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Triangle of numbers n!(n-1)!...(n-k+1)!/(1!2!...k!).
17

%I #60 Apr 13 2022 01:14:45

%S 1,1,1,1,2,1,1,6,6,1,1,24,72,24,1,1,120,1440,1440,120,1,1,720,43200,

%T 172800,43200,720,1,1,5040,1814400,36288000,36288000,1814400,5040,1,1,

%U 40320,101606400,12192768000,60963840000,12192768000,101606400,40320,1

%N Triangle of numbers n!(n-1)!...(n-k+1)!/(1!2!...k!).

%C Product of all matrix elements of n X k matrix M(i,j) = i+j (i=1..n-k, j=1..k). - _Peter Luschny_, Nov 26 2012

%C These are the generalized binomial coefficients associated to the sequence A000178. - _Tom Edgar_, Feb 13 2014

%H G. C. Greubel, <a href="/A009963/b009963.txt">Rows n = 0..50 of the triangle, flattened</a>

%F T(n,k) = T(n-1,k-1)*A008279(n,n-k) = A000178(n)/(A000178(k)*A000178(n-k)) i.e., a "supercombination" of "superfactorials". - _Henry Bottomley_, May 22 2002

%F Equals ConvOffsStoT transform of the factorials starting (1, 2, 6, 24, ...); e.g., ConvOffs transform of (1, 2, 6, 24) = (1, 24, 72, 24, 1). Note that A090441 = ConvOffsStoT transform of the factorials, A000142. - _Gary W. Adamson_, Apr 21 2008

%F Asymptotic: T(n,k) ~ exp((3/2)*k^2 - zeta'(-1) + 3/4 - (3/2)*n*k)*(1+n)^((1/2)*n^2 + n + 5/12)*(1+k)^(-(1/2)*k^2 - k - 5/12)*(1 + n - k)^(-(1/2)*n^2 + n*k - (1/2)*k^2 - n + k - 5/12)/(sqrt(2*Pi). - _Peter Luschny_, Nov 26 2012

%F T(n,k) = (n-k)!*C(n-1,k-1)*T(n-1,k-1) + k!*C(n-1,k)*T(n-1,k) where C(i,j) is given by A007318. - _Tom Edgar_, Feb 13 2014

%F T(n,k) = Product_{i=1..k} (n+1-i)!/i!. - _Alois P. Heinz_, Jun 07 2017

%F T(n,k) = BarnesG(n+2)/(BarnesG(k+2)*BarnesG(n-k+2)). - _G. C. Greubel_, Jan 04 2022

%e Rows start:

%e 1;

%e 1, 1;

%e 1, 2, 1;

%e 1, 6, 6, 1;

%e 1, 24, 72, 24, 1;

%e 1, 120, 1440, 1440, 120, 1; etc.

%t (* First program *)

%t row[n_]:= Table[Product[i+j, {i,1,n-k}, {j,1,k}], {k,0,n}];

%t Array[row, 9, 0] // Flatten (* _Jean-François Alcover_, Jun 01 2019, after _Peter Luschny_ *)

%t (* Second program *)

%t T[n_, k_]:= BarnesG[n+2]/(BarnesG[k+2]*BarnesG[n-k+2]);

%t Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jan 04 2022 *)

%o (Sage)

%o def A009963_row(n):

%o return [mul(mul(i+j for j in (1..k)) for i in (1..n-k)) for k in (0..n)]

%o for n in (0..7): A009963_row(n) # _Peter Luschny_, Nov 26 2012

%o (Sage)

%o def triangle_to_n_rows(n): #changing n will give you the triangle to row n.

%o N=[[1]+n*[0]]

%o for i in [1..n]:

%o N.append([])

%o for j in [0..n]:

%o if i>=j:

%o N[i].append(factorial(i-j)*binomial(i-1,j-1)*N[i-1][j-1]+factorial(j)*binomial(i-1,j)*N[i-1][j])

%o else:

%o N[i].append(0)

%o return [[N[i][j] for j in [0..i]] for i in [0..n]]

%o # _Tom Edgar_, Feb 13 2014

%o (Magma)

%o A009963:= func< n,k | (1/Factorial(n+1))*(&*[ Factorial(n-j+1)/Factorial(j): j in [0..k]]) >;

%o [A009963(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Jan 04 2022

%Y Cf. A000178, A007318, A060854, A090441.

%Y Central column is A079478.

%Y Columns include A010796, A010797, A010798, A010799, A010800.

%Y Row sums give A193520.

%K nonn,tabl

%O 0,5

%A _N. J. A. Sloane_