1,k-1)*C(n,k+1)*C(n+1,k) = C(n-1,k)*C(n+1,k+1)*C(n,k-1).
T(n,k) = 1/2*(n^3-n)*A056939(n-2,k-1), for n >= 2 and 1 <= k <= n-1.
Row sums are A197209.

A342972 Triangle T(n,k) read by rows: T(n,k) = Product_{j=0..n-1} binomial(n+j,k)/binomial(k+j,k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 10, 10, 1, 1, 35, 105, 35, 1, 1, 126, 1176, 1176, 126, 1, 1, 462, 13860, 41580, 13860, 462, 1, 1, 1716, 169884, 1557270, 1557270, 169884, 1716, 1, 1, 6435, 2147145, 61408347, 184225041, 61408347, 2147145, 6435, 1

Offset: 0

Seiichi Manyama, Apr 01 2021

Triangle read by rows: T(n,k) = generalized binomial coefficients (n,k)n where (n,k)_m is Product{j=1..k} binomial(n-j+m,m)/binomial(j-1+m,m).

			Triangle begins:
  1;
  1,    1;
  1,    3,       1;
  1,   10,      10,        1;
  1,   35,     105,       35,         1;
  1,  126,    1176,     1176,       126,        1;
  1,  462,   13860,    41580,     13860,      462,       1;
  1, 1716,  169884,  1557270,   1557270,   169884,    1716,    1;
  1, 6435, 2147145, 61408347, 184225041, 61408347, 2147145, 6435, 1;

Seiichi Manyama, Rows n = 0..50, flattened

Row sums gives A342967.

Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 1,...,12: A007318 (Pascal), A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

T[n_, k_] := Product[Binomial[n + i, k]/Binomial[k + i, k], {i, 0, n - 1}]; Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Amiram Eldar, Apr 01 2021 *)

T(n, k) = prod(j=0, n-1, binomial(n+j, k)/binomial(k+j, k));

T(n, k) = prod(j=0, k-1, binomial(2*n-1, n+j)/binomial(2*n-1, j));

f(n, k, m) = prod(j=1, k, binomial(n-j+m, m)/binomial(j-1+m, m));
T(n, k) = f(n, k, n);

T(n,k) = Product_{j=0..k-1} binomial(2*n-1,n+j)/binomial(2*n-1,j).

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)

A087647 Triangle of 3-Narayana numbers, N(n,k), for n >= 1, 0 <= k <= 2n-2.

Original entry on oeis.org

1, 1, 3, 1, 1, 10, 20, 10, 1, 1, 22, 113, 190, 113, 22, 1, 1, 40, 400, 1456, 2212, 1456, 400, 40, 1, 1, 65, 1095, 7095, 20760, 29484, 20760, 7095, 1095, 65, 1, 1, 98, 2541, 26180, 127435, 320034, 433092, 320034, 127435, 26180, 2541, 98, 1, 1, 140, 5250, 79870

Offset: 1

Robert A. Sulanke (sulanke(AT)math.boisestate.edu), Sep 23 2003

N(n,k) counts ballot sequences for three candidates having length 3n, ending in a tie and having k instances of a vote for a weaker candidate being followed immediately by a vote for a stronger one.

Equivalently, N(n,k) counts the lattice paths P := p_1p_2...p_{3n} using the steps X := (1,0,0), Y := (0,1,0) and Z := (0,0,1), running from (0,0,0) to (n,n,n), lying in {(x,y,z) : 0<=x<=y<=z } and satisfying #{i : p_ip_{i+1} in {XY,XZ,YZ} } = k.

			1;
1,3,1;
1,10,20,10,1;
1,22,113,190,113,22,1;
1,40,400,1456,2212,1456,400,40,1;
1,65,1095,7095,20760,29484,20760,7095,1095,65,1;
1,98,2541,26180,127435,320034,433092,320034,127435,26180,2541,98,1

Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
R. A. Sulanke, Counting Lattice Paths by Narayana Polynomials Electronic J. Combinatorics 7, No. 1, R40, 1-9, 2000.
R. A. Sulanke, Generalizing Narayana and Schroeder Numbers to Higher Dimensions, Electron. J. Combin. 11 (2004), Research Paper 54, 20 pp.
R. A. Sulanke, Three-dimensional Narayana and Schröder numbers, Theoretical Computer Science, Volume 346, Issues 2-3, 28 November 2005, Pages 455-468.

Cf. A001263 (Narayana numbers), A005789 (= Sum[N(n, k), {k, 0, 2n-2}], 3-dimensional Catalan numbers), A056939 (antichains in the poset 3*m*n).

seq( seq( add(2*(-1)^(k-j)*binomial(3*n+1, k-j)* binomial(n+j,n)*binomial(n+j+1,n)*binomial(n+j+2,n)/(n+1)^2/(n+2), j = 0 .. k), k = 0 .. 2*n-2), n = 1 ..7 );

For 0<=k<=2n-2, N(n, k) := Sum[2*(-1)^(k-j)*C(3*n+1, k-j)*C(n+j, n)*C(n+j+1, n)*C(n+j+2, n)/(n+1)^2/(n+2), {j, 0, k}] = Sum[(-1)^(k-j)*C(3*n+1, k-j)*a(n, j), {j, 0, k}] where a(m, n) is an entry in the triangle of A056939.

Recurrence: If N_n(t) := Sum[t^k*N(n, k), {k, 0, 2n-2}] then (3n-4)(n+2)(n+1)^2 N_n(t) = (3n-2)(n+1)( 4(1+t+t^2) - 5(1+7t+t^2)n +3(1+7t+t^2)n^2 ) N_{n-1}(t) - (n-2)( -12 +29n -30n^2 +9n^3)(1-t)^4 N_{n-2}(t) +(3n-1)(n-2)(n-3)(n-4) (1-t)^6 N_{n-3}(t).

A095266 A sequence generated from the Narayana triangle considered as a matrix, or from Pascal's triangle.

Original entry on oeis.org

1, 42, 303, 1144, 3105, 6906, 13447, 23808, 39249, 61210, 91311, 131352, 183313, 249354, 331815, 433216, 556257, 703818, 878959, 1084920, 1325121, 1603162, 1922823, 2288064, 2703025, 3172026, 3699567, 4290328, 4949169, 5681130

Offset: 1

Gary W. Adamson, May 31 2004

A095267 has the same recursion rule but is derived from the matrix derived from A056939 (a type of generalized Narayana triangle).

			a(7) = 23808 = 5*a(6) - 10*a(5) + 10*a(4) - 5*a(3) + a(2) = 5*13447 - 10*6906 + 10*3105 - 5*1144 + 303.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A095267, A001263.

a[n_] := (MatrixPower[{{1, 0, 0, 0, 0}, {1, 1, 0, 0, 0}, {1, 3, 1, 0, 0}, {1, 6, 6, 1, 0}, {1, 10, 20, 10, 1}}, n].{{1}, {0}, {0}, {0}, {0}})[[5, 1]]; Table[ a[n], {n, 30}] (* Robert G. Wilson v, Jun 05 2004 *)

a(n+6) = 5*a(n+5) - 10*a(n+4) + 10*a(n+3) - 5*a(n+2) + a(n), where the multipliers with changed signs are found in the characteristic polynomial of the generating matrix M: x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1. Let M be the 5th-order Matrix M, having Narayana triangle (A001263) rows (fill in with zeros): [1 0 0 0 0 / 1 1 0 0 0 / 1 3 1 0 0 / 1 6 6 1 0 / 1 10 20 10 1]. Then M^n *[1 0 0 0 0] = [1 n A000326(n) A005915(n) a(n)] where A000326 = the pentagonal numbers and A005915 = the hex prism numbers.
From Colin Barker, Oct 21 2012: (Start)
a(n) = (n*(-8 + 25*n - 30*n^2 + 15*n^3))/2.
G.f.: -x*(39*x^3 + 103*x^2 + 37*x + 1)/(x-1)^5. (End)

Edited and corrected by Robert G. Wilson v, Jun 05 2004

Typo in recurrence fixed by Colin Barker, Oct 21 2012

cp's OEIS Frontend

A342889 Triangle read by rows: T(n,k) = generalized binomial coefficients (n,k)_10 (n >= 0, 0 <= k <= n).

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A342890 Triangle read by rows: T(n,k) = generalized binomial coefficients (n,k)_11 (n >= 0, 0 <= k <= n).

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A342891 Triangle read by rows: T(n,k) = generalized binomial coefficients (n,k)_12 (n >= 0, 0 <= k <= n).

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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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A120258 Triangle of central coefficients of generalized Pascal-Narayana triangles.

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A197208 Triangular array: T(n,k) = sqrt(C(n-1,k-1)*C(n-1,k)*C(n,k+1)* C(n+1,k+1)*C(n+1,k)*C(n,k-1)), where C(n,k) = binomial(n,k).

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A342972 Triangle T(n,k) read by rows: T(n,k) = Product_{j=0..n-1} binomial(n+j,k)/binomial(k+j,k).

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A197208 Triangular array: T(n,k) = sqrt(C(n-1,k-1)C(n-1,k)C(n,k+1)* C(n+1,k+1)C(n+1,k)C(n,k-1)), where C(n,k) = binomial(n,k).