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%I #25 Feb 20 2024 11:49:06
%S 1,1,2,1,3,6,1,6,12,24,1,10,30,60,120,1,20,90,180,360,720,1,35,210,
%T 630,1260,2520,5040,1,70,560,2520,5040,10080,20160,40320,1,126,1680,
%U 7560,22680,45360,90720,181440,362880,1,252,4200,25200,113400,226800,453600,907200,1814400,3628800
%N Triangular array of multinomial coefficients: T(n,k) = n!/(n(1)!*n(2)!* ... *n(k)!), where n(i) = floor((n + i - 1)/k) for i = 1 .. k.
%C T(n,k) is the number of permutations p of [n] such that p(i)<p(i+k) for i in [n-k]. T(4,2) = 6: 1234, 1243, 1324, 2134, 2143, 3142. - _Alois P. Heinz_, Feb 09 2023
%H Clark Kimberling, <a href="/A248686/b248686.txt">Table of n, a(n) for n = 1..5000</a>
%e First seven rows:
%e 1
%e 1 2
%e 1 3 6
%e 1 6 12 24
%e 1 10 30 60 120
%e 1 20 90 180 360 720
%e 1 35 210 630 1260 2520 5040
%e ...
%e Writing floor as [ ], the numbers comprising row 4 are
%e T(4,1) = 4!/[4/1]! = 24/24 = 1
%e T(4,2) = 4!/([4/2]![5/2]!) = 24/(2*2) = 6
%e T(4,3) = 4!/([4/3]![5/3]![6/3]! = 24/(1*1*2) = 12
%e T(4,4) = 4!/([4/4]![5/4]![6/4]![7/4]!) = 24/(1*1*1*1) = 24.
%p T:= (n, k)-> combinat[multinomial](n, floor((n+i)/k)$i=0..k-1):
%p seq(seq(T(n, k), k=1..n), n=1..10); # _Alois P. Heinz_, Feb 09 2023
%t f[n_, k_] := f[n, k] = n!/Product[Floor[(n + i)/k]!, {i, 0, k - 1}]
%t t = Table[f[n, k], {n, 0, 10}, {k, 1, n}];
%t u = Flatten[t] (* A248686 sequence *)
%t TableForm[t] (* A248686 array *)
%t Table[Sum[f[n, k], {k, 1, n}], {n, 1, 22}] (* A248687 *)
%Y Main diagonal is A000142.
%Y T(2n,n) gives A000680.
%Y Row sums give A248687.
%Y Cf. A333706.
%K nonn,tabl,easy
%O 1,3
%A _Clark Kimberling_, Oct 11 2014
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