|
|
A162995
|
|
A scaled version of triangle A162990.
|
|
5
|
|
|
1, 3, 1, 12, 4, 1, 60, 20, 5, 1, 360, 120, 30, 6, 1, 2520, 840, 210, 42, 7, 1, 20160, 6720, 1680, 336, 56, 8, 1, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 1814400, 604800, 151200, 30240, 5040, 720, 90, 10, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
We get this scaled version of triangle A162990 by dividing the coefficients in the left hand columns by their 'top-values' and then taking the square root.
|
|
LINKS
|
|
|
FORMULA
|
a(n,m) = (n+1)!/(m+1)! for n = 1, 2, 3, ..., and m = 1, 2, ..., n.
|
|
EXAMPLE
|
The first few rows of the triangle are:
[1]
[3, 1]
[12, 4, 1]
[60, 20, 5, 1]
|
|
MAPLE
|
a := proc(n, m): (n+1)!/(m+1)! end: seq(seq(a(n, m), m=1..n), n=1..9); # Johannes W. Meijer, revised Nov 23 2012
|
|
MATHEMATICA
|
Table[(n+1)!/(m+1)!, {n, 10}, {m, n}] (* Paolo Xausa, Mar 31 2024 *)
|
|
PROG
|
(Haskell)
a162995 n k = a162995_tabl !! (n-1) !! (k-1)
a162995_row n = a162995_tabl !! (n-1)
a162995_tabl = map fst $ iterate f ([1], 3)
where f (row, i) = (map (* i) row ++ [1], i + 1)
|
|
CROSSREFS
|
A056542(n) equals the row sums for n>=1.
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|