%I #42 Dec 28 2022 04:08:50

%S 1,1,6,1,20,1680,1,70,34650,63063000,1,252,756756,11732745024,

%T 623360743125120,1,924,17153136,2308743493056,1370874167589326400,

%U 2670177736637149247308800,1,3432,399072960,472518347558400,3177459078523411968000,85722533226982363751829504000,7363615666157189603982585462030336000

%N Triangle read by rows: T(n, k) = (n*k)!/(n!)^k.

%C T(m,n) is the number of ways to distribute n*m different toys among m different kids so that each kid gets exactly n toys. For example, with n=3 and m=2, the 6 different toys, t1, t2, t3, t4, t5 and t6, can be distributed in exactly 20 ways among the 2 kids, k1 and k2, since there are C(6,3)=20 ways to choose the three toys for k1, with the other three toys going to k2. The proof for the general case is based on the identity C(n*m,n)*C(n*m-n,n)*...*C(n*m-n*(m-1),n) = (n*m)!/(n!)^m. - _Dennis P. Walsh_, Apr 12 2018

%H Seiichi Manyama, <a href="/A120666/b120666.txt">Rows n = 1..26, flattened</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MacdonaldsConstant-TermConjecture.html">Macdonald's Constant-Term Conjecture</a>

%F T(n, k) = (k*n)!/(n!)^k.

%e Triangle begins:

%e 1;

%e 1, 6;

%e 1, 20, 1680;

%e 1, 70, 34650, 63063000;

%e 1, 252, 756756, 11732745024, 623360743125120;

%p T:= (m, n)-> (n*m)!/(m!)^n:

%p seq(seq(T(m, n), n=1..m), m=1..7); # _Alois P. Heinz_, Apr 12 2018

%t Table[(n*k)!/(n!)^k, {n,10}, {k,n}]//Flatten

%o (Magma) [Factorial(n*k)/(Factorial(n))^k: k in [1..n], n in [1..10]]; // _G. C. Greubel_, Dec 26 2022

%o (SageMath)

%o def A120666(n,k): return gamma(n*k+1)/(factorial(n))^k

%o flatten([[A120666(n,k) for k in range(1,n+1)] for n in range(1,11)]) # _G. C. Greubel_, Dec 26 2022

%Y Cf. A000984, A006480, A034841, A089759, A187783.

%K nonn,tabl

%O 1,3

%A _Roger L. Bagula_, Aug 11 2006

%E Edited by _N. J. A. Sloane_, Jun 17 2007

%E Offset corrected by _Alois P. Heinz_, Apr 12 2018

%E New name using formula by _Joerg Arndt_, Apr 15 2018