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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.
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%I #28 Nov 11 2024 21:20:21

%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,changed

%O 1,3

%A _Clark Kimberling_, Oct 11 2014