%I #21 Mar 25 2015 08:37:13
%S 1,0,1,1,0,2,5,4,1,0,12,40,51,31,9,1,0,144,564,904,769,376,106,16,1,0,
%T 2880,12576,23300,24080,15345,6273,1650,270,25,1,0,86400,408960,
%U 840216,991276,748530,381065,133848,32523,5370,575,36,1,0,3628800,18299520
%N Triangle read by rows, T(n,k) = sum(j=0..k-1, S(n+1,j+1)*S(n,k-j)) where S denotes the Stirling cycle numbers A132393, T(0,0)=1, n>=0, 0<=k<=2n.
%C These are also the coefficients of the polynomials interpolating the sequence k -> n!*((n+k)!/k!)*binomial(n+k-1,k-1) (for fixed n>=0). Divided by n! these polynomials generate the rows of Lah numbers L(n+k, k) = ((n+k)!/k!)* binomial(n+k-1,k-1).
%F T(n, n) = A187235(n) for n>=1 (after the explicit formula of _Vaclav Kotesovec_).
%e [1]
%e [0, 1, 1]
%e [0, 2, 5, 4, 1]
%e [0, 12, 40, 51, 31, 9, 1]
%e [0, 144, 564, 904, 769, 376, 106, 16, 1]
%e [0, 2880, 12576, 23300, 24080, 15345, 6273, 1650, 270, 25, 1]
%e For example in the case n=3 the polynomial (k^6+9*k^5+31*k^4+51*k^3+40*k^2+12*k)/3! generates the Lah numbers 0, 24, 240, 1200, 4200, 11760, 28224, ... (A253285).
%p # This is a special case of the recurrence given in A246117.
%p t := proc(n,k) option remember; if n=0 and k=0 then 1 elif
%p k <= 0 or k>n then 0 else iquo(n,2)*t(n-1,k)+t(n-1,k-1) fi end:
%p A254881 := (n,k) -> t(2*n,k):
%p seq(print(seq(A254881(n,k), k=0..2*n)), n=0..5);
%p # Illustrating the comment:
%p restart: with(PolynomialTools): with(CurveFitting): for N from 0 to 5 do
%p CoefficientList(PolynomialInterpolation([seq([k,N!*((N+k)!/k!)*binomial(N+k-1,k-1)], k=0..2*N)], n), n) od;
%t Flatten[{1,Table[Table[Sum[Abs[StirlingS1[n+1,j+1]] * Abs[StirlingS1[n,k-j]],{j,0,k-1}],{k,0,2*n}],{n,1,10}]}] (* _Vaclav Kotesovec_, Feb 10 2015 *)
%o (Sage)
%o def T(n,k):
%o if n == 0: return 1
%o return sum(stirling_number1(n+1,j+1)*stirling_number1(n,k-j) for j in range(k))
%o for n in range (6): [T(n,k) for k in (0..2*n)]
%Y Cf. A246117, A254882, A187235.
%Y The sequences A000012, A002378, A083374, A253285 are the Lah number rows generated by the polynomials divided by n! for n=0, 1, 2, 3 respectivly.
%K nonn,tabf
%O 0,6
%A _Peter Luschny_, Feb 10 2015