A248686 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.
1, 1, 2, 1, 3, 6, 1, 6, 12, 24, 1, 10, 30, 60, 120, 1, 20, 90, 180, 360, 720, 1, 35, 210, 630, 1260, 2520, 5040, 1, 70, 560, 2520, 5040, 10080, 20160, 40320, 1, 126, 1680, 7560, 22680, 45360, 90720, 181440, 362880, 1, 252, 4200, 25200, 113400, 226800, 453600, 907200, 1814400, 3628800
Offset: 1
Examples
First seven rows: 1 1 2 1 3 6 1 6 12 24 1 10 30 60 120 1 20 90 180 360 720 1 35 210 630 1260 2520 5040 ... Writing floor as [ ], the numbers comprising row 4 are T(4,1) = 4!/[4/1]! = 24/24 = 1 T(4,2) = 4!/([4/2]![5/2]!) = 24/(2*2) = 6 T(4,3) = 4!/([4/3]![5/3]![6/3]!) = 24/(1*1*2) = 12 T(4,4) = 4!/([4/4]![5/4]![6/4]![7/4]!) = 24/(1*1*1*1) = 24.
Links
- Clark Kimberling, Table of n, a(n) for n = 1..5000
Programs
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Maple
T:= (n, k)-> combinat[multinomial](n, floor((n+i)/k)$i=0..k-1): seq(seq(T(n, k), k=1..n), n=1..10); # Alois P. Heinz, Feb 09 2023
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Mathematica
f[n_, k_] := f[n, k] = n!/Product[Floor[(n + i)/k]!, {i, 0, k - 1}] t = Table[f[n, k], {n, 0, 10}, {k, 1, n}]; u = Flatten[t] (* A248686 sequence *) TableForm[t] (* A248686 array *) Table[Sum[f[n, k], {k, 1, n}], {n, 1, 22}] (* A248687 *)
Comments
Alois P. Heinz, Feb 09 2023