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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.

A131980 A coefficient tree from the list partition transform relating A000129, A000142, A000165, A110327, and A110330.

Original entry on oeis.org

1, 2, 6, 2, 24, 24, 120, 240, 24, 720, 2400, 720, 5040, 25200, 15120, 720, 40320, 282240, 282240, 40320, 362880, 3386880, 5080320, 1451520, 40320, 3628800, 43545600, 91445760, 43545600, 3628800, 39916800, 598752000, 1676505600, 1197504000, 199584000, 3628800
Offset: 0

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Author

Tom Copeland, Oct 30 2007, Nov 29 2007, Nov 30 2007

Keywords

Comments

Construct the infinite array of polynomials
a(0,t) = 1
a(1,t) = 2
a(2,t) = 6 + 2 t
a(3,t) = 24 + 24 t
a(4,t) = 120 + 240 t + 24 t^2
a(5,t) = 720 + 2400 t + 720 t^2
a(6,t) = 5040 + 25200 t + 15120 t^2 + 720 t^3
This array is the reciprocal array of the following array b(n,t) under the list partition transform and its associated operations described in A133314.
b(0,t) = 1, b(1,t) = -2, b(2,t) = -2*(t-1), b(n,t) = 0 for n>2.
Then A000165(n) = a(n,1).
Lower triangular matrix A110327 = binomial(n,k)*a(n-k,2).
n! * A000129(n+1) = a(n,2) = A110327(n,0).
A110330 = matrix inverse of binomial(n,k)*a(n-k,2) = binomial(n,k)*b(n-k,2).
A000142(n+1) = a(n,0).
From Peter Bala, Sep 09 2013: (Start)
Let {P(n,x)}n>=0 be a polynomial sequence. Koutras has defined generalized Eulerian numbers associated with the sequence P(n,x) as the coefficients A(n,k) in the expansion of P(n,x) in a series of factorials of degree n, namely P(n,x) = Sum_{k=0..n} A(n,k)* binomial(x+n-k,n). The choice P(n,x) = x^n produces the classical Eulerian numbers of A008292. Let now P(n,x) = x*(x + 1)*...*(x + n - 1) denote the n-th rising factorial polynomial. Then the present table is the generalized Eulerian numbers associated with the polynomial sequence P(n,2*x). See A228955 for the generalized Eulerian numbers associated with the polynomial sequence P(n,2*x + 1). (End)

Examples

			Triangle begins as:
        1;
        2;
        6,        2;
       24,       24;
      120,      240,       24;
      720,     2400,      720;
     5040,    25200,    15120,      720;
    40320,   282240,   282240,    40320;
   362880,  3386880,  5080320,  1451520,   40320;
  3628800, 43545600, 91445760, 43545600, 3628800;
		

Crossrefs

Cf. A228955.

Programs

  • GAP
    Flat(List([0..10], n-> List([0..Int(n/2)], k-> Factorial(n)*Binomial(n+1, 2*k+1) ))); # G. C. Greubel, Dec 30 2019
  • Magma
    [Factorial(n)*Binomial(n+1, 2*k+1): k in [0..Floor(n/2)], n in [0..10]]; // G. C. Greubel, Dec 30 2019
    
  • Maple
    for n from 0 to 10 do
    seq( n!*binomial(n+1,2*k+1), k = 0..floor(n/2) )
    end do; # Peter Bala, Sep 09 2013
  • Mathematica
    Table[n!*Binomial[n+1, 2*k+1], {n,0,10}, {k,0,Floor[n/2]}]//Flatten (* G. C. Greubel, Dec 30 2019 *)
  • PARI
    T(n,k) = n!*binomial(n+1, 2*k+1);
    for(n=0,10, for(k=0, n\2, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 30 2019
    
  • Sage
    [[factorial(n)*binomial(n+1, 2*k+1) for k in (0..floor(n/2))] for n in (0..10)] # G. C. Greubel, Dec 30 2019
    

Formula

E.g.f. for the polynomials b(.,t), introduced above, is 1 - 2x - (t-1) * x^2; therefore e.g.f. for the polynomials a(.,t), which are the row polynomials of this array, is 1 / ( 1 - 2x - (t-1) * x^2 ) = (t-1) / ( t - ( 1 + x*(t-1) )^2 ).
Also, a(n,t) = (1 - t*u^2)^(n+1) (D_u)^n [ 1 / (1 - t*u^2) ] with eval. at u = 1/t. Compare A076743.
a(n,t) = n!*Sum_{k>=0} binomial(n+1,2k+1) * t^k = n!*Sum_{k>=0} A034867(n,k) * t^k.
Additional relations are given by formulas in A133314.
From Peter Bala, Sep 09 2013: (Start)
Recurrence equation: T(n+1,k) = (n+2 +2*k)T(n,k) + (n +2 -2*k)T(n,k-1).
Let P(n,x) = x*(x + 1)*...*(x + n - 1) denote the n-th rising factorial.
T(n,k) = Sum_{j=0..k+1} (-1)^(k+1-j)*binomial(n+1,k+1-j)*P(n,2*j) for n >= 1.
The row polynomial a(n,t) satisfies t*a(n,t)/(1 - t)^(n+1) = Sum_{j>=1} P(n,2*j)*t^j. For example, for n = 3 we have t*(24 + 24*t)/(1 - t)^4 = 2*3*4*t + (4*5*6)*t^2 + (6*7*8)*t^3 + ..., while for n = 4 we have t*(120 + 240*t + 24*t^2)/(1 - t)^5 = (2*3*4*5)*t + (4*5*6*7)*t^2 + (6*7*8*9)*t^3 + .... (End)

Extensions

Removed erroneous and duplicate statements. - Tom Copeland, Dec 03 2013