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.

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A181543 Triangle of cubed binomial coefficients, T(n,k) = C(n,k)^3, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 27, 27, 1, 1, 64, 216, 64, 1, 1, 125, 1000, 1000, 125, 1, 1, 216, 3375, 8000, 3375, 216, 1, 1, 343, 9261, 42875, 42875, 9261, 343, 1, 1, 512, 21952, 175616, 343000, 175616, 21952, 512, 1, 1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1
Offset: 0

Views

Author

Paul D. Hanna, Oct 30 2010

Keywords

Comments

Diagonal of rational function R(x,y,z,t) = 1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z) with respect to x, y, z, i.e., T(n,k) = [(xyz)^n*t^k] R(x,y,z,t). - Gheorghe Coserea, Jul 01 2018

Examples

			Triangle begins:
  1;
  1,   1;
  1,   8,     1;
  1,  27,    27,      1;
  1,  64,   216,     64,       1;
  1, 125,  1000,   1000,     125,       1;
  1, 216,  3375,   8000,    3375,     216,      1;
  1, 343,  9261,  42875,   42875,    9261,    343,     1;
  1, 512, 21952, 175616,  343000,  175616,  21952,   512,   1;
  1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1;
  ...
		

Crossrefs

Cf. A000172 (row sums), A181545 (antidiagonal sums), A002897, A181544, A248658.
Variants: A008459, A007318.

Programs

  • Maple
    T:= (n, k)-> binomial(n, k)^3:
    seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Jan 06 2021
  • Mathematica
    Flatten[Table[Binomial[n,k]^3,{n,0,10},{k,0,n}]] (* Harvey P. Dale, May 23 2011 *)
  • PARI
    T(n,k)=binomial(n,k)^3
    for(n=0,10,for(k=0,n,print1(T(n,k),", "));print())
    
  • PARI
    T(n,k)=polcoeff(polcoeff(sum(m=0,n,(3*m)!/m!^3*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(3*m+1)),n,x),k,y)
    for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Nov 04 2010
    
  • PARI
    diag(expr, N=22, var=variables(expr)) = {
      my(a = vector(N));
      for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
      for (n = 1, N, a[n] = expr;
        for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
      return(a);
    };
    x='x; y='y; z='z; t='t;
    concat(apply(Vec, diag(1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z), 10, [x, y, z]))) \\ Gheorghe Coserea, Jul 01 2018

Formula

Row sums equal A000172, the Franel numbers.
Central terms are A002897(n) = C(2n,n)^3.
Antidiagonal sums equal A181545;
The g.f. of the antidiagonal sums is Sum_{n>=0} (3n)!/(n!)^3 * x^(3n)/(1-x-x^2)^(3n+1).
G.f. for column k: [Sum_{j=0..2k} A181544(k,j)*x^j]/(1-x)^(3k+1), where the row sums of A181544 equals De Bruijn's s(3,n) = (3n)!/(n!)^3.
G.f.: A(x,y) = Sum_{n>=0} (3n)!/n!^3 * x^(2n)*y^n/(1-x-x*y)^(3n+1). - Paul D. Hanna, Nov 04 2010

A132813 Triangle read by rows: A001263 * A127648 as infinite lower triangular matrices.

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 12, 18, 4, 1, 20, 60, 40, 5, 1, 30, 150, 200, 75, 6, 1, 42, 315, 700, 525, 126, 7, 1, 56, 588, 1960, 2450, 1176, 196, 8, 1, 72, 1008, 4704, 8820, 7056, 2352, 288, 9, 1, 90, 1620, 10080, 26460, 31752, 17640, 4320, 405, 10
Offset: 0

Views

Author

Gary W. Adamson, Sep 01 2007

Keywords

Comments

Also T(n,k) = binomial(n-1, k-1)*binomial(n, k-1), related to Narayana polynomials (see Sulanke reference). - Roger L. Bagula, Apr 09 2008
h-vector for cluster complex associated to the root system B_n. See p. 8, Athanasiadis and C. Savvidou. - Tom Copeland, Oct 19 2014

Examples

			First few rows of the triangle are:
  1;
  1,  2;
  1,  6,   3;
  1, 12,  18,   4;
  1, 20,  60,  40,   5;
  1, 30, 150, 200,  75,   6;
  1, 42, 315, 700, 525, 126, 7;
  ...
		

Crossrefs

Family of polynomials (see A062145): A008459 (c=1), this sequence (c=2), A062196 (c=3), A062145 (c=4), A062264 (c=5), A062190 (c=6).
Columns: A000012 (k=0), A002378 (k=1), A006011 (k=2), 4*A006542 (k=3), 5*A006857 (k=4), 6*A108679 (k=5), 7*A134288 (k=6), 8*A134289 (k=7), 9*A134290 (k=8), 10*A134291 (k=9).
Diagonals: A000027 (k=n), A002411 (k=n-1), A004302 (k=n-2), A108647 (k=n-3), A134287 (k=n-4).
Main diagonal: A000894.
Sums: (-1)^floor((n+1)/2)*A001405 (signed row), A001700 (row), A203611 (diagonal).
Cf. A103371 (mirrored).

Programs

  • GAP
    Flat(List([0..10],n->List([0..n], k->(k+1)*Binomial(n+1,k+1)*Binomial(n+1,k)/(n+1)))); # Muniru A Asiru, Feb 26 2019
    
  • Haskell
    a132813 n k = a132813_tabl !! n !! k
    a132813_row n = a132813_tabl !! n
    a132813_tabl = zipWith (zipWith (*)) a007318_tabl $ tail a007318_tabl
    -- Reinhard Zumkeller, Apr 04 2014
    
  • Magma
    /* triangle */ [[(k+1)*Binomial(n+1,k+1)*Binomial(n+1,k)/(n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 19 2014
    
  • Maple
    P := (n, x) -> hypergeom([1-n, -n], [1], x): for n from 1 to 9 do PolynomialTools:-CoefficientList(simplify(P(n,x)),x) od; # Peter Luschny, Nov 26 2014
  • Mathematica
    T[n_,k_]=Binomial[n-1,k-1]*Binomial[n,k-1]; Table[Table[T[n,k],{k,1,n}],{n,1,11}]; Flatten[%] (* Roger L. Bagula, Apr 09 2008 *)
    P[n_, x_] := HypergeometricPFQ[{1-n, -n}, {1}, x]; Table[CoefficientList[P[n, x], x], {n, 1, 10}] // Flatten (* Jean-François Alcover, Nov 27 2014, after Peter Luschny *)
  • PARI
    tabl(nn) = {for (n = 1, nn, for (k = 1, n, print1(binomial(n-1, k-1)*binomial(n, k-1) , ", ");););} \\ Michel Marcus, Feb 12 2014
    
  • SageMath
    def A132813(n,k): return binomial(n,k)*binomial(n+1,k)
    print(flatten([[A132813(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Mar 12 2025

Formula

T(n,k) = (k+1)*binomial(n+1,k+1)*binomial(n+1,k)/(n+1), n >= k >= 0.
From Roger L. Bagula, May 14 2010: (Start)
T(n, m) = coefficients(p(x,n)), where
p(x,n) = (1-x)^(2*n)*Sum_{k >= 0} binomial(k+n-1, k)*binomial(n+k, k)*x^k,
or p(x,n) = (1-x)^(2*n)*Hypergeometric2F1([n, n+1], [1], x). (End)
T(n,k) = binomial(n,k) * binomial(n+1,k). - Reinhard Zumkeller, Apr 04 2014
These are the coefficients of the polynomials Hypergeometric2F1([1-n,-n], [1], x). - Peter Luschny, Nov 26 2014
G.f.: A(x,y) = A281260(x,y)/(1-A281260(x,y))/x. - Vladimir Kruchinin, Oct 10 2020

A145905 Square array read by antidiagonals: Hilbert transform of triangle A060187.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 9, 5, 1, 1, 27, 25, 7, 1, 1, 81, 125, 49, 9, 1, 1, 243, 625, 343, 81, 11, 1, 1, 729, 3125, 2401, 729, 121, 13, 1, 1, 2187, 15625, 16807, 6561, 1331, 169, 15, 1, 1, 6561, 78125, 117649, 59049, 14641, 2197, 225, 17, 1, 1, 19683, 390625, 823543
Offset: 0

Views

Author

Peter Bala, Oct 27 2008

Keywords

Comments

Definition of the Hilbert transform of a triangular array:
For many square arrays in the database the entries in a row are polynomial in the column index, of degree d say and hence the row generating function has the form P(x)/(1-x)^(d+1), where P is some polynomial function. Often the array whose rows are formed from the coefficients of these P polynomials is of independent interest. This suggests the following definition.
Let [L(n,k)]n,k>=0 be a lower triangular array and let R(n,x) := sum {k = 0 .. n} L(n,k)*x^k, denote the n-th row generating polynomial of L. Then we define the Hilbert transform of L, denoted Hilb(L), to be the square array whose n-th row, n >= 0, has the generating function R(n,x)/(1-x)^(n+1).
In this particular case, L is the array A060187, the array of Eulerian numbers of type B, whose row polynomials are the h-polynomials for permutohedra of type B. The Hilbert transform is an infinite Vandermonde matrix V(1,3,5,...).
We illustrate the Hilbert transform with a few examples:
(1) The Delannoy number array A008288 is the Hilbert transform of Pascal's triangle A007318 (view as the array of coefficients of h-polynomials of n-dimensional cross polytopes).
(2) The transpose of the array of nexus numbers A047969 is the Hilbert transform of the triangle of Eulerian numbers A008292 (best viewed in this context as the coefficients of h-polynomials of n-dimensional permutohedra of type A).
(3) The sequence of Eulerian polynomials begins [1, x, x + x^2, x + 4*x^2 + x^3, ...]. The coefficients of these polynomials are recorded in triangle A123125, whose Hilbert transform is A004248 read as square array.
(4) A108625, the array of crystal ball sequences for the A_n lattices, is the Hilbert transform of A008459 (viewed as the triangle of coefficients of h-polynomials of n-dimensional associahedra of type B).
(5) A142992, the array of crystal ball sequences for the C_n lattices, is the Hilbert transform of A086645, the array of h-vectors for type C root polytopes.
(6) A108553, the array of crystal ball sequences for the D_n lattices, is the Hilbert transform of A108558, the array of h-vectors for type D root polytopes.
(7) A086764, read as a square array, is the Hilbert transform of the rencontres numbers A008290.
(8) A143409 is the Hilbert transform of triangle A073107.

Examples

			Triangle A060187 (with an offset of 0) begins
1;
1, 1;
1, 6, 1;
so the entries in the first three rows of the Hilbert transform of
A060187 come from the expansions:
Row 0: 1/(1-x) = 1 + x + x^2 + x^3 + ...;
Row 1: (1+x)/(1-x)^2 = 1 + 3*x + 5*x^2 + 7*x^3 + ...;
Row 2: (1+6*x+x^2)/(1-x)^3 = 1 + 9*x + 25*x^2 + 49*x^3 + ...;
The array begins
n\k|..0....1.....2.....3......4
================================
0..|..1....1.....1.....1......1
1..|..1....3.....5.....7......9
2..|..1....9....25....49.....81
3..|..1...27...125...343....729
4..|..1...81...625..2401...6561
5..|..1..243..3125.16807..59049
...
		

Crossrefs

Cf. A008292, A039755, A052750 (first superdiagonal), A060187, A114172, A145901.

Programs

  • Maple
    T:=(n,k) -> (2*k + 1)^n: seq(seq(T(n-k,k),k = 0..n),n = 0..10);

Formula

T(n,k) = (2*k + 1)^n, (see equation 4.10 in [Franssens]). This array is the infinite Vandermonde matrix V(1,3,5,7, ....) having a LDU factorization equal to A039755 * diag(2^n*n!) * transpose(A007318).

A192721 The number of pairs of permutations in the product group S_n X S_n with k common descents, n >= 1 and 0 <= k <= n-1.

Original entry on oeis.org

1, 3, 1, 19, 16, 1, 211, 299, 65, 1, 3651, 7346, 3156, 246, 1, 90921, 237517, 160322, 28722, 917, 1, 3081513, 9903776, 9302567, 2864912, 245407, 3424, 1, 136407699, 520507423, 632274183, 288196659, 46261609, 2041965, 12861, 1
Offset: 1

Views

Author

Peter Bala, Jul 11 2011

Keywords

Comments

Let S_n denote the symmetric group on {1,2,...,n}. A permutation p_1p_2...p_n in S_n has a descent at position i (1 <= i <= n-1) if p_i > p_(i+1). The Eulerian numbers A008292 (with an offset of 0 in the column indexing) enumerate permutations by descents. We define a pair of permutations p_1p_2...p_n and q_1q_2...q_n to have a common descent at position i (1 <= i <= n-1) if both p_i > p_(i+1) and q_i > q_(i+1). For example, the permutations (3241) and (4231) in S_4 have common descents at positions i = 1 and i = 3. The table entry T(n,k) gives the number of pairs of permutations in the Cartesian product S_n x S_n with k common descents.
The generalized Stirling numbers associated with this triangle is A061691. See also A192722.

Examples

			The triangle begins
n/k|.....0.......1.......2......3....4.....5
============================================
..1|.....1
..2|.....3.......1
..3|....19......16.......1
..4|...211.....299......65......1
..5|..3651....7346....3156....246....1
..6|.90921..237517..160322..28722..917.....1
..
Row 3 entries T(3,0) = 19, T(3,1) = 16 and T(3,2) = 1 can be read from the following table:
============================================
Number of common descents in S_3 x S_3
============================================
.
...|.123...132...213...231...312...321
======================================
123|..0.....0.....0.....0.....0.....0
132|..0.....1.....0.....1.....0.....1
213|..0.....0.....1.....0.....1.....1
231|..0.....1.....0.....1.....0.....1
312|..0.....0.....1.....0.....1.....1
321|..0.....1.....1.....1.....1.....2
Matrix identity A192721 * A007318 = row reverse of A192722:
/...1................\ /..1..............\
|...3.....1...........||..1....1..........|
|..19....16.....1.....||..1....2....1.....|
|.211...299....65....1||..1....3....3....1|
|.....................||..................|
=
/...1...................\
|...4......1.............|
|..36.....18......1......|
|.576....432.....68.....1|
|........................|
		

Crossrefs

Cf. A000275 (first column), A001044 (row sums), A008292, A008459, A061691, A192722.

Programs

  • Maple
    #A192721
    #J = sum {n>=0} z^n/n!^2
    J := unapply(BesselJ(0, 2*I*sqrt(z)),z):
    G := (1-x)/(-x + J(z*(x-1))):
    Gser := simplify(series(G, z = 0, 12)):
    for n from 1 to 10 do
    P[n] := n!^2*sort(coeff(Gser, z, n)) od:
    for n from 1 to 10 do seq(coeff(P[n],x,k), k = 0..n-1) od;
    # gives sequence in triangular form
  • Mathematica
    max = 9; j[z_] := BesselJ[0, 2 I*Sqrt[z]]; g = (1 - x)/(-x + j[z*(x - 1)]); gser = Series[g, {z, 0, max}]; p[n_] := n!^2 Coefficient[ gser, z, n]; a[n_, k_] := Coefficient[ p[n], x, k]; Flatten[ Table[ a[n, k], {n, 1, max-1}, {k, 0, n-1}]] (* Jean-François Alcover, Dec 13 2011, after Maple *)

Formula

Generating function (Carlitz et al. 1976): Let J(z) = sum {n>=0} z^n/n!^2. Then (1-x)/(J(z*(x-1))-x) = 1 + sum {n>=1} (sum {k = 0..n-1} T(n,k)*x^k)*z^n/n!^2 = 1 + z + (3+x)*z^2/2!^2 + (19+16*x+x^2)*z^3/3!^2 + .... Define a polynomial sequence {p(n,x) }n>=0 by means of the generating function J(z)^x = sum {n>=0} p(n,x)*z^n/n!^2. The generalized Eulerian polynomials associated with the sequence {p(n,x)} as defined by [Koutras, 1994] are the polynomials sum {k = 0..n-1} T(n,k)*x^(n-k).
Relations with other sequences: The first column of the array (x*I-A008459)^-1 (I the identity matrix) is a sequence of rational functions whose numerator polynomials are the row generating polynomials for the present triangle. The change of variable x -> (x+1)/x followed by z -> x*z transforms the above bivariate generating function (1-x)/(J(z*(x-1))-x) into 1/(1+x-x*J(z)), which is the generating function for A192722. Equivalently, if we postmultiply the present triangle by Pascal's triangle A007318 we obtain the row reversed form of A192722: A192721 * A007318 = row reverse of A192722.
Row n sum = n!^2 = A001044(n).
First column [1,3,19,211,3651,...] = A000275 (apart from initial term).

A059797 Second in a series of arrays counting standard tableaux by partition type.

Original entry on oeis.org

2, 5, 5, 9, 16, 9, 14, 35, 35, 14, 20, 64, 90, 64, 20, 27, 105, 189, 189, 105, 27, 35, 160, 350, 448, 350, 160, 35, 44, 231, 594, 924, 924, 594, 231, 44, 54, 320, 945, 1728, 2100, 1728, 945, 320, 54, 65, 429, 1430, 3003, 4290, 4290, 3003, 1430, 429, 65
Offset: 0

Views

Author

Alford Arnold, Feb 22 2001

Keywords

Comments

The first array in the series is Pascal's triangle, A007318. The initial partition for each subsequent array in the series is chosen as described in A053445. When cells are squared, as in A008459, row sums yield 1, 2, 6, 24, ... (A000142). E.g., (1 + 16 + 36 + 16 + 1) + (25 + 25) = 70 + 50 = 120 using row five from A007318 and row two from this array.
Number of lattice paths from (0,0) to (n,k) in exactly n+k+2 steps. Moves allowed: (0,1), (0,-1), (-1,0) and (1,0). Paths must stay in Quadrant I but may touch the axes. - Juan Luis Vargas-Molina, Mar 03 2014

Examples

			a(5) = 16 because we can write T(2,2) = T(1,2) + T(1,1) + A007318(4,2) = 5 + 5 + 6.
   2;
   5,   5;
   9,  16,   9;
  14,  35,  35,  14;
  20,  64,  90,  64,  20;
  27, 105, 189, 189, 105,  27;
T(n,k) as a grid for the first few lattice points. Where T(0,0)=2 since there are two ways to get to (0,0) with "0+0+2" steps under the move restrictions.
     k = 0  1   2   3    4    5
T(0,k) = 2  5   9   14   20   27
T(1,k) = 5  16  35  64   105  160
T(2,k) = 9  35  90  189  350  594
T(3,k) = 14 64  189 448  924  1728
T(4,k) = 20 105 350 924  2100 4290
T(5,k) = 27 160 594 1728 4290 9504
- _Juan Luis Vargas-Molina_, Mar 03 2014
		

References

  • Stanton and White, Constructive Combinatorics, 1986, pp. 84, 91.

Crossrefs

Programs

  • Maple
    A059797 := proc(n,k) option remember; if n <0 or k<0 or k>n then 0; else procname(n-1,k-1)+procname(n-1,k)+binomial(n+2,k+1) ; end if; end proc:
    seq(seq( A059797(n,k),k=0..n),n=0..15) ; # R. J. Mathar, Jan 27 2011
  • Mathematica
    t[n_, k_] /; (n < 0 || k < 0 || k > n) = 0; t[n_, k_] := t[n, k] = t[n - 1, k - 1] + t[n - 1, k] + Binomial[n + 2, k + 1]; Flatten[ Table[ t[n, k], {n, 0, 9}, {k, 0, n}]] (* Jean-François Alcover, Dec 20 2011, after R. J. Mathar *)
    T[n_, k_] := (k + 1)(n + k +4) FactorialPower[k + n + 2, n]/(n! + (n + 1)!)
    MatrixForm[Table[T[n, k], {n, 0, 10}, {k, 0, 10}]] (* Juan Luis Vargas-Molina, Mar 03 2014 *)

Formula

T(row, col) = T(row-1, col-1) + T(row-1, col) + A007318(row+2, col+1). - Corrected by R. J. Mathar, Jan 27 2011
T(n,k) = (k+1)(n+k+4)FallingFactorial(n+k+2,n)/((n+1)!+n!) n,k >= 0. - Juan Luis Vargas-Molina, Mar 03 2014

A108558 Symmetric triangle, read by rows, where row n equals the (n+1)-th differences of the crystal ball sequence for D_n lattice, for n>=0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 9, 9, 1, 1, 20, 54, 20, 1, 1, 35, 180, 180, 35, 1, 1, 54, 447, 852, 447, 54, 1, 1, 77, 931, 2863, 2863, 931, 77, 1, 1, 104, 1724, 7768, 12550, 7768, 1724, 104, 1, 1, 135, 2934, 18186, 43128, 43128, 18186, 2934, 135, 1, 1, 170, 4685, 38200, 124850, 183356, 124850, 38200, 4685, 170, 1
Offset: 0

Views

Author

Paul D. Hanna, Jun 10 2005

Keywords

Comments

Row n equals the (n+1)-th differences of row n of the square array A108553. G.f. of row n equals: (1-x)^(n+1)*CBD_n(x), where CBD_n denotes the g.f. of the crystal ball sequence for D_n lattice.
From Peter Bala, Oct 23 2008: (Start)
Let D_n be the root lattice generated as a monoid by {+-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(D_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(D_n) [Ardila et al.]. See A108556 for the corresponding array of f-vectors for these type D_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A086645 for the array of h-vectors associated with type C_n polytopes.
The Hilbert transform of this array (as defined in A145905) equals A108553.
(End)

Examples

			G.f.s of initial rows of square array A108553 are:
  (1)/(1-x),
  (1 + x)/(1-x)^2,
  (1 + 2*x + x^2)/(1-x)^3,
  (1 + 9*x + 9*x^2 + x^3)/(1-x)^4,
  (1 + 20*x + 54*x^2 + 20*x^3 + x^4)/(1-x)^5,
  (1 + 35*x + 180*x^2 + 180*x^3 + 35*x^4 + x^5)/(1-x)^6.
Triangle begins:
  1;
  1,   1;
  1,   2,    1;
  1,   9,    9,     1;
  1,  20,   54,    20,      1;
  1,  35,  180,   180,     35,      1;
  1,  54,  447,   852,    447,     54,      1;
  1,  77,  931,  2863,   2863,    931,     77,     1;
  1, 104, 1724,  7768,  12550,   7768,   1724,   104,    1;
  1, 135, 2934, 18186,  43128,  43128,  18186,  2934,  135,   1;
  1, 170, 4685, 38200, 124850, 183356, 124850, 38200, 4685, 170, 1;
  ...
		

Crossrefs

Cf. A108553, A008353, A108558, A008459, A086645, A108556. Row n equals (n+1)-th differences of: A001844 (row 2), A005902 (row 3), A007204 (row 4), A008356 (row 5), A008358 (row 6), A008360 (row 7), A008362 (row 8), A008377 (row 9), A008379 (row 10).
T(2n,n) gives A305693.

Programs

  • Mathematica
    T[1, 0] = T[1, 1]=1; T[n_, k_] := Binomial[2n, 2k] - 2n Binomial[n-2, k-1];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 25 2018 *)
  • PARI
    T(n,k)=if(n
    				

Formula

T(n, k) = C(2*n, 2*k) - 2*n*C(n-2, k-1) for n>1, with T(0, 0)=1, T(1, 0)=T(1, 1)=1. Row sums equal A008353: 2^(n-1)*(2^n-n) for n>1.
From Peter Bala, Oct 23 2008: (Start)
O.g.f. : rational function N(x,z)/D(x,z), where N(x,z) = 1 - 3*(1 + x)*z + (3 + 2*x + 3*x^2)*z^2 - (1 + x)*(1 - 8*x + x^2)z^3 - 8*x*(1 + x^2)*z^4 + 2*x*(1 + x)*(1 - x)^2*z^5 and D(x,z) = ((1 - z)^2 - 2*x*z*(1 + z) + x^2*z^2)*(1 - z*(1 + x))^2.
For n >= 2, the row n generating polynomial equals 1/2*[(1 + sqrt(x))^(2n) + (1 - sqrt(x))^(2n)] - 2*n*x*(1 + x)^(n-2).
(End)

A169714 The function W_5(2n) (see Borwein et al. reference for definition).

Original entry on oeis.org

1, 5, 45, 545, 7885, 127905, 2241225, 41467725, 798562125, 15855173825, 322466645545, 6687295253325, 140927922498025, 3010302779775725, 65046639827565525, 1419565970145097545, 31249959913055650125, 693192670456484513025
Offset: 0

Views

Author

N. J. A. Sloane, Apr 17 2010

Keywords

Comments

Row sums of the fourth power of A008459. - Peter Bala, Mar 05 2013

Crossrefs

Programs

Formula

Sum_{n>=0} a(n)*x^n/n!^2 = (Sum_{n>=0} x^n/n!^2)^5 = BesselI(0, 2*sqrt(x))^5. - Peter Bala, Mar 05 2013
D-finite with recurrence: n^4*a(n) = (35*n^4 - 70*n^3 + 63*n^2 - 28*n + 5)*a(n-1) - (n-1)^2*(259*n^2 - 518*n + 285)*a(n-2) + 225*(n-2)^2*(n-1)^2*a(n-3). - Vaclav Kotesovec, Mar 09 2014
a(n) ~ 5^(2*n+5/2) / (16 * Pi^2 * n^2). - Vaclav Kotesovec, Mar 09 2014

A169715 The function W_6(2n) (see Borwein et al. reference for definition).

Original entry on oeis.org

1, 6, 66, 996, 18306, 384156, 8848236, 218040696, 5651108226, 152254667436, 4229523740916, 120430899525096, 3499628148747756, 103446306284890536, 3102500089343886696, 94219208840385966096, 2892652835496484004226, 89662253086458906345036
Offset: 0

Views

Author

N. J. A. Sloane, Apr 17 2010

Keywords

Comments

Row sums of the fifth power of A008459. - Peter Bala, Mar 05 2013
a(n)/6^(2n) is the probability that two throws of n 6-sided dice will give the same result - Henry Bottomley, Aug 30 2016

Crossrefs

Programs

  • Maple
    W := proc(n,s)
        local a,ai ;
        if s = 0 then
            return 1;
        end if;
        a := 0 ;
        for ai in combinat[partition](s/2) do
            if nops(ai) <= n then
                af := [op(ai),seq(0,i=1+nops(ai)..n)] ;
                a := a+combinat[numbperm](af)*(combinat[multinomial](s/2,op(ai)))^2 ;
            end if ;
        end do;
        a ;
    end proc:
    A169715 := proc(n)
        W(6,2*n) ;
    end proc: # R. J. Mathar, Mar 27 2012
  • Mathematica
    a[n_] := SeriesCoefficient[BesselI[0, 2*Sqrt[x]]^6, {x, 0, n}]*n!^2; Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Dec 30 2013, after Peter Bala *)
    max = 17; Total /@ MatrixPower[Table[Binomial[n, k]^2, {n, 0, max}, {k, 0, max}], 5] (* Jean-François Alcover, Mar 24 2015, after Peter Bala *)

Formula

Sum_{n>=0} a(n)*x^n/n!^2 = (Sum_{n>=0} x^n/n!^2)^6 = BesselI(0, 2*sqrt(x))^6. - Peter Bala, Mar 05 2013
Recurrence: n^5*a(n) = 2*(2*n-1)*(14*n^4 - 28*n^3 + 28*n^2 - 14*n + 3)*a(n-1) - 4*(n-1)^3*(196*n^2 - 392*n + 255)*a(n-2) + 1152*(n-2)^2*(n-1)^2*(2*n-3)*a(n-3). - Vaclav Kotesovec, Mar 09 2014
a(n) ~ 3^(2*n+3) * 4^(n-1) / (Pi*n)^(5/2). - Vaclav Kotesovec, Mar 09 2014

A067804 Triangle read by rows: T(n,k) is the number of walks (each step +-1) of length 2n which have a cumulative value of 0 last at step 2k.

Original entry on oeis.org

1, 2, 2, 6, 4, 6, 20, 12, 12, 20, 70, 40, 36, 40, 70, 252, 140, 120, 120, 140, 252, 924, 504, 420, 400, 420, 504, 924, 3432, 1848, 1512, 1400, 1400, 1512, 1848, 3432, 12870, 6864, 5544, 5040, 4900, 5040, 5544, 6864, 12870, 48620, 25740, 20592, 18480
Offset: 0

Views

Author

Henry Bottomley, Feb 07 2002

Keywords

Comments

Since the triangle is symmetric, the probability that a one-dimensional random walk returns to the origin at all in the steps m through to 2m is 1/2 (for m odd).
Diagonal sums are A106183. - Paul Barry, Apr 24 2005

Examples

			Triangle begins:
    1;
    2,   2;
    6,   4,   6;
   20,  12,  12,  20;
   70,  40,  36,  40,  70;
  252, 140, 120, 120, 140, 252;
  ...
For a walk of length 4 (=2*2), 6 are only ever 0 at step 0, 4 are zero at step 2 but not step 4 and 6 are 0 at step 4.
For n=3,k=2, T(3,2)=12 since there are 12 monotonic paths from (0,0) to (2,2) and then on to (3,3). Using E for eastward steps and N for northward steps, the 12 paths are given by EENNNE, ENENNE, ENNENE, NNEENE, NENENE, NEENNE, EENNEN, ENENEN, ENNEEN, NNEEEN, NENEEN, NEENEN. - _Dennis P. Walsh_, Mar 23 2012
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 79, ex. 3f.

Crossrefs

Columns include A000984, A028329. Central diagonal is A002894.

Programs

  • Magma
    /* As triangle */ [[Binomial(2*k, k)*Binomial(2*n-2*k, n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 19 2015
  • Mathematica
    Table[Table[Binomial[2k,k]Binomial[2(n-k),n-k],{k,0,n}],{n,0,10}]//Grid  (* Geoffrey Critzer, Jun 30 2013 *)
    T[ n_, k_] := SeriesCoefficient[ D[ InverseJacobiSN[2 x, m] / 2, x], {x, 0, 2 n}, {m, 0, k}]; (* Michael Somos, May 06 2017 *)
  • PARI
    T(n, k) = binomial(2*k, k) * binomial(2*n-2*k, n-k) /* Michael Somos, Aug 20 2005 */
    

Formula

T(n, k) = C(2k, k)*C(2n-2k, n-k) = C(2n, n)*C(n, k)^2/C(2n, 2k) = A000984(k)*A000984(n-k) = A000984(n)*A008459(n, k)/A007318(2n, 2k).
Row sums are 4^n = A000302(n).
G.f.: A(x,y) = 1/sqrt((1-4*x)*(1-4*x*y)). - Vladeta Jovovic, Dec 12 2003
Sum{k>=0} T(n, k)*(-3)^k = (-4)^n * A002426(n). Sum_{k>=0} T(n, k)/(2*k+1) = 2^(4*n)/((2*n+1)*C(2*n, n)). - Philippe Deléham, Dec 31 2003
O.g.f.: A(x,y) = 1 + x*d/dx(log(B(x,y))), where B(x,y) is the o.g.f. of A120406. - Peter Bala, Jul 17 2015

A110608 Number triangle T(n,k) = binomial(n,k)*binomial(2n,n-k).

Original entry on oeis.org

1, 2, 1, 6, 8, 1, 20, 45, 18, 1, 70, 224, 168, 32, 1, 252, 1050, 1200, 450, 50, 1, 924, 4752, 7425, 4400, 990, 72, 1, 3432, 21021, 42042, 35035, 12740, 1911, 98, 1, 12870, 91520, 224224, 244608, 127400, 31360, 3360, 128, 1, 48620, 393822, 1145664, 1559376
Offset: 0

Views

Author

Paul Barry, Jul 30 2005

Keywords

Comments

First column is A000984. Second column is A110609 = n^2*A000108. Row sums are A005809.

Examples

			Triangle begin
n\k|   0     1     2    3   4   5
---------------------------------
0  |   1
1  |   2     1
2  |   6     8     1
3  |  20    45    18    1
4  |  70   224   168   32   1
5  | 252  1050  1200  450  50   1
...
		

Crossrefs

Cf. A000108, A000984, A005809 (row sums), A008459, A110609 (column 2), A120986.

Programs

  • Mathematica
    Flatten[Table[Table[Binomial[n,k]Binomial[2n,n-k],{k,0,n}],{n,0,10}]] (* Harvey P. Dale, Aug 10 2011 *)
  • Maxima
    B(x,y):=(sqrt(-x*(4*x^2*y^3+(-12*x^2-8*x)*y^2+(12*x^2-20*x+4)*y-4*x^2+x))/(2*3^(3/2))-(x*(18*y+9)-2)/54)^(1/3);
    C(x,y):=-B(x,y)-(x*(3*y-3)+1)/(9*B(x,y))-1/3;
    A(x,y):=x*diff(C(x,y),x)*(-1/C(x,y)+1/(1+C(x,y)));
    taylor(A(x,y),x,0,7,y,0,7); /* Vladimir Kruchinin, Sep 24 2018 */
  • PARI
    for(n=0,10, for(k=0,n, print1(binomial(n,k)*binomial(2*n,n-k), ", "))) \\ G. C. Greubel, Sep 01 2017
    

Formula

From Peter Bala, Oct 13 2015: (Start)
n-th row polynomial R(n,t) = [x^n] ( (1 + t*x)*(1 + x)^2 )^n.
Cf. A008459, whose n-th row polynomial is equal to [x^n] ( (1 + t*x)*(1 + x) )^n.
exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (2 + t)*x + (5 + 6*t + t^2)*x^2 + ... is the o.g.f. for A120986. (End)
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