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.

Showing 1-2 of 2 results.

A248045 (2*(n-1))! * (2*n-1)! / (n * (n-1)!^3).

Original entry on oeis.org

1, 6, 120, 4200, 211680, 13970880, 1141620480, 111307996800, 12614906304000, 1629845894476800, 236475822507724800, 38072607423743692800, 6735922851893114880000, 1299070835722243584000000, 271245990498804460339200000, 60962536364606302461235200000
Offset: 1

Views

Author

Reinhard Zumkeller, Sep 30 2014

Keywords

Comments

Central terms in triangles of Lah numbers: a(n) = - A008297(2*n-1,n) = A105278(2*n-1,n) = A000891(n-1)*A000142(n) = A000894(n-1)*A000142(n-1).
a(n) = n * A204515(n-1). - Reinhard Zumkeller, Oct 19 2014

Crossrefs

Cf. A187535 (Central Lah numbers).

Programs

  • Haskell
    a248045 n = a000891 (n - 1) * a000142 n

Formula

n*a(n) = 4*(2*n-1)*(2*n-3)*a(n-1). - R. J. Mathar, Oct 07 2014

A247500 Triangle read by rows: T(n, k) = n!*binomial(n + 1, k)/(k + 1)!, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 12, 6, 1, 24, 60, 40, 10, 1, 120, 360, 300, 100, 15, 1, 720, 2520, 2520, 1050, 210, 21, 1, 5040, 20160, 23520, 11760, 2940, 392, 28, 1, 40320, 181440, 241920, 141120, 42336, 7056, 672, 36, 1, 362880, 1814400, 2721600, 1814400, 635040, 127008, 15120, 1080, 45, 1
Offset: 0

Views

Author

Peter Luschny, Oct 17 2014

Keywords

Comments

An alternative definition would have been: (n-k)!*N(n,k) where N(n,k) are the little Narayana numbers A090181(n,k). This adds a first column (1,0,0,...) to the triangle and amounts to (Gamma(n)*Gamma(n+1))/(Gamma(k)*Gamma(k+1)*Gamma(n-k+2)). - Peter Luschny, Jun 18 2015
From Peter Bala, Sep 03 2023: (Start)
Let E(y) = Sum_{n >= 0} y^n/(n+1)!. Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence n!*(n+1)! as defined in Wang and Wang.
Let B(y) = Sum_{n >= 0} y^n/(n!*(n+1)!) = 1/sqrt(y)*BesselI(1,2*sqrt(y)). A generating function for the triangle is E(y)*B(x*y) = 1 + (1 + x)*y/(1!*2!) + (2 + 3*x + x^2)*y^2/(2!*3!) + (6 + 12*x + 6*x^2 + x^3)*y^3/(3!*4!) + .... Cf. A105278 with a generating function exp(y)*B(x*y).
The n-th power of this array has a generating function E(y)^n*B(x*y). In particular, the matrix inverse has a generating function B(x*y)/E(y). (End)

Examples

			Triangle begins:
                      1;
                   1,    1;
                2,    3,    1;
             6,   12,    6,    1;
         24,   60,   40,   10,    1;
     120,  360,  300,  100,   15,    1;
  720, 2520, 2520, 1050,  210,   21,    1;
		

Crossrefs

Cf. A247499 (row sums), A008297.
Cf. A204515 (central terms), A105278, A004736.

Programs

  • Haskell
    a247500 n k = a247500_tabl !! n !! k
    a247500_row n = a247500_tabl !! n
    a247500_tabl = zipWith (zipWith div) a105278_tabl a004736_tabl
    -- Reinhard Zumkeller, Oct 19 2014
  • Magma
    /* triangle */ [[Factorial(n)/Factorial(k) * Binomial(n+2, k+1) /(n+2): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 18 2014
    
  • Maple
    T := (n,k) -> ((k+1)*(n+1)*GAMMA(n+1)^2)/(GAMMA(k+2)^2*GAMMA(n-k+2));
    A247500 := (n, k) -> (n!/(k+1)!)*binomial(n + 1, k):
  • Mathematica
    Table[((k + 1) (n + 1) Gamma[n + 1]^2)/(Gamma[k + 2]^2*
    Gamma[n - k + 2]), {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jun 19 2015 *)

Formula

T(n, k) = ((k+1)*(n+1)*Gamma(n+1)^2)/(Gamma(k+2)^2 *Gamma(n-k+2)). (original name)
T(n, k) = (n!/k!)*C(n+2, k+1)/(n+2).
T(n, 0) = A000142(n).
T(n, n-1) = A000217(n).
T(n+1, 1) = A001710(n+2).
Sum_{k=0..n} T(n, k) = A247499(n).
L(n+1, k+1) = T(n-1, k)*P(n) for n>=1 and 0<=k<=n; here L(n,k) denote the unsigned Lah numbers and P(n) the pronic numbers. - Peter Luschny, Oct 18 2014
T(n,k) = A105278(n+1,k+1) / (n+1-k), k=0..n. - Reinhard Zumkeller, Oct 19 2014
From Peter Bala, May 24 2023: (Start)
Triangle equals A164652 * A008277 (assuming the same offset for the three triangles).
This is equivalent to the Stirling number identity Sum_{i = 0..n} (n+1)!/(i+1)!* binomial(n,i)*Stirling1(i+1,k) = (-1)^(n+k+1)*Stirling1(n+1,k) for n, k >= 0. (End)

Extensions

Name updated by Peter Luschny, Jan 09 2022
Showing 1-2 of 2 results.