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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A002895 Domb numbers: number of 2n-step polygons on diamond lattice.

Original entry on oeis.org

1, 4, 28, 256, 2716, 31504, 387136, 4951552, 65218204, 878536624, 12046924528, 167595457792, 2359613230144, 33557651538688, 481365424895488, 6956365106016256, 101181938814289564, 1480129751586116848, 21761706991570726096, 321401321741959062016
Offset: 0

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Author

Keywords

Comments

a(n) is the (2n)th moment of the distance from the origin of a 4-step random walk in the plane. - Peter M.W. Gill (peter.gill(AT)nott.ac.uk), Mar 03 2004
Row sums of the cube of A008459. - Peter Bala, Mar 05 2013
Conjecture: Let D(n) be the (n+1) X (n+1) Hankel-type determinant with (i,j)-entry equal to a(i+j) for all i,j = 0..n. Then the number D(n)/12^n is always a positive odd integer. - Zhi-Wei Sun, Aug 14 2013
It appears that the expansions exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 4*x + 22*x^2 + 152*x^3 + 1241*x^4 + ... and exp( Sum_{n >= 1} 1/4*a(n)*x^n/n ) = 1 + x + 4*x^2 + 25*x^3 + 199*x^4 + ... have integer coefficients. See A267219. - Peter Bala, Jan 12 2016
This is one of the Apéry-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Named after the British-Israeli theoretical physicist Cyril Domb (1920-2012). - Amiram Eldar, Mar 20 2021

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.

Programs

  • Maple
    A002895 := n -> add(binomial(n,k)^2*binomial(2*n-2*k,n-k)*binomial(2*k,k), k=0..n): seq(A002895(n), n=0..25); # Wesley Ivan Hurt, Dec 20 2015
    A002895 := n -> binomial(2*n,n)*hypergeom([1/2, -n, -n, -n],[1, 1, 1/2 - n], 1):
    seq(simplify(A002895(n)), n=0..19); # Peter Luschny, May 23 2017
  • Mathematica
    Table[Sum[Binomial[n,k]^2 Binomial[2n-2k,n-k]Binomial[2k,k],{k,0,n}], {n,0,30}] (* Harvey P. Dale, Aug 15 2011 *)
    a[n_] = Binomial[2*n, n]*HypergeometricPFQ[{1/2, -n, -n, -n}, {1, 1, 1/2-n}, 1]; (* or *) a[n_] := SeriesCoefficient[BesselI[0, 2*Sqrt[x]]^4, {x, 0, n}]*n!^2; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Dec 30 2013, after Vladeta Jovovic *)
    max = 19; Total /@ MatrixPower[Table[Binomial[n, k]^2, {n, 0, max}, {k, 0, max}], 3] (* Jean-François Alcover, Mar 24 2015, after Peter Bala *)
  • PARI
    C=binomial;
    a(n) = sum(k=0,n, C(n,k)^2 * C(2*n-2*k,n-k) * C(2*k,k) );
    /* Joerg Arndt, Apr 19 2013 */
    
  • Python
    from math import comb
    def A002895(n): return (sum(comb(n,k)**2*comb(n-k<<1,n-k)*comb(m:=k<<1,k) for k in range(n+1>>1))<<1) + (0 if n&1 else comb(n,n>>1)**4) # Chai Wah Wu, Jun 17 2025

Formula

a(n) = Sum_{k=0..n} binomial(n, k)^2 * binomial(2n-2k, n-k) * binomial(2k, k).
D-finite with recurrence: n^3*a(n) = 2*(2*n-1)*(5*n^2-5*n+2)*a(n-1) - 64*(n-1)^3*a(n-2). - Vladeta Jovovic, Jul 16 2004
Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0, 2*sqrt(x))^4. - Vladeta Jovovic, Aug 01 2006
G.f.: hypergeom([1/6, 1/3],[1],108*x^2/(1-4*x)^3)^2/(1-4*x). - Mark van Hoeij, Oct 29 2011
From Zhi-Wei Sun, Mar 20 2013: (Start)
Via the Zeilberger algorithm, Zhi-Wei Sun proved that:
(1) 4^n*a(n) = Sum_{k = 0..n} (binomial(2k,k)*binomial(2(n-k),n-k))^3/ binomial(n,k)^2,
(2) a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*binomial(2k,n)*binomial(2k,k)* binomial(2(n-k),n-k). (End)
a(n) ~ 2^(4*n+1)/((Pi*n)^(3/2)). - Vaclav Kotesovec, Aug 20 2013
G.f. y=A(x) satisfies: 0 = x^2*(4*x - 1)*(16*x - 1)*y''' + 3*x*(128*x^2 - 30*x + 1)*y'' + (448*x^2 - 68*x + 1)*y' + 4*(16*x - 1)*y. - Gheorghe Coserea, Jun 26 2018
a(n) = Sum_{p+q+r+s=n} (n!/(p!*q!*r!*s!))^2 with p,q,r,s >= 0. See Verrill, p. 5. - Peter Bala, Jan 06 2020
From Peter Bala, Jul 25 2024: (Start)
a(n) = 2*Sum_{k = 1..n} (k/n)*binomial(n, k)^2*binomial(2*n-2*k, n-k)* binomial(2*k, k) for n >= 1.
a(n-1) = (1/2)*Sum_{k = 1..n} (k/n)^3*binomial(n, k)^2*binomial(2*n-2*k, n-k)* binomial(2*k, k) for n >= 1. Cf. A081085. (End)

Extensions

More terms from Vladeta Jovovic, Mar 11 2003

A063007 T(n,k) = binomial(n,k)*binomial(n+k,k), 0 <= k <= n, triangle read by rows.

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 12, 30, 20, 1, 20, 90, 140, 70, 1, 30, 210, 560, 630, 252, 1, 42, 420, 1680, 3150, 2772, 924, 1, 56, 756, 4200, 11550, 16632, 12012, 3432, 1, 72, 1260, 9240, 34650, 72072, 84084, 51480, 12870, 1, 90, 1980, 18480, 90090, 252252, 420420, 411840, 218790, 48620
Offset: 0

Views

Author

Henry Bottomley, Jul 02 2001

Keywords

Comments

T(n,k) is the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's Cluster algebra of finite type B_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of triangle A008459 (squares of binomial coefficients). For example, x^2+6*x+6 = y^2+4*y+1. - Paul Boddington, Mar 07 2003
T(n,k) is the number of lattice paths from (0,0) to (n,n) using steps E=(1,0), N=(0,1) and D=(1,1) (i.e., bilateral Schroeder paths), having k N=(0,1) steps. E.g. T(2,0)=1 because we have DD; T(2,1) = 6 because we have NED, NDE, EDN, END, DEN and DNE; T(2,2)=6 because we have NNEE, NENE, NEEN, EENN, ENEN and ENNE. - Emeric Deutsch, Apr 20 2004
Another version of [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] DELTA [0, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...] = 1; 1, 0; 1, 2, 0; 1, 6, 6, 0; 1, 12, 30, 20, 0; ..., where DELTA is the operator defined in A084938. - Philippe Deléham Apr 15 2005
Terms in row n are the coefficients of the Legendre polynomial P(n,2x+1) with increasing powers of x.
From Peter Bala, Oct 28 2008: (Start)
Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type B_n (a cyclohedron) [Fomin & Reading, p.60]. See A008459 for the corresponding h-vectors for associahedra of type B_n and A001263 and A033282 respectively for the h-vectors and f-vectors for associahedra of type A_n.
An alternative description of this triangle in terms of f-vectors is as follows. Let A_n be the root lattice generated as a monoid by {e_i - e_j: 0 <= i,j <= n+1}. Let P(A_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the f-vectors of a unimodular triangulation of P(A_n) [Ardila et al.]. A008459 is the corresponding array of h-vectors for these type A_n polytopes. See A127674 (without the signs) for the array of f-vectors for type C_n polytopes and A108556 for the array of f-vectors associated with type D_n polytopes.
The S-transform on the ring of polynomials is the linear transformation of polynomials that is defined on the basis monomials x^k by S(x^k) = binomial(x,k) = x(x-1)...(x-k+1)/k!. Let P_n(x) denote the S-transform of the n-th row polynomial of this array. In the notation of [Hetyei] these are the Stirling polynomials of the type B associahedra. The first few values are P_1(x) = 2*x + 1, P_2(x) = 3*x^2 + 3*x + 1 and P_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials P_n(-x) satisfy a Riemann hypothesis. See A142995 for further details. The sequence of values P_n(k) for k = 0,1,2,3, ... produces the n-th row of A108625. (End)
This is the row reversed version of triangle A104684. - Wolfdieter Lang, Sep 12 2016
T(n, k) is also the number of (n-k)-dimensional faces of a convex n-dimensional Lipschitz polytope of real functions f defined on the set X = {1, 2, ..., n+1} which satisfy the condition f(n+1) = 0 (see Gordon and Petrov). - Stefano Spezia, Sep 25 2021
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial ((x+1)*(x+2)*(x+3)*...*(x+n) / n!)^2 in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 09 2022
Chapoton's observation above is correct: the precise expansion is ((x+1)*(x+2)*(x+3)*...*(x+n)/ n!)^2 = Sum_{k = 0..n} (-1)^k*T(n,n-k)*binomial(x+2*n-k, 2*n-k), as can be verified using the WZ algorithm. For example, n = 3 gives ((x+1)*(x+2)*(x+3)/3!)^2 = 20*binomial(x+6,6) - 30*binomial(x+5,5) + 12*binomial(x+4,4) - binomial(x+3,3). - Peter Bala, Jun 24 2023

Examples

			The triangle T(n, k) starts:
  n\k 0  1    2     3     4      5      6      7      8     9
  0:  1
  1:  1  2
  2:  1  6    6
  3:  1 12   30    20
  4:  1 20   90   140    70
  5:  1 30  210   560   630    252
  6:  1 42  420  1680  3150   2772    924
  7:  1 56  756  4200 11550  16632  12012   3432
  8:  1 72 1260  9240 34650  72072  84084  51480  12870
  9:  1 90 1980 18480 90090 252252 420420 411840 218790 48620
... reformatted by _Wolfdieter Lang_, Sep 12 2016
From _Petros Hadjicostas_, Jul 11 2020: (Start)
Its inverse (from Table II, p. 92, in Ser's book) is
   1;
  -1/2,  1/2;
   1/3, -1/2,    1/6;
  -1/4,  9/20,  -1/4,   1/20;
   1/5, -2/5,    2/7,  -1/10,  1/70;
  -1/6,  5/14, -25/84,  5/36, -1/28,  1/252;
   1/7, -9/28,  25/84, -1/6,   9/154, -1/84, 1/924;
   ... (End)
		

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 366.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, Table I, p. 92.
  • D. Zagier, Integral solutions of Apery-like recurrence equations, in: Groups and Symmetries: from Neolithic Scots to John McKay, CRM Proc. Lecture Notes 47, Amer. Math. Soc., Providence, RI, 2009, pp. 349-366.

Crossrefs

See A331430 for an essentially identical triangle, except with signed entries.
Columns include A000012, A002378, A033487 on the left and A000984, A002457, A002544 on the right.
Main diagonal is A006480.
Row sums are A001850. Alternating row sums are A033999.
Cf. A033282 (f-vectors type A associahedra), A108625, A080721 (f-vectors type D associahedra).
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

Programs

  • Haskell
    a063007 n k = a063007_tabl !! n !! k
    a063007_row n = a063007_tabl !! n
    a063007_tabl = zipWith (zipWith (*)) a007318_tabl a046899_tabl
    -- Reinhard Zumkeller, Nov 18 2014
    
  • Magma
    /* As triangle: */ [[Binomial(n,k)*Binomial(n+k,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 03 2015
  • Maple
    p := (n,x) -> orthopoly[P](n,1+2*x): seq(seq(coeff(p(n,x),x,k), k=0..n), n=0..9);
  • Mathematica
    Flatten[Table[Binomial[n, k]Binomial[n + k, k], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Dec 24 2011 *)
    Table[CoefficientList[Hypergeometric2F1[-n, n + 1, 1, -x], x], {n, 0, 9}] // Flatten
    (* Peter Luschny, Mar 09 2018 *)
  • PARI
    {T(n, k) = local(t); if( n<0, 0, t = (x + x^2)^n; for( k=1, n, t=t'); polcoeff(t, k) / n!)} /* Michael Somos, Dec 19 2002 */
    
  • PARI
    {T(n, k) = binomial(n, k) * binomial(n+k, k)} /* Michael Somos, Sep 22 2013 */
    
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, (n+k)! / (k!^2 * (n-k)!))} /* Michael Somos, Sep 22 2013 */
    

Formula

T(n, k) = (n+k)!/(k!^2*(n-k)!) = T(n-1, k)*(n+k)/(n-k) = T(n, k-1)*(n+k)*(n-k+1)/k^2 = T(n-1, k-1)*(n+k)*(n+k-1)/k^2.
binomial(x, n)^2 = Sum_{k>=0} T(n,k) * binomial(x, n+k). - Michael Somos, May 11 2012
T(n, k) = A109983(n, k+n). - Michael Somos, Sep 22 2013
G.f.: G(t, z) = 1/sqrt(1-2*z-4*t*z+z^2). Row generating polynomials = P_n(1+2z), i.e., T(n, k) = [z^k] P_n(1+2*z), where P_n are the Legendre polynomials. - Emeric Deutsch, Apr 20 2004
Sum_{k>=0} T(n, k)*A000172(k) = Sum_{k>=0} T(n, k)^2 = A005259(n). - Philippe Deléham, Jun 08 2005
1 + z*d/dz(log(G(t,z))) = 1 + (1 + 2*t)*z + (1 + 8*t + 8*t^2)*z^2 + ... is the o.g.f. for a signed version of A127674. - Peter Bala, Sep 02 2015
If R(n,t) denotes the n-th row polynomial then x^3 * exp( Sum_{n >= 1} R(n,t)*x^n/n ) = x^3 + (1 + 2*t)*x^4 + (1 + 5*t + 5*t^2)*x^5 + (1 + 9*t + 21*t^2 + 14*t^3)*x^6 + ... is an o.g.f for A033282. - Peter Bala, Oct 19 2015
P(n,x) := 1/(1 + x)*Integral_{t = 0..x} R(n,t) dt are (modulo differences of offset) the row polynomials of A033282. - Peter Bala, Jun 23 2016
From Peter Bala, Mar 09 2018: (Start)
R(n,x) = Sum_{k = 0..n} binomial(2*k,k)*binomial(n+k,n-k)*x^k.
R(n,x) = Sum_{k = 0..n} binomial(n,k)^2*x^k*(1 + x)^(n-k).
n*R(n,x) = (1 + 2*x)*(2*n - 1)*R(n-1,x) - (n - 1)*R(n-2,x).
R(n,x) = (-1)^n*R(n,-1 - x).
R(n,x) = 1/n! * (d/dx)^n ((x^2 + x)^n). (End)
The row polynomials are R(n,x) = hypergeom([-n, n + 1], [1], -x). - Peter Luschny, Mar 09 2018
T(n,k) = C(n+1,k)*A009766(n,k). - Bob Selcoe, Jan 18 2020 (Connects this triangle with the Catalan triangle. - N. J. A. Sloane, Jan 18 2020)
If we let A(n,k) = (-1)^(n+k)*(2*k+1)*(n*(n-1)*...*(n-(k-1)))/((n+1)*...*(n+(k+1))) for n >= 0 and k = 0..n, and we consider both T(n,k) and A(n,k) as infinite lower triangular arrays, then they are inverses of one another. (Empty products are by definition 1.) See the example below. The rational numbers |A(n,k)| appear in Table II on p. 92 in Ser's (1933) book. - Petros Hadjicostas, Jul 11 2020
From Peter Bala, Nov 28 2021: (Start)
Row polynomial R(n,x) = Sum_{k >= n} binomial(k,n)^2 * x^(k-n)/(1+x)^(k+1) for x > -1/2.
R(n,x) = 1/(1 + x)^(n+1) * hypergeom([n+1, n+1], [1], x/(1 + x)).
R(n,x) = (1 + x)^n * hypergeom([-n, -n], [1], x/(1 + x)).
R(n,x) = hypergeom([(n+1)/2, -n/2], [1], -4*x*(1 + x)).
If we set R(-1,x) = 1, we can run the recurrence n*R(n,x) = (1 + 2*x)*(2*n - 1)*R(n-1,x) - (n - 1)*R(n-2,x) backwards to give R(-n,x) = R(n-1,x).
R(n,x) = [t^n] ( (1 + t)*(1 + x*(1 + t)) )^n. (End)
n*T(n,k) = (2*n-1)*T(n-1,k) + (4*n-2)*T(n-1,k-1) - (n-1)*T(n-2,k). - Fabián Pereyra, Jun 30 2022
From Peter Bala, Oct 07 2024: (Start)
n-th row polynomial R(n,x) = Sum_{k = 0..n} binomial(n, k) * x^k o (1 + x)^(n-k), where o denotes the black diamond product of power series as defined by Dukes and White (see Bala, Section 4.4, exercise 3).
Denote this triangle by T. Then T * transpose(T) = A143007, the square array of crystal ball sequences for the A_n X A_n lattices.
Let S denote the triangle ((-1)^(n+k)*T(n, k))n,k >= 0, a signed version of this triangle. Then S^(-1) * T = A007318, Pascal's triangle; it appears that T * S^(-1) = A110098.
T = A007318 * A115951. (End)

A108625 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the A_n lattice.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 13, 19, 7, 1, 1, 21, 55, 37, 9, 1, 1, 31, 131, 147, 61, 11, 1, 1, 43, 271, 471, 309, 91, 13, 1, 1, 57, 505, 1281, 1251, 561, 127, 15, 1, 1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1, 1, 91, 1405, 6637, 12559, 11253, 5321, 1415, 217, 19, 1
Offset: 0

Views

Author

Paul D. Hanna, Jun 12 2005

Keywords

Comments

Compare to the corresponding array A108553 of crystal ball sequences for D_n lattice.
From Peter Bala, Jul 18 2008: (Start)
Row reverse of A099608.
This array has a remarkable relationship with the constant zeta(2). The row, column and diagonal entries of the array occur in series acceleration formulas for zeta(2).
For the entries in row n we have zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k)). For example, n = 4 gives zeta(2) = 2*(1 - 1/4 + 1/9 - 1/16) + 1/(1*21) + 1/(4*21*131) + 1/(9*131*471) + ... . See A142995 for further details.
For the entries in column k we have zeta(2) = (1 + 1/4 + 1/9 + ... + 1/k^2) + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k)). For example, k = 4 gives zeta(2) = (1 + 1/4 + 1/9 + 1/16) + 2*(1/(1*9) - 1/(4*9*61) + 1/(9*61*309) - ... ). See A142999 for further details.
Also, as consequence of Apery's proof of the irrationality of zeta(2), we have a series acceleration formula along the main diagonal of the table: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n,n)*T(n-1,n-1)) = 5*(1/3 - 1/(2^2*3*19) + 1/(3^2*19*147) - ...).
There also appear to be series acceleration results along other diagonals. For example, for the main subdiagonal, calculation supports the result zeta(2) = 2 - Sum_{n >= 1} (-1)^(n+1)*(n^2+(2*n+1)^2)/(n^2*(n+1)^2*T(n,n-1)*T(n+1,n)) = 2 - 10/(2^2*7) + 29/(6^2*7*55) - 58/(12^2*55*471) + ..., while for the main superdiagonal we appear to have zeta(2) = 1 + Sum_{n >= 1} (-1)^(n+1)*((n+1)^2 + (2*n+1)^2)/(n^2*(n+1)^2*T(n-1,n)*T(n,n+1)) = 1 + 13/(2^2*5) - 34/(6^2*5*37) + 65/(12^2*37*309) - ... .
Similar series acceleration results hold for Apery's constant zeta(3) involving the crystal ball sequences for the product lattices A_n x A_n; see A143007 for further details. Similar results also hold between the constant log(2) and the crystal ball sequences of the hypercubic lattices A_1 x...x A_1 and between log(2) and the crystal ball sequences for lattices of type C_n ; see A008288 and A142992 respectively for further details. (End)
This array is the Hilbert transform of triangle A008459 (see A145905 for the definition of the Hilbert transform). - Peter Bala, Oct 28 2008

Examples

			Square array begins:
  1,   1,    1,     1,      1,       1,       1, ... A000012;
  1,   3,    5,     7,      9,      11,      13, ... A005408;
  1,   7,   19,    37,     61,      91,     127, ... A003215;
  1,  13,   55,   147,    309,     561,     923, ... A005902;
  1,  21,  131,   471,   1251,    2751,    5321, ... A008384;
  1,  31,  271,  1281,   4251,   11253,   25493, ... A008386;
  1,  43,  505,  3067,  12559,   39733,  104959, ... A008388;
  1,  57,  869,  6637,  33111,  124223,  380731, ... A008390;
  1,  73, 1405, 13237,  79459,  350683, 1240399, ... A008392;
  1,  91, 2161, 24691, 176251,  907753, 3685123, ... A008394;
  1, 111, 3191, 43561, 365751, 2181257, ...      ... A008396;
  ...
As a triangle:
  [0]  1
  [1]  1,  1
  [2]  1,  3,   1
  [3]  1,  7,   5,    1
  [4]  1, 13,  19,    7,    1
  [5]  1, 21,  55,   37,    9,    1
  [6]  1, 31, 131,  147,   61,   11,   1
  [7]  1, 43, 271,  471,  309,   91,  13,   1
  [8]  1, 57, 505, 1281, 1251,  561, 127,  15,  1
  [9]  1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1
       ...
Inverse binomial transform of rows yield rows of triangle A063007:
  1;
  1,  2;
  1,  6,   6;
  1, 12,  30,  20;
  1, 20,  90, 140,  70;
  1, 30, 210, 560, 630, 252; ...
Product of the g.f. of row n and (1-x)^(n+1) generates the symmetric triangle A008459:
  1;
  1,  1;
  1,  4,   1;
  1,  9,   9,   1;
  1, 16,  36,  16,  1;
  1, 25, 100, 100, 25, 1;
  ...
		

Crossrefs

Rows include: A003215 (row 2), A005902 (row 3), A008384 (row 4), A008386 (row 5), A008388 (row 6), A008390 (row 7), A008392 (row 8), A008394 (row 9), A008396 (row 10).
Cf. A063007, A099601 (n-th term of A_{2n} lattice), A108553.
Cf. A008459 (h-vectors type B associahedra), A145904, A145905.
Cf. A005258 (main diagonal), A108626 (antidiagonal sums).

Programs

  • Magma
    T:= func< n,k | (&+[Binomial(n,j)^2*Binomial(n+k-j,k-j): j in [0..k]]) >; // array
    A108625:= func< n,k | T(n-k,k) >; // antidiagonals
    [A108625(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Maple
    T := (n,k) -> binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1):
    seq(seq(simplify(T(n,k)),k=0..n),n=0..10); # Peter Luschny, Feb 10 2018
  • Mathematica
    T[n_, k_]:= HypergeometricPFQ[{-n, -k, n+1}, {1, 1}, 1] (* Michael Somos, Jun 03 2012 *)
  • PARI
    T(n,k)=sum(i=0,k,binomial(n,i)^2*binomial(n+k-i,k-i))
    
  • SageMath
    def T(n,k): return sum(binomial(n,j)^2*binomial(n+k-j, k-j) for j in range(k+1)) # array
    def A108625(n,k): return T(n-k, k) # antidiagonals
    flatten([[A108625(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n, k) = Sum_{i=0..k} C(n, i)^2 * C(n+k-i, k-i).
G.f. for row n: (Sum_{i=0..n} C(n, i)^2 * x^i)/(1-x)^(n+1).
Sum_{k=0..n} T(n-k, k) = A108626(n) (antidiagonal sums).
From Peter Bala, Jul 23 2008 (Start):
O.g.f. row n: 1/(1 - x)*Legendre_P(n,(1 + x)/(1 - x)).
G.f. for square array: 1/sqrt((1 - x)*((1 - t)^2 - x*(1 + t)^2)) = (1 + x + x^2 + x^3 + ...) + (1 + 3*x + 5*x^2 + 7*x^3 + ...)*t + (1 + 7*x + 19*x^2 + 37*x^3 + ...)*t^2 + ... . Cf. A142977.
Main diagonal is A005258.
Recurrence relations:
Row n entries: (k+1)^2*T(n,k+1) = (2*k^2+2*k+n^2+n+1)*T(n,k) - k^2*T(n,k-1), k = 1,2,3,... ;
Column k entries: (n+1)^2*T(n+1,k) = (2*k+1)*(2*n+1)*T(n,k) + n^2*T(n-1,k), n = 1,2,3,... ;
Main diagonal entries: (n+1)^2*T(n+1,n+1) = (11*n^2+11*n+3)*T(n,n) + n^2*T(n-1,n-1), n = 1,2,3,... .
Series acceleration formulas for zeta(2):
Row n: zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k));
Column k: zeta(2) = 1 + 1/2^2 + 1/3^2 + ... + 1/k^2 + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k));
Main diagonal: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,n-1)*T(n,n)).
Conjectural result for superdiagonals: zeta(2) = 1 + 1/2^2 + ... + 1/k^2 + Sum_{n >= 1} (-1)^(n+1) * (5*n^2 + 6*k*n + 2*k^2)/(n^2*(n+k)^2*T(n-1,n+k-1)*T(n,n+k)), k = 0,1,2... .
Conjectural result for subdiagonals: zeta(2) = 2*(1 - 1/2^2 + ... + (-1)^(k+1)/k^2) + (-1)^k*Sum_{n >= 1} (-1)^(n+1)*(5*n^2 + 4*k*n + k^2)/(n^2*(n+k)^2*T(n+k-1,n-1)*T(n+k,n)), k = 0,1,2... .
Conjectural congruences: the main superdiagonal numbers S(n) := T(n,n+1) appear to satisfy the supercongruences S(m*p^r - 1) = S(m*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers m and r. If p is prime of the form 4*n + 1 we can write p = a^2 + b^2 with a an odd number. Then calculation suggests the congruence S((p-1)/2) == 2*a^2 (mod p). (End)
From Michael Somos, Jun 03 2012: (Start)
T(n, k) = hypergeom([-n, -k, n + 1], [1, 1], 1).
T(n, n-1) = A208675(n).
T(n+1, n) = A108628(n). (End)
T(n, k) = binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1). - Peter Luschny, Feb 10 2018
From Peter Bala, Jun 23 2023: (Start)
T(n, k) = Sum_{i = 0..k} (-1)^i * binomial(n, i)*binomial(n+k-i, k-i)^2.
T(n, k) = binomial(n+k, k)^2 * hypergeom([-n, -k, -k], [-n - k, -n - k], 1). (End)
From Peter Bala, Jun 28 2023; (Start)
T(n,k) = the coefficient of (x^n)*(y^k)*(z^n) in the expansion of 1/( (1 - x - y)*(1 - z ) - x*y*z ).
T(n,k) = B(n, k, n) in the notation of Straub, equation 24.
The supercongruences T(n*p^r, k*p^r) == T(n*p^(r-1), k*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and k.
The formula T(n,k) = hypergeom([n+1, -n, -k], [1, 1], 1) allows the table indexing to be extended to negative values of n and k; clearly, we find that T(-n,k) = T(n-1,k) for all n and k. It appears that T(n,-k) = (-1)^n*T(n,k-1) for n >= 0, while T(n,-k) = (-1)^(n+1)*T(n,k-1) for n <= -1 [added Sep 10 2023: these follow from the identities immediately below]. (End)
T(n,k) = Sum_{i = 0..n} (-1)^(n+i) * binomial(n, i)*binomial(n+i, i)*binomial(k+i, i) = (-1)^n * hypergeom([n + 1, -n, k + 1], [1, 1], 1). - Peter Bala, Sep 10 2023
From G. C. Greubel, Oct 05 2023: (Start)
Let t(n,k) = T(n-k, k) (antidiagonals).
t(n, k) = Hypergeometric3F2([k-n, -k, n-k+1], [1,1], 1).
T(n, 2*n) = A363867(n).
T(3*n, n) = A363868(n).
T(2*n, 2*n) = A363869(n).
T(n, 3*n) = A363870(n).
T(2*n, 3*n) = A363871(n). (End)
T(n, k) = Sum_{i = 0..n} binomial(n, i)*binomial(n+i, i)*binomial(k, i). - Peter Bala, Feb 26 2024
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n, k) = A005259(n), the Apéry numbers associated with zeta(3). - Peter Bala, Jul 18 2024
From Peter Bala, Sep 21 2024: (Start)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*T(n, k) = binomial(2*n, n) = A000984(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n-1, n-k) = A376458(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(i, k) = A143007(n, i). (End)
From Peter Bala, Oct 12 2024: (Start)
The square array = A063007 * transpose(A007318).
Conjecture: for positive integer m, Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * T(m*n, k) = ((m+1)*n)!/( ((m-1)*n)!*n!^2) (verified up to m = 10 using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). (End)

A086645 Triangle read by rows: T(n, k) = binomial(2n, 2k).

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 15, 15, 1, 1, 28, 70, 28, 1, 1, 45, 210, 210, 45, 1, 1, 66, 495, 924, 495, 66, 1, 1, 91, 1001, 3003, 3003, 1001, 91, 1, 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1, 1, 153, 3060, 18564, 43758, 43758, 18564, 3060, 153, 1, 1, 190, 4845, 38760
Offset: 0

Views

Author

Philippe Deléham, Jul 26 2003

Keywords

Comments

Terms have the same parity as those of Pascal's triangle.
Coefficients of polynomials (1/2)*((1 + x^(1/2))^(2n) + (1 - x^(1/2))^(2n)).
Number of compositions of 2n having k parts greater than 1; example: T(3, 2) = 15 because we have 4+2, 2+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2, 3+3. - Philippe Deléham, May 18 2005
Number of binary words of length 2n - 1 having k runs of consecutive 1's; example: T(3,2) = 15 because we have 00101, 01001, 01010, 01011, 01101, 10001, 10010, 10011, 10100, 10110, 10111, 11001, 11010, 11011, 11101. - Philippe Deléham, May 18 2005
Let M_n be the n X n matrix M_n(i, j) = T(i, j-1); then for n > 0, det(M_n) = A000364(n), Euler numbers; example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385 = A000364(4). - Philippe Deléham, Sep 04 2005
Equals ConvOffsStoT transform of the hexagonal numbers, A000384: (1, 6, 15, 28, 45, ...); e.g., ConvOffs transform of (1, 6, 15, 28) = (1, 28, 70, 28, 1). - Gary W. Adamson, Apr 22 2008
From Peter Bala, Oct 23 2008: (Start)
Let C_n be the root lattice generated as a monoid by {+-2*e_i: 1 <= i <= n; +-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(C_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(C_n) [Ardila et al.]. See A127674 for (a signed version of) the corresponding array of f-vectors for these type C_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes.
The Hilbert transform of this triangle is A142992 (see A145905 for the definition of this term).
(End)
Diagonal sums: A108479. - Philippe Deléham, Sep 08 2009
Coefficients of Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
Generalized Narayana triangle for 4^n (or cosh(2x)). - Paul Barry, Sep 28 2010
Coefficients of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A086646(n,k). - R. J. Mathar, Mar 12 2013
Let E(y) = Sum_{n>=0} y^n/(2*n)! = cosh(sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence (2*n)! as defined in Wang and Wang. Cf. A103327. - Peter Bala, Aug 06 2013
Row 6, (1,66,495,924,495,66,1), plays a role in expansions of powers of the Dedekind eta function. See the Chan link, p. 534, and A034839. - Tom Copeland, Dec 12 2016

Examples

			From _Peter Bala_, Oct 23 2008: (Start)
The triangle begins
n\k|..0.....1.....2.....3.....4.....5.....6
===========================================
0..|..1
1..|..1.....1
2..|..1.....6.....1
3..|..1....15....15.....1
4..|..1....28....70....28.....1
5..|..1....45...210...210....45.....1
6..|..1....66...495...924...495....66.....1
...
(End)
From _Peter Bala_, Aug 06 2013: (Start)
Viewed as the generalized Riordan array (cosh(sqrt(y)), y) with respect to the sequence (2*n)! the column generating functions begin
1st col: cosh(sqrt(y)) = 1 + y/2! + y^2/4! + y^3/6! + y^4/8! + ....
2nd col: 1/2!*y*cosh(sqrt(y)) = y/2! + 6*y^2/4! + 15*y^3/6! + 28*y^4/8! + ....
3rd col: 1/4!*y^2*cosh(sqrt(y)) = y^2/4! + 15*y^3/6! + 70*y^4/8! + 210*y^5/10! + .... (End)
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.

Crossrefs

Cf. A008459, A108558, A127674, A142992. - Peter Bala, Oct 23 2008
Cf. A103327 (binomial(2n+1, 2k+1)), A103328 (binomial(2n, 2k+1)), A091042 (binomial(2n+1, 2k)). -Wolfdieter Lang, Jan 06 2013
Cf. A086646 (unsigned matrix inverse), A103327.
Cf. A034839.

Programs

  • Magma
    /* As triangle: */ [[Binomial(2*n, 2*k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 14 2016
  • Maple
    A086645:=(n,k)->binomial(2*n,2*k): seq(seq(A086645(n,k),k=0..n),n=0..12);
  • Mathematica
    Table[Binomial[2 n, 2 k], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 13 2016 *)
  • Maxima
    create_list(binomial(2*n,2*k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    {T(n, k) = binomial(2*n, 2*k)};
    
  • PARI
    {T(n, k) = sum( i=0, min(k, n-k), 4^i * binomial(n, 2*i) * binomial(n - 2*i, k-i))}; /* Michael Somos, May 26 2005 */
    

Formula

T(n, k) = (2*n)!/((2*(n-k))!*(2*k)!) row sums = A081294. COLUMNS: A000012, A000384
Sum_{k>=0} T(n, k)*A000364(k) = A000795(n) = (2^n)*A005647(n).
Sum_{k>=0} T(n, k)*2^k = A001541(n). Sum_{k>=0} T(n, k)*3^k = 2^n*A001075(n). Sum_{k>=0} T(n, k)*4^k = A083884(n). - Philippe Deléham, Feb 29 2004
O.g.f.: (1 - z*(1+x))/(x^2*z^2 - 2*x*z*(1+z) + (1-z)^2) = 1 + (1 + x)*z +(1 + 6*x + x^2)*z^2 + ... . - Peter Bala, Oct 23 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A081294(n), A001541(n), A090965(n), A083884(n), A099140(n), A099141(n), A099142(n), A165224(n), A026244(n) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 08 2009
Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
From Paul Barry, Sep 28 2010: (Start)
G.f.: 1/(1-x-x*y-4*x^2*y/(1-x-x*y)) = (1-x*(1+y))/(1-2*x*(1+y)+x^2*(1-y)^2);
E.g.f.: exp((1+y)*x)*cosh(2*sqrt(y)*x);
T(n,k) = Sum_{j=0..n} C(n,j)*C(n-j,2*(k-j))*4^(k-j). (End)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k) - T(n-2,k-2), with T(0,0)=T(1,0)=T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 26 2013
From Peter Bala, Sep 22 2021: (Start)
n-th row polynomial R(n,x) = (1-x)^n*T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A008459.
R(n,x) = Sum_{k = 0..n} binomial(n,2*k)*(4*x)^k*(1 + x)^(n-2*k).
R(n,x) = n*Sum_{k = 0..n} (n+k-1)!/((n-k)!*(2*k)!)*(4*x)^k*(1-x)^(n-k) for n >= 1. (End)

A062190 Coefficient triangle of certain polynomials N(5; m,x).

Original entry on oeis.org

1, 1, 6, 1, 14, 21, 1, 24, 84, 56, 1, 36, 216, 336, 126, 1, 50, 450, 1200, 1050, 252, 1, 66, 825, 3300, 4950, 2772, 462, 1, 84, 1386, 7700, 17325, 16632, 6468, 792, 1, 104, 2184, 16016, 50050, 72072, 48048, 13728, 1287, 1, 126, 3276, 30576, 126126, 252252, 252252, 123552, 27027, 2002
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

The e.g.f. of the m-th (unsigned) column sequence without leading zeros of the generalized (a=5) Laguerre triangle L(5; n+m,m)= A062138(n+m,m), n >= 0, is N(5; m,x)/(1-x)^(2*(m+3)), with the row polynomials N(5; m,x) := Sum_{k=0..m} a(m,k)*x^k.

Examples

			Triangle begins as:
  1;
  1,   6;
  1,  14,   21;
  1,  24,   84,    56;
  1,  36,  216,   336,    126;
  1,  50,  450,  1200,   1050,    252;
  1,  66,  825,  3300,   4950,   2772,     462;
  1,  84, 1386,  7700,  17325,  16632,    6468,    792;
  1, 104, 2184, 16016,  50050,  72072,   48048,  13728,   1287;
  1, 126, 3276, 30576, 126126, 252252,  252252, 123552,  27027,  2002;
  1, 150, 4725, 54600, 286650, 756756, 1051050, 772200, 289575, 50050, 3003;
		

Crossrefs

Family of polynomials (see A062145): A008459 (c=1), A132813 (c=2), A062196 (c=3), A062145 (c=4), A062264 (c=5), this sequence (c=6).
Columns k: A028557 (k=1), A104676 (k=2), A104677 (k=3), A104678 (k=4), A104679 (k=5), A104680 (k=6).
Diagonals: A000389 (k=n), A027818 (k=n-1), A104670 (k=n-2), A104671 (k=n-3), A104672 (k=n-4), A104673 (k=n-5), A104674 (k=n-6).
Cf. A003516 (row sums), A113894 (main diagonal).

Programs

  • Magma
    A062190:= func< n,k | Binomial(n,k)*Binomial(n+5,k) >;
    [A062190(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 28 2025
    
  • Maple
    A062190 := proc(m,k)
        add( (binomial(m, j)*(2*m+5-j)!/((m+5)!*(m-j)!))*(x^(m-j))*(1-x)^j,j=0..m) ;
        coeftayl(%,x=0,k) ;
    end proc: # R. J. Mathar, Nov 29 2015
  • Mathematica
    NN[5, m_, x_] := x^m*(2*m+5)!*Hypergeometric2F1[-m, -m, -2*m-5, (x-1)/x]/((m+5)!*m!); Table[CoefficientList[NN[5, m, x], x], {m, 0, 8}] // Flatten (* Jean-François Alcover, Sep 18 2013 *)
    A062190[n_,k_]:= Binomial[n,k]*Binomial[n+5,k];
    Table[A062190[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 28 2025 *)
  • SageMath
    def A062190(n,k): return binomial(n,k)*binomial(n+5,k)
    print(flatten([[A062190(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Feb 28 2025

Formula

T(m, k) = [x^k]N(5; m, x), with N(5; m, x) = ((1-x)^(2*(m+3)))*(d^m/dx^m)(x^m/(m!*(1-x)^(m+6))).
N(5; m, x) = Sum_{j=0..m} ((binomial(m, j)*(2*m+5-j)!/((m+5)!*(m-j)!))*(x^(m-j))*(1-x)^j).
N(5; m, x)= x^m*(2*m+5)! * 2F1(-m, -m; -2*m-5; (x-1)/x)/((m+5)!*m!). - Jean-François Alcover, Sep 18 2013
T(n, k) = binomial(n, k)*binomial(n+5, k). - G. C. Greubel, Feb 28 2025

A103371 Number triangle T(n,k) = C(n,n-k)*C(n+1,n-k).

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 12, 1, 5, 40, 60, 20, 1, 6, 75, 200, 150, 30, 1, 7, 126, 525, 700, 315, 42, 1, 8, 196, 1176, 2450, 1960, 588, 56, 1, 9, 288, 2352, 7056, 8820, 4704, 1008, 72, 1, 10, 405, 4320, 17640, 31752, 26460, 10080, 1620, 90, 1, 11, 550, 7425, 39600, 97020
Offset: 0

Views

Author

Paul Barry, Feb 03 2005

Keywords

Comments

Columns include A000027, A002411, A004302, A108647, A134287. Row sums are C(2n+1,n+1) or A001700.
T(n-1,k-1) is the number of ways to put n identical objects into k of altogether n distinguishable boxes. See the partition array A035206 from which this triangle arises after summing over all entries related to partitions with fixed part number k.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - Abdullahi Umar, Oct 02 2008
The o.g.f. of the (n+1)-th diagonal is given by G(n, x) = (n+1)*Sum_{k=1..n} A001263(n, k)*x^(k-1) / (1 - x)^(2*n+1), for n >= 1 and for n = 0 it is G(0, x) = 1/(1-x). - Wolfdieter Lang, Jul 31 2017

Examples

			The triangle T(n, k) begins:
n\k  0   1    2     3     4     5     6    7  8 9 ...
0:   1
1:   2   1
2:   3   6    1
3:   4  18   12     1
4:   5  40   60    20     1
5:   6  75  200   150    30     1
6:   7 126  525   700   315    42     1
7:   8 196 1176  2450  1960   588    56    1
8:   9 288 2352  7056  8820  4704  1008   72  1
9:  10 405 4320 17640 31752 26460 10080 1620 90 1
...  reformatted. - _Wolfdieter Lang_, Jul 31 2017
From _R. J. Mathar_, Mar 29 2013: (Start)
The matrix inverse starts
       1;
      -2,       1;
       9,      -6,      1;
     -76,      54,    -12,      1;
    1055,    -760,    180,    -20,   1;
  -21906,   15825,  -3800,    450, -30,   1;
  636447, -460026, 110775, -13300, 945, -42, 1; (End)
O.g.f. of 4th diagonal [4, 40,200, ...] is G(3, x) = 4*(1 + 3*x + x^2)/(1 - x)^7, from the n = 3 row [1, 3, 1] of A001263. See a comment above. - _Wolfdieter Lang_, Jul 31 2017
		

Crossrefs

Cf. A007318, A000894 (central terms), A132813 (mirrored).

Programs

  • Haskell
    a103371 n k = a103371_tabl !! n !! k
    a103371_row n = a103371_tabl !! n
    a103371_tabl = map reverse a132813_tabl
    -- Reinhard Zumkeller, Apr 04 2014
    
  • Magma
    /* As triangle */ [[Binomial(n,n-k)*Binomial(n+1,n-k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Aug 01 2017
    
  • Maple
    A103371 := (n,k) -> binomial(n,k)^2*(n+1)/(k+1);
    seq(print(seq(A103371(n, k), k=0..n)), n=0..7); # Peter Luschny, Oct 19 2011
  • Mathematica
    Flatten[Table[Binomial[n,n-k]Binomial[n+1,n-k],{n,0,10},{k,0,n}]] (* Harvey P. Dale, May 26 2014 *)
    CoefficientList[Series[Series[E^(x(1+y))(BesselI[0,2*x*Sqrt[y]]+BesselI[1,2*x*Sqrt[y]]/Sqrt[y]),{x,0,8}],{y,0,8}],{x,y}]*Range[0,8]! (* Natalia L. Skirrow, Apr 14 2025 *)
  • Maxima
    create_list(binomial(n,k)*binomial(n+1,k+1),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    for(n=0,10, for(k=0,n, print1(binomial(n,k)*binomial(n+1,k+1), ", "))) \\ G. C. Greubel, Nov 09 2018

Formula

Number triangle T(n, k) = C(n, n-k)*C(n+1, n-k) = C(n, k)*C(n+1, k+1); Column k of this triangle has g.f. Sum_{j=0..k} (C(k, j)*C(k+1, j) * x^(k+j))/(1-x)^(2*k+2); coefficients of the numerators are the rows of the reverse triangle C(n, k)*C(n+1, k).
T(n,k) = C(n, k)*Sum_{j=0..(n-k)} C(n-j, k). - Paul Barry, Jan 12 2006
T(n,k) = (n+1-k)*N(n+1,k+1), with N(n,k):=A001263(n,k), the Narayana triangle (with offset [1,1]).
O.g.f.: ((1-(1-y)*x)/sqrt((1-(1+y)*x)^2-4*x^2*y) -1)/2, (from o.g.f. of A001263, Narayana triangle). - Wolfdieter Lang, Nov 13 2007
From Peter Bala, Jan 24 2008: (Start)
Matrix product of A007318 and A122899.
O.g.f. for row n: (1-x)^n*P(n,1,0,(1+x)/(1-x)) = 1/(2*x)*(1-x)^(n+1)*( Legendre_P(n+1,(1+x)/(1-x)) - Legendre_P(n,(1+x)/(1-x)) ), where P(n,a,b,x) denotes the Jacobi polynomial.
O.g.f. for column k: x^k/(1-x)^(k+2)*P(k,0,1,(1+x)/(1-x)). Compare with A008459. (End)
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,1). (Cf. A194595, A197653, A197654). - Peter Luschny, Oct 20 2011
T(n,k) = A003056(n+1,k+1)*C(n,k)^2/(k+1). - Peter Luschny, Oct 29 2011
T(n,k) = A007318(n, k)*A135278(n, k), n >= k >= 0. - Wolfdieter Lang, Jul 31 2017
From Natalia L. Skirrow, Apr 14 2025: (Start)
T(n,k) = A008459(n,k) + n*N(n,k+1).
E.g.f.: e^(x*(1+y))*(I_0(2*x*sqrt(y)) + I_1(2*x*sqrt(y))/sqrt(y)), where I_n is the modified Bessel function of the first kind. (The I_0 contributes A008459(n,k), the I_1 contributes n*N(n,k+1))
O.g.f. for row n: (n+1)*2F1(-n,-n;2;y) = (n+1)*2F1(2+n,2+n;2;y)*(1-y)^(2*(n+1)) (by Euler's hypergeometric transformation); (n+1)*2F1(2+n,2+n;2;y) is the o.g.f. for row n of (k+n+1)!^2/(k!*(k+1)!*n!*(n+1)!), which is column n+1 of A132812.
O.g.f. for column k: 2F1(1+k,2+k;1;x)*x^k = 2F1(-k,-1-k;1;x)*x^k/(1-x)^(2+2*k). 2F1(-k,-1-k;1;x) is the kth row of A132813, the reflection of the kth row of this triangle.
O.g.f. for diagonal d (beginning at a(d,0)): (d+1)*x^d*2F1(d+1,d+2;2;x*y). 2F1(d+1,d+2;2;x) = 2F1(1-d,-d;2;x)/(1-x)^(2*d+1), numerator being the o.g.f. of row d of the Narayana triangle.
These respectively yield:
T(n,k) = Sum_{i=0..n+k} C(2*(n+1),i)*(-1)^i*A132812(n+1+k-i,n+1),
T(d+k,k) = Sum_{i=0..k} C(d-i+1+2*k,d-i)*T(k,k-i),
T(d+k,k) = Sum_{i=0..d} C(k-i + 2*d,k-i)*N(d,i+1)*(d+1).
E.g.f. for column k: 1F1(2+k;1;x)*x^k/k!.
E.g.f. for diagonal d: (d+1)*x^d*1F1(d+2;2;x*y)/d!. (End)

A062196 Triangle read by rows, T(n, k) = binomial(n, k)*binomial(n + 2, k).

Original entry on oeis.org

1, 1, 3, 1, 8, 6, 1, 15, 30, 10, 1, 24, 90, 80, 15, 1, 35, 210, 350, 175, 21, 1, 48, 420, 1120, 1050, 336, 28, 1, 63, 756, 2940, 4410, 2646, 588, 36, 1, 80, 1260, 6720, 14700, 14112, 5880, 960, 45, 1, 99, 1980, 13860, 41580, 58212, 38808, 11880, 1485, 55
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

Also the coefficient triangle of certain polynomials N(2; m,x) := Sum_{k=0..m} T(m,k)*x^k. The e.g.f. of the m-th (unsigned) column sequence without leading zeros of the generalized (a=2) Laguerre triangle L(2; n+m,m) = A062139(n+m,m), n >= 0, is N(2; m,x)/(1-x)^(3+2*m), with the row polynomials N(2; m,x).

Examples

			Triangle starts:
  n\k 0...1.....2......3..... 4.....;
  [0] 1;
  [1] 1,  3;
  [2] 1,  8,    6;
  [3] 1, 15,   30,    10;
  [4] 1, 24,   90,    80,    15;
  [5] 1, 35,  210,   350,   175,    21;
  [6] 1, 48,  420,  1120,  1050,   336,    28;
  [7] 1, 63,  756,  2940,  4410,  2646,   588,    36;
  [8] 1, 80, 1260,  6720, 14700, 14112,  5880,   960,   45;
  [9] 1, 99, 1980, 13860, 41580, 58212, 38808, 11880, 1485, 55.
		

Crossrefs

Family of polynomials (see A062145): A008459 (c=1), A132813 (c=2), this sequence (c=3), A062145 (c=4), A062264 (c=5), A062190 (c=6).
Sums include: A001791 (row), (-1)^n*A089849(n+1) (alternating sign row).
Diagonals: A000217 (k=n), A002417 (k=n-1), A001297 (k=n-2), A105946 (k=n-3), A105947 (k=n-4), A105948 (k=n-5), A107319 (k=n-6).
Columns: A005563 (k=1), A033487 (k=2), A027790 (k=3), A107395 (k=4), A107396 (k=5), A107397 (k=6), A107398 (k=7), A107399 (k=8).

Programs

  • Magma
    A062196:= func;
    [A062196(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 21 2025
    
  • Maple
    T := (n, k) -> binomial(n, k)*binomial(n + 2, k);
    seq(seq(T(n, k), k=0..n), n=0..9); # Peter Luschny, Sep 30 2021
  • Mathematica
    A062196[n_, k_]:= Binomial[n, k]*Binomial[n+2, k];
    Table[A062196[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 21 2025 *)
  • SageMath
    def A062196(n,k): return binomial(n,k)*binomial(n+2,k)
    print(flatten([[A062196(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Feb 21 2025

Formula

T(m, k) = [x^k] N(2; m, x), where N(2; m, x) = ((1-x)^(3+2*m))*(d^m/dx^m)(x^m/(m!*(1-x)^(m+3))).
N(2; m, x) = Sum_{j=0..m} ((binomial(m, j)*(2*m+2-j)!/((m+2)!*(m-j)!)*(x^(m-j)))*(1-x)^j).
T(n,m) = binomial(n, m)*(binomial(n+1, m) + binomial(n+1, m-1)). - Vladimir Kruchinin, Apr 06 2018
From G. C. Greubel, Feb 21 2025: (Start)
T(2*n, n) = (n+1)^2*A000108(n)*A000108(n+1).
T(2*n-1, n) = (4*n^2 - 1)*A000108(n-1)*A000108(n), n >= 1.
T(2*n+1, n) = (1/2)*binomial(n+2,2)*A000108(n+1)*A000108(n+2). (End)

Extensions

New name by Peter Luschny, Sep 30 2021

A062145 Triangle read by rows: T(n, k) = [z^k] P(n, z) where P(n, z) = Sum_{k=0..n} binomial(n, k) * Pochhammer(n - k + c, k) * z^k / k! and c = 4.

Original entry on oeis.org

1, 1, 4, 1, 10, 10, 1, 18, 45, 20, 1, 28, 126, 140, 35, 1, 40, 280, 560, 350, 56, 1, 54, 540, 1680, 1890, 756, 84, 1, 70, 945, 4200, 7350, 5292, 1470, 120, 1, 88, 1540, 9240, 23100, 25872, 12936, 2640, 165, 1, 108, 2376, 18480, 62370, 99792, 77616, 28512, 4455, 220
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

Coefficient triangle of certain polynomials N(3; m,x).

Examples

			As a square array:
    1,    1,     1,     1,     1,     1,    1,  1, ... A000012;
    4,   10,    18,    28,    40,    54,   70, 88, ... A028552;
   10,   45,   126,   280,   540,   945, 1540, ....... A105938;
   20,  140,   560,  1680,  4200,  9240, ............. A105939;
   35,  350,  1890,  7350, 23100, 62370, ............. A027803;
   56,  756,  5292, 25872, 99792, .................... A105940;
   84, 1470, 12936, 77616, ........................... A105942;
  120, 2640, 28512, .................................. A105943;
  165, 4455, 57015, .................................. A105944;
  ....;
As a triangle:
  1;
  1,   4;
  1,  10,   10;
  1,  18,   45,    20;
  1,  28,  126,   140,    35;
  1,  40,  280,   560,   350,    56;
  1,  54,  540,  1680,  1890,   756,    84;
  1,  70,  945,  4200,  7350,  5292,  1470,   120;
  1,  88, 1540,  9240, 23100, 25872, 12936,  2640,  165;
  1, 108, 2376, 18480, 62370, 99792, 77616, 28512, 4455, 220;
  ....;
		

Crossrefs

Family of polynomials: A008459 (c=1), A132813 (c=2), A062196 (c=3), this sequence (c=4), A062264 (c=5), A062190 (c=6).
Columns: A028552 (k=1), A105938 (k=2), A105939 (k=3), A027803 (k=4), A105940 (k=5), A105942 (k=6), A105943 (k=7), A105944 (k=8).
Diagonals: A000292 (k=n), A027800 (k=n-1), A107417 (k=n-2), A107418 (k=n-3), A107419 (k=n-4), A107420 (k=n-5), A107421 (k=n-6), A107422 (k=n-7).
Sums: A002054 (row).

Programs

  • Magma
    A062145:= func< n,k | Binomial(n,k)*Binomial(n+3,k) >;
    [A062145(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 07 2025
    
  • Mathematica
    NN[3, m_, x_] := x^m*(2*m+3)!*Hypergeometric2F1[-m, -m, -2*m-3, (x-1)/x]/( (m+3)!*m!); Table[CoefficientList[NN[3, m, x], x], {m, 0, 9}] // Flatten (* Jean-François Alcover, Sep 18 2013 *)
    P[c_, n_, z_] := Sum[Binomial[n, k] Pochhammer[n-k+c, k] z^k /k!, {k,0,n}];
    CL[c_] := Table[CoefficientList[P[c, n, z], z], {n, 0, 5}] // TableForm
    CL[4]  (* Peter Luschny, Feb 12 2024 *)
    A062145[n_,k_]:= Binomial[n,k]*Binomial[n+3,k];
    Table[A062145[n,k], {n,0,12},{k,0,n}]//Flatten (* G. C. Greubel, Mar 07 2025 *)
  • SageMath
    def A062145(n,k): return binomial(n,k)*binomial(n+3,k)
    print(flatten([[A062145(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Mar 07 2025

Formula

The e.g.f. of the m-th (unsigned) column sequence without leading zeros of the generalized (a=3) Laguerre triangle L(3; n+m, m) = A062137(n+m, m), n >= 0, is N(3; m, x)/(1-x)^(2*(m+2)), with the row polynomials N(3; m, x) := Sum_{k=0..m} a(m, k)*x^k.
N(3; m, x) := ((1-x)^(2*(m+2)))*(d^m/dx^m)(x^m/(m!*(1-x)^(m+4))); a(m, k) = [x^k]N(3; m, x).
N(3; m, x) = Sum_{j=0..m} ((binomial(m, j)*(2*m+3-j)!/((m+3)!*(m-j)!))*(x^(m-j))*(1-x)^j).
N(3; m, x)= x^m*(2*m+3)! * 2F1(-m, -m; -2*m-3; (x-1)/x)/((m+3)!*m!). - Jean-François Alcover, Sep 18 2013
From G. C. Greubel, Mar 07 2025 : (Start)
T(n, k) = binomial(n, k)*binomial(n+3, k).
T(2*n, n) = (1/2)*(n+1)^2*A000108(n)*A000108(n+2).
Sum_{k=0..n} (-1)^k*T(n, k) = (-1)^floor((n+2)/2)*(A047074(n+3) - A047074(n+ 2)). (End)

Extensions

New name by Peter Luschny, Feb 12 2024
More terms from G. C. Greubel, Mar 07 2025

A001249 Squares of tetrahedral numbers: a(n) = binomial(n+3,n)^2.

Original entry on oeis.org

1, 16, 100, 400, 1225, 3136, 7056, 14400, 27225, 48400, 81796, 132496, 207025, 313600, 462400, 665856, 938961, 1299600, 1768900, 2371600, 3136441, 4096576, 5290000, 6760000, 8555625, 10732176, 13351716, 16483600, 20205025, 24601600, 29767936, 35808256
Offset: 0

Views

Author

Keywords

Comments

Total area of all square and rectangular regions from an n+1 X n+1 grid. E.g., n = 2, there are 9 individual squares, 4 2 X 2's and 1 3 X 3, total area 9 + 16 + 9 = 34. The rectangular regions include 6 2 X 1's, 6 1 X 2's, 3 3 X 1's, 3 1 X 3's, 2 3 X 2's and 2 2 X 3's, total area 12 + 12 + 9 + 9 + 12 + 12 = 66, hence a(2) = 34 + 66 = 100. - Jon Perry, Jul 29 2003 [Index/grid size adjusted by Rick L. Shepherd, Jun 27 2017]
Number of 3 X 3 submatrices of an n+3 X n+3 matrix. - Rick L. Shepherd, Jun 27 2017
The inverse binomial transform gives row n=2 of A087107. - R. J. Mathar, Aug 31 2022

Crossrefs

Cf. A000290, A000292, A006542, A033455, A108674 (first diffs.), A086020 (partial sums).
Third column of triangle A008459.

Programs

Formula

From R. J. Mathar, Aug 19 2008: (Start)
a(n) = (A000292(n+1))^2.
O.g.f.: (1+x)(x^2+8x+1)/(1-x)^7. (End)
a(n) = C(n+4, 3)*C(n+4, 4)/(n+4) + A001303(n) = C(n+4, 3)*C(n+3, 3)/4 + A001303(n) = C(n+4, 6) + 3*C(n+5, 6) + C(n+6,6) + A001303(n). - Gary Detlefs, Aug 07 2013
-n^2*a(n) + (n+3)^2*a(n-1) = 0. - R. J. Mathar, Aug 15 2013
a(n) = 9*A040977(n-1) + A000579(n+6) + A000579(n+3). - R. J. Mathar, Aug 15 2013
a(n) = (n+3)*C(n+2, 2)*C(n+3, 3)/3. - Gary Detlefs, Jan 06 2014
a(n) = A000290(n+1)*A000290(n+2)*A000290(n+3)/36. - Bruno Berselli, Nov 12 2014
G.f. 2F1(4,4;1;x). - R. J. Mathar, Aug 09 2015
E.g.f.: exp(x)*(1 + 15*x + 69*x^2/2! + 147*x^3/3! + 162*x^4/4! + 90*x^5/5! + 20*x^6/6!). Computed from the o.g.f with the formulas (23) - (25) of the W. Lang link given in A060187. - Wolfdieter Lang, Jul 27 2017
From Amiram Eldar, Jan 24 2022: (Start)
Sum_{n>=0} 1/a(n) = 9*Pi^2 - 351/4.
Sum_{n>=0} (-1)^n/a(n) = 63/4 - 3*Pi^2/2. (End)
a(n) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+21*a(n-5)-7*a(n-6)+a(n-7). - Wesley Ivan Hurt, Aug 29 2022
a(n) = a(n-1)+A000217(n+1)*A000330(n+1). - J. M. Bergot, Aug 29 2022
a(n) = A002415(n+2)^2 - 20*A006857(n-1). - Yasser Arath Chavez Reyes, Nov 08 2024

A062264 Coefficient triangle of certain polynomials N(4; m,x).

Original entry on oeis.org

1, 1, 5, 1, 12, 15, 1, 21, 63, 35, 1, 32, 168, 224, 70, 1, 45, 360, 840, 630, 126, 1, 60, 675, 2400, 3150, 1512, 210, 1, 77, 1155, 5775, 11550, 9702, 3234, 330, 1, 96, 1848, 12320, 34650, 44352, 25872, 6336, 495, 1, 117, 2808, 24024, 90090, 162162, 144144, 61776, 11583, 715
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

The e.g.f. of the m-th (unsigned) column sequence without leading zeros of the generalized (a=4) Laguerre triangle L(4; n+m,m) = A062140(n+m,m), n >= 0, is N(4; m,x)/(1-x)^(5+2*m), with the row polynomials N(4; m,x) := Sum_{k=0..m} T(m,k)*x^k.

Examples

			Triangle begins as:
  1;
  1,   5;
  1,  12,   15;
  1,  21,   63,    35;
  1,  32,  168,   224,     70;
  1,  45,  360,   840,    630,    126;
  1,  60,  675,  2400,   3150,   1512,    210;
  1,  77, 1155,  5775,  11550,   9702,   3234,    330;
  1,  96, 1848, 12320,  34650,  44352,  25872,   6336,    495;
  1, 117, 2808, 24024,  90090, 162162, 144144,  61776,  11583,   715;
  1, 140, 4095, 43680, 210210, 504504, 630630, 411840, 135135, 20020, 1001;
		

Crossrefs

Family of polynomials (see A062145): A008459 (c=1), A132813 (c=2), A062196 (c=3), A062145 (c=4), this sequence (c=5), A062190 (c=6).
Columns: A028347 (k=2), A104473 (k=3), A104474 (k=4), A104475 (k=5), A027814 (k=6), A103604 (k=7), A104476 (k=8), A104478 (k=9).
Diagonals: A000332 (k=n), A027810 (k=n-1), A105249 (k=n-2), A105250 (k=n-3), A105251 (k=n-4), A105252 (k=n-5), A105253 (k=n-6), A105254 (k=n-7).
Sums: A002694 (row).

Programs

  • Magma
    A062264:= func< n,k | Binomial(n,k)*Binomial(n+4,k) >;
    [A062264(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 03 2025
    
  • Mathematica
    A062264[n_, k_]:= Binomial[n,k]*Binomial[n+4,k];
    Table[A062264[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 03 2025 *)
  • SageMath
    def A062264(n,k): return binomial(n,k)*binomial(n+4,k)
    print(flatten([[A062264(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Mar 03 2025

Formula

T(m, k) = [x^k] N(4; m, x), with N(4; m, x) = ((1-x)^(2*m+5))*(d^m/dx^m)((x^m)/(m!*(1-x)^(m+5))).
N(4; m, x) = Sum_{j=0..m} (binomial(m, j)*(2*m+4-j)!/((m+4)!*(m-j)!)*(x^(m-j))*(1-x)^j).
From G. C. Greubel, Mar 03 2025: (Start)
T(n, k) = binomial(n,k)*binomial(n+4,k).
Sum_{k=0..n} (-1)^k*T(n, k) = (1/4)*( (1+(-1)^n)*(-1)^((n+2)/2)*(n^2 + 5*n - 2)*Catalan((n+2)/2)/(n+1) + 8*(1-(-1)^n)*(-1)^((n+1)/2)*Catalan((n+1)/2) ). (End)
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