%I #43 Sep 14 2023 18:08:12
%S 1,1,1,2,3,1,6,12,6,1,24,60,40,10,1,120,360,300,100,15,1,720,2520,
%T 2520,1050,210,21,1,5040,20160,23520,11760,2940,392,28,1,40320,181440,
%U 241920,141120,42336,7056,672,36,1,362880,1814400,2721600,1814400,635040,127008,15120,1080,45,1
%N Triangle read by rows: T(n, k) = n!*binomial(n + 1, k)/(k + 1)!, 0 <= k <= n.
%C 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
%C From _Peter Bala_, Sep 03 2023: (Start)
%C 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.
%C 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).
%C 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)
%H Reinhard Zumkeller, <a href="/A247500/b247500.txt">Rows n = 0..125 of triangle, flattened</a>
%H W. Wang and T. Wang, <a href="http://dx.doi.org/10.1016/j.disc.2007.12.037">Generalized Riordan arrays</a>, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.
%F T(n, k) = ((k+1)*(n+1)*Gamma(n+1)^2)/(Gamma(k+2)^2 *Gamma(n-k+2)). (original name)
%F T(n, k) = (n!/k!)*C(n+2, k+1)/(n+2).
%F T(n, 0) = A000142(n).
%F T(n, n-1) = A000217(n).
%F T(n+1, 1) = A001710(n+2).
%F Sum_{k=0..n} T(n, k) = A247499(n).
%F 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
%F T(n,k) = A105278(n+1,k+1) / (n+1-k), k=0..n. - _Reinhard Zumkeller_, Oct 19 2014
%F From _Peter Bala_, May 24 2023: (Start)
%F Triangle equals A164652 * A008277 (assuming the same offset for the three triangles).
%F 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)
%e Triangle begins:
%e 1;
%e 1, 1;
%e 2, 3, 1;
%e 6, 12, 6, 1;
%e 24, 60, 40, 10, 1;
%e 120, 360, 300, 100, 15, 1;
%e 720, 2520, 2520, 1050, 210, 21, 1;
%p T := (n,k) -> ((k+1)*(n+1)*GAMMA(n+1)^2)/(GAMMA(k+2)^2*GAMMA(n-k+2));
%p A247500 := (n, k) -> (n!/(k+1)!)*binomial(n + 1, k):
%t Table[((k + 1) (n + 1) Gamma[n + 1]^2)/(Gamma[k + 2]^2*
%t Gamma[n - k + 2]), {n, 0, 9}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Jun 19 2015 *)
%o (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
%o (Haskell)
%o a247500 n k = a247500_tabl !! n !! k
%o a247500_row n = a247500_tabl !! n
%o a247500_tabl = zipWith (zipWith div) a105278_tabl a004736_tabl
%o -- _Reinhard Zumkeller_, Oct 19 2014
%Y Cf. A247499 (row sums), A008297.
%Y Cf. A000142, A000217, A001710.
%Y Cf. A204515 (central terms), A105278, A004736.
%Y Cf. A090181, A001263.
%K nonn,tabl
%O 0,4
%A _Peter Luschny_, Oct 17 2014
%E Name updated by _Peter Luschny_, Jan 09 2022