A002695 -id:A002695 - cp's OEIS frontend

A331514 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/(1 - 2kx + ((k-2)*x)^2)^(3/2).

1, 1, 0, 1, 3, -6, 1, 6, 6, 0, 1, 9, 30, 10, 30, 1, 12, 66, 140, 15, 0, 1, 15, 114, 450, 630, 21, -140, 1, 18, 174, 1000, 2955, 2772, 28, 0, 1, 21, 246, 1850, 8430, 18963, 12012, 36, 630, 1, 24, 330, 3060, 18855, 69384, 119812, 51480, 45, 0

Offset: 0

Seiichi Manyama, Jan 19 2020

			Square array begins:
    1,  1,    1,     1,     1,      1, ...
    0,  3,    6,     9,    12,     15, ...
   -6,  6,   30,    66,   114,    174, ...
    0, 10,  140,   450,  1000,   1850, ...
   30, 15,  630,  2955,  8430,  18855, ...
    0, 21, 2772, 18963, 69384, 187425, ...

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

Columns k=1..5 give A000217(n+1), A002457, A002695(n+1), A331515, A331516.

Cf. A307883, A331511.

T[n_, k_] = 1/2 * Sum[If[k == 2 && n == j - 1, 1, (k - 2)^(n + 1 - j)] * j * Binomial[n + 1, j] * Binomial[n + 1 + j, j], {j, 1, n + 1}]; Table[Table[T[n, k - n], {n, 0, k}], {k, 0, 9}] //Flatten (* Amiram Eldar, Jan 20 2020 *)

T(n,k) = (1/2)*sum(j=1,n+1,(k-2)^(n+1-j)*j*binomial(n+1,j)*binomial(n+1+j,j));
matrix(7, 7, n, k, T(n-1, k-1)) \\ Michel Marcus, Jan 20 2020

T(n,k) = (1/2) * Sum_{j=1..n+1} (k-2)^(n+1-j) * j * binomial(n+1,j) * binomial(n+1+j,j).
n * T(n,k) = k * (2*n+1) * T(n-1,k) - (k-2)^2 * (n+1) * T(n-2,k) for n > 1.
T(n,k) = ((n+2)/2) * Sum_{j=0..n} (k-1)^j * binomial(n+1,j) * binomial(n+1,j+1).
T(n,k) = Sum_{j=0..n} (k/2)^j * (-(k-2)^2/(2*k))^(n-j) * (2*j+1) * binomial(2*j,j) * binomial(j,n-j) for k > 0. - Seiichi Manyama, Aug 20 2025

A132818 The matrix product A127773 * A001263 of infinite lower triangular matrices.

Original entry on oeis.org

1, 3, 3, 6, 18, 6, 10, 60, 60, 10, 15, 150, 300, 150, 15, 21, 315, 1050, 1050, 315, 21, 28, 588, 2940, 4900, 2940, 588, 28, 36, 1008, 7056, 17640, 17640, 7056, 1008, 36, 45, 1620, 15120, 52920, 79380, 52920, 15120, 1620, 45, 55, 2475, 29700, 138600, 291060

Offset: 1

Gary W. Adamson, Sep 02 2007

			First few rows of the triangle are:
1;
3, 3;
6, 18, 6;
10, 60, 60, 10;
15, 150, 300, 150, 15;
21, 315, 1050, 1050, 315, 21;
...

Cf. A127773, A001263, A002457 (row sums), A006472. A002695, A008459.

A132818 := proc(n,k)
    (n+1-k)*binomial(n+1,k)*binomial(n,k-1)/2 ;
end proc: # R. J. Mathar, Jul 29 2015

T(n,k) = A000217(n) * A001263(n,k).
Let a(n) = A006472(n), the 'triangular' factorial numbers. Then a(n)/(a(k)*a(n-k)) produces the present triangle (with a different offset). - Peter Bala, Dec 07 2011
T(n,k) = 1/2*(n+1-k)*C(n+1,k)*C(n,k-1), for n,k >= 1. O.g.f.: x*y/((1-x-x*y)^2 - 4*x^2*y)^(3/2) = x*y + x^2*(3*y + 3*y^2) + x^3*(6*y + 18*y^2 + 6*y^3) + .... Cf. A008459 with o.g.f.: x*y/((1-x-x*y)^2 - 4*x^2*y)^(1/2). Sum {k = 1..n-1} T(n,k)*2^(n-k) = A002695(n). - Peter Bala, Apr 10 2012

Corrected by R. J. Mathar, Jul 29 2015

A277660 2nd-order coefficients of the 1/N-expansion of traces of negative powers of complex Wishart matrices with parameter c=2.

Original entry on oeis.org

0, 0, 2, 30, 310, 2730, 21980, 167076, 1220100, 8650620, 59958030, 408172050, 2738441706, 18151701750, 119100934680, 774719545320, 5001728701800, 32081745977496, 204596905143930, 1298154208907430, 8199305968563710, 51576591659861730, 323239814342259892, 2019025558874685900

Offset: 0

Fabio Deelan Cunden, Oct 26 2016

These numbers provide the 2nd order of the 1/N-expansion of traces of powers of a random time-delay matrix without time-reversal symmetry. (The 0th order is instead given by the Large Schröder numbers A006318.)

G. C. Greubel, Table of n, a(n) for n = 0..1000
F. D. Cunden, F. Mezzadri, N. Simm and P. Vivo, Large-N expansion for the time-delay matrix of ballistic chaotic cavities, J. Math. Phys. 57, 111901 (2016).
J. Kuipers, M. Sieber and D. Savin, Efficient semiclassical approach for time delays, New J. Phys. 16 (2014), 123018.

Cf. A001850, A002695, A277661, A277662, A277663, A277664, A277665.

a := proc(n) option remember; if n = 1 then 0 elif n = 2 then 2 else (3*(2*n - 1)*a(n-1) - (n + 1)*a(n-2))/(n - 2) fi; end:
seq(a(n), n = 1..25); # Peter Bala, Sep 28 2024

a[n_] := If[n<2, 0, 2 GegenbauerC[n-2, 5/2, 3]]; a /@ Range[0, 20] (* Andrey Zabolotskiy, Oct 27 2016 *)
CoefficientList[Series[(2 x^2) / (x^2 - 6 x + 1)^(5/2), {x, 0, 25}], x] (* Vincenzo Librandi, Oct 30 2016 *)

x='x+O('x^50); concat([0,0], Vec((2*x^2)/(x^2-6*x+1)^(5/2))) \\ G. C. Greubel, Jun 05 2017

G.f.: (2*x^2)/(x^2-6*x+1)^(5/2).
a(n) = 2*C_(n-2)^(5/2)(3) for n >= 2, where C_n^(m)(x) is the Gegenbauer polynomial. - Andrey Zabolotskiy, Oct 26 2016
a(n) ~ (1 + sqrt(2))^(2*n+1) * n^(3/2) / (3*2^(13/4)*sqrt(Pi)). - Vaclav Kotesovec, Oct 27 2016, simplified Aug 27 2025
From Peter Bala, Sep 20 2024: (Start)
a(n) = (1/6) * Sum_{k = 0..n} k*(k - 1)*binomial(n, k)*binomial(n+k, k).
a(n) = (1/12)*n*(n + 1)*(n - 1)*(n + 2)*hypergeom([n+3, -n+2], [3], -1).
a(n) = (2/3) * d^2/dx^2(Legendre_P(n, x)) at x = 3.
a(n) = (1/12)*n*(n + 1)*A001850(n) - (1/2)*A002695(n).
P-recursive: (n - 2)*a(n) = 3*(2*n - 1)*a(n-1) - (n + 1)*a(n-2) with a(1) = 0 and a(2) = 2. (End)

a(9)-a(22) from Andrey Zabolotskiy, Oct 26 2016

a(23) from Fabio Deelan Cunden, Oct 29 2016

A277652 Numerators of factorial moments of order 2 for the number of parts in dissections of rooted and convex polygons.

Original entry on oeis.org

0, 0, 4, 40, 312, 2212, 14920, 97632, 626080, 3957448, 24747948, 153483720, 945638232, 5795135820, 35357242128, 214919392128, 1302250826880, 7869116134672, 47437683195220, 285373276253352, 1713562776624952, 10272384482513140, 61489533128765784, 367581030765071200

Offset: 0

Ricardo Gómez Aíza, Oct 26 2016

a(n)/A001003(n) is the factorial moment of order two for the number of parts in a (uniform) random (rooted) dissection of a convex (n+2)-gon.

			A convex 3-gon is a triangle. There is only one dissection of a rooted triangle, with one single part. The factorial moment of order two is therefore 0 and hence a(1) = 0.
A convex 4-gon is a quadrilateral. There are three dissections of a rooted quadrilateral, two with two parts and one with one part. Then the expectation of the number of parts is 5/3, and the expectation of the number of parts squared is 9/3, hence the factorial moment of order two is 9/3 - 5/3 = 4/3. The second Schröder number is A001003(2) = 3, therefore a(2) = 4.

Robert Israel, Table of n, a(n) for n = 0..1300
Ricardo Gómez Aíza, RNA structures and dissections of polygons: an invitation to analytic combinatorics, Misc. Mat. 60 (2015) 105-130 (In Spanish)

Denominators are the Schröder numbers A001003.

Cf. A002695, A035029.

s := (z^2-6*z+1)^(1/2): g := z/s^3-(1/s-(z+1-s)/(4*z))/2: ser := series(g,z,30):
seq(coeff(ser,z,n), n=0..23); # Peter Luschny, Nov 17 2016

CoefficientList[Series[z/Sqrt[(z^2 - 6*z + 1)^3] - (1/Sqrt[z^2 - 6*z + 1] - (z + 1 - Sqrt[z^2 - 6*z + 1])/(4*z))/2, {z, 0, 20}], z]

a(n) = A002695(n) - A035029(n-1), n >= 1.
G.f.: (z/sqrt(z^2 - 6*z + 1)^3) - (1/sqrt(z^2 - 6*z + 1) - (z + 1 - sqrt(z^2 - 6*z + 1))/(4*z))/2.
D-finite with recurrence (-n^3-5*n^2-6*n)*a(n)+(6*n^3+27*n^2+35*n+12)*a(n+1)+(-n^3-4*n^2-3*n)*a(n+2) = 0. - Robert Israel, Nov 18 2016

A376467 Triangular array read by rows: A063007 * A007318.

Original entry on oeis.org

1, 3, 2, 13, 18, 6, 63, 132, 90, 20, 321, 900, 930, 420, 70, 1683, 5910, 8190, 5600, 1890, 252, 8989, 37926, 65940, 60480, 30870, 8316, 924, 48639, 239624, 501228, 577080, 395010, 160776, 36036, 3432, 265729, 1497096, 3660300, 5072760, 4358970, 2378376, 804804, 154440, 12870, 1462563, 9274410, 25951860, 42060480, 43513470, 29801772, 13513500, 3912480, 656370, 48620

Offset: 0

Peter Bala, Sep 30 2024

Note that the n-th row generating polynomial of A063007 is equal to P(n,2*x + 1), where P(n,x) denotes the n-th Legendre polynomial.

The matrix product A063007 * A007318^(-1) is equal to a signed version of A063007 and A007318^(-1) * A063007 = A115951.

			Triangle begins
 n\k|     0       1       2       3       4       5      6     7
 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |     1
  1 |     3       2
  2 |    13      18       6
  3 |    63     132      90      20
  4 |   321     900     930     420      70
  5 |  1683    5910    8190    5600    1890     252
  6 |  8989   37926   65940   60480   30870    8316    924
  7 | 48639  239624  501228  577080  395010  160776  36036  3432
  ...

A000984 (main diagonal), A002457( (1/3)*first subdiagonal ), A001850 (Column 0), A002695 ( (1/2)*Column 1 ), A277660 ( (1/3)*Column 2 ), A006442 (row sums).

Cf. A063007, A007318, A115951, A118384.

A376467 := proc(n, k); add(binomial(n, j)*binomial(n+j, j)*binomial(j, k), j = k..n) end:
seq(print(seq(A376467(n, k) , k = 0..n)), n = 0..10);

T(n, k) = Sum_{j = k..n} binomial(n, j)*binomial(n+j, j)*binomial(j, k).
(n - k)*T(n, k) = 3*(2*n - 1)*T(n-1, k) - (n + k - 1)*T(n-2, k).
T(n, k) = (1/k!) * (d/dx)^k (P(n, 2*x+1)) evaluated at x = 1, where P(n,x) denotes the n-th Legendre polynomial.
G.f. for triangle: 1/sqrt(1 - 6*t + t^2 - 4*t*x) = 1 + (3 + 2*x)*t + (13 + 18*x + 6*x^2)*t^2 + ....
G.f. for column k: binomial(2*k, k) * x^k/(1 - 6*x + x^2)^(k+1/2).
T(n, k) is divisible by binomial(2*k, k) and the array ( T(n, k)/binomial(2*k, k) )n,k >= 0 is the Riordan array (1/sqrt(1 - 6*x + x^2), x/(1 - 6*x + x^2)).
T(n, k) is divisible by binomial(n+k, k) and the array ( T(n, k)/binomial(n+k, k) )n,k >= 0 is the Riordan array A118384.
T(n, n) = binomial(2*n, n); T(n, n-1) = 3*(2*n-1)!/(n-1)!^2 = 3 * A002457(n-1) for n >= 1.
The n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*(1 + x)^k = P(n, 2*x+3) = hypergeom([-n, n+1], [1], -1-x).
Recurrence: n*R(n, x) = (2*x + 3)*(2*n - 1)*R(n-1, x) - (n - 1)*R(n-2, x) with R(0, x) = 1.
If we set R(-1,x) = 1, we can run the recurrence backwards to give R(-n, x) = Sum_{k = 0..n} binomial(-n, k)*binomial(-n+k, k)*(1 + x)^k = R(n-1, x).
R(n, x) = (-1)^n * R(n, -x-3).
R(n, x) = 1/n! * (d/dx)^n( ((1 + x)*(2 + x))^n ).
R(1, x) = 3 + 2*x divides R(2*n+1, x) in the polynomial ring Z[x].

A351385 Triangle read by rows: T(n,k) = Sum_{j=k..n} binomial(n + j, n)*binomial(n, j)/(j + 1).

Original entry on oeis.org

1, 2, 1, 6, 5, 2, 22, 21, 15, 5, 90, 89, 79, 49, 14, 394, 393, 378, 308, 168, 42, 1806, 1805, 1784, 1644, 1224, 594, 132, 8558, 8557, 8529, 8277, 7227, 4917, 2145, 429, 41586, 41585, 41549, 41129, 38819, 31889, 19877, 7865, 1430, 206098, 206097, 206052, 205392, 200772, 182754, 140712, 80652, 29172, 4862

Offset: 0

David Callan, Feb 09 2022

T(n,k) is the number of central Delannoy paths of steps E = (1,0), N = (0,1), D = (1,1) from the origin to (n,n) with k E steps above the diagonal line y=x. For example, T(3,1) = 5 counts ENNE, NEEN, NED, NDE, DNE. That the titular sum counts these paths is a consequence of the following equidistribution result: among the central Delannoy n-paths with j E steps, the statistic "number of E steps above y=x" is uniformly distributed over {0,1,...,j}. So, for k <= j <= n, there are binomial(n + j, n) binomial(n, j)/(j + 1) central Delannoy n-paths with j E steps, k of which are above y = x.

			Triangle begins:
   n
  [0]  1;
  [1]  2,  1;
  [2]  6,  5,  2;
  [3] 22, 21, 15,  5;
  [4] 90, 89, 79, 49, 14;
      ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Columns k=0..1 give: A006318, A035011.
Main diagonal gives A000108.
Row sums give A001850.
Cf. A001003, A002695, A080243, A088617 gives summands in title.

Flatten[Table[
  Sum[Binomial[n + j, n] Binomial[n, j]/(j + 1), {j, k, n}], {n, 0,
   10}, {k, 0, n}]]

T(n,k)={sum(j=k, n, binomial(n+j, n)*binomial(n,j)/(j+1))} \\ Andrew Howroyd, Feb 09 2022

G.f.: 2/(sqrt(1 - 6*x + x^2) + sqrt(1 - 2*x + x^2 - 4*x*y)).
From Alois P. Heinz, Feb 09 2022: (Start)
Sum_{k=0..n} k * T(n,k) = A002695(n).
Sum_{k=0..n} (-1)^k * T(n,k) = A001003(n).
Sum_{k=0..n} (-1)^k * T(n,n-k) = A080243(n). (End)

A110221 Triangle read by rows: T(n,k) (0<=k<=floor(n/2)) is the number of Delannoy paths of length n, having k ED's.

Original entry on oeis.org

1, 3, 11, 2, 45, 18, 195, 120, 6, 873, 720, 90, 3989, 4110, 870, 20, 18483, 22806, 6930, 420, 86515, 124264, 49560, 5320, 70, 408105, 668520, 331128, 52920, 1890, 1936881, 3562830, 2111760, 456120, 29610, 252, 9238023, 18850590, 13020480, 3575880

Offset: 0

Emeric Deutsch, Jul 20 2005

A Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1).

Row n has 1+floor(n/2) terms. Row sums are the central Delannoy numbers (A001850). Column 0 yields A026375. Sum(k*T(n,k),k=0..floor(n/2))=2*A002695(n).

			T(2,1)=2 because we have NED and EDN.
Triangle begins:
1;
3;
11,2;
45,18;
195,120,6;

Robert A. Sulanke, Objects Counted by the Central Delannoy Numbers, Journal of Integer Sequences, Volume 6, 2003, Article 03.1.5.

Cf. A001850, A026375, A002695.

R:=(1-z-sqrt(1-6*z+5*z^2-4*z^2*t))/2/z/(1-z+t*z): G:=1/(1-z-2*t*z^2*R-2*z*R+2*z^2*R): Gser:=simplify(series(G,z=0,15)): P[0]:=1: for n from 1 to 12 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 12 do seq(coeff(t*P[n],t^k),k=1..1+floor(n/2)) od; # yields sequence in triangular form

G.f.: 1/(1-z-2tz^2*R-2zR+2z^2*R), where R=[1-z-sqrt(1-6z+5z^2-4tz^2)]/[2z(1-z+tz)].

cp's OEIS Frontend

A331514 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/(1 - 2*k*x + ((k-2)*x)^2)^(3/2).

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A132818 The matrix product A127773 * A001263 of infinite lower triangular matrices.

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A277660 2nd-order coefficients of the 1/N-expansion of traces of negative powers of complex Wishart matrices with parameter c=2.

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A277652 Numerators of factorial moments of order 2 for the number of parts in dissections of rooted and convex polygons.

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A376467 Triangular array read by rows: A063007 * A007318.

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A351385 Triangle read by rows: T(n,k) = Sum_{j=k..n} binomial(n + j, n)*binomial(n, j)/(j + 1).

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A110221 Triangle read by rows: T(n,k) (0<=k<=floor(n/2)) is the number of Delannoy paths of length n, having k ED's.

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A331514 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/(1 - 2kx + ((k-2)*x)^2)^(3/2).