OFFSET
0,2
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).
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)
LINKS
M. V. Koutras, Eulerian numbers associated with sequences of polynomials, The Fibonacci Quarterly, 32 (1994), 44-57.
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)
EXAMPLE
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;
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 *)
PROG
(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
(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
(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
(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
CROSSREFS
KEYWORD
easy,nonn,tabf
AUTHOR
Tom Copeland, Oct 30 2007, Nov 29 2007, Nov 30 2007
EXTENSIONS
Removed erroneous and duplicate statements. - Tom Copeland, Dec 03 2013
STATUS
approved