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A coefficient tree from the list partition transform relating A000129, A000142, A000165, A110327, and A110330.
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%I #34 Feb 20 2024 01:05:29

%S 1,2,6,2,24,24,120,240,24,720,2400,720,5040,25200,15120,720,40320,

%T 282240,282240,40320,362880,3386880,5080320,1451520,40320,3628800,

%U 43545600,91445760,43545600,3628800,39916800,598752000,1676505600,1197504000,199584000,3628800

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

%C Construct the infinite array of polynomials

%C a(0,t) = 1

%C a(1,t) = 2

%C a(2,t) = 6 + 2 t

%C a(3,t) = 24 + 24 t

%C a(4,t) = 120 + 240 t + 24 t^2

%C a(5,t) = 720 + 2400 t + 720 t^2

%C a(6,t) = 5040 + 25200 t + 15120 t^2 + 720 t^3

%C 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.

%C b(0,t) = 1, b(1,t) = -2, b(2,t) = -2*(t-1), b(n,t) = 0 for n>2.

%C Then A000165(n) = a(n,1).

%C Lower triangular matrix A110327 = binomial(n,k)*a(n-k,2).

%C n! * A000129(n+1) = a(n,2) = A110327(n,0).

%C A110330 = matrix inverse of binomial(n,k)*a(n-k,2) = binomial(n,k)*b(n-k,2).

%C A000142(n+1) = a(n,0).

%C From _Peter Bala_, Sep 09 2013: (Start)

%C 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)

%H M. V. Koutras, <a href="http://www.fq.math.ca/Scanned/32-1/koutras.pdf">Eulerian numbers associated with sequences of polynomials</a>, The Fibonacci Quarterly, 32 (1994), 44-57.

%F 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 ).

%F 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.

%F a(n,t) = n!*Sum_{k>=0} binomial(n+1,2k+1) * t^k = n!*Sum_{k>=0} A034867(n,k) * t^k.

%F Additional relations are given by formulas in A133314.

%F From _Peter Bala_, Sep 09 2013: (Start)

%F Recurrence equation: T(n+1,k) = (n+2 +2*k)T(n,k) + (n +2 -2*k)T(n,k-1).

%F Let P(n,x) = x*(x + 1)*...*(x + n - 1) denote the n-th rising factorial.

%F 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.

%F 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)

%e Triangle begins as:

%e 1;

%e 2;

%e 6, 2;

%e 24, 24;

%e 120, 240, 24;

%e 720, 2400, 720;

%e 5040, 25200, 15120, 720;

%e 40320, 282240, 282240, 40320;

%e 362880, 3386880, 5080320, 1451520, 40320;

%e 3628800, 43545600, 91445760, 43545600, 3628800;

%p for n from 0 to 10 do

%p seq( n!*binomial(n+1,2*k+1), k = 0..floor(n/2) )

%p end do; # _Peter Bala_, Sep 09 2013

%t Table[n!*Binomial[n+1, 2*k+1], {n,0,10}, {k,0,Floor[n/2]}]//Flatten (* _G. C. Greubel_, Dec 30 2019 *)

%o (PARI) T(n,k) = n!*binomial(n+1, 2*k+1);

%o for(n=0,10, for(k=0, n\2, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Dec 30 2019

%o (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

%o (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

%o (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

%Y Cf. A228955.

%K easy,nonn,tabf

%O 0,2

%A _Tom Copeland_, Oct 30 2007, Nov 29 2007, Nov 30 2007

%E Removed erroneous and duplicate statements. - _Tom Copeland_, Dec 03 2013