# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/

Search: id:a248686
Showing 1-1 of 1

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

# Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE