OFFSET
1,4
COMMENTS
REFERENCES
Solution to Problem 1831 by J. W. Grossman. Mathematics Magazine, 83, No. 5, 2010, pp. 392-393.
LINKS
J. W. Grossman, Solution to Problem 1831 proposed by Emeric Deutsch, Mathematics Magazine, 83, No. 5, 2010, pp. 392-393.
FORMULA
T(n,1)=(n-1)!; T(n,k)=n!/((n-k+1)(n-k+2)) for 2 <= k <= n.
EXAMPLE
T(4,3)=4 because we have (132)(4), (13)(24), (123)(4), (13)(2)(4).
Triangle starts:
1;
1, 1;
2, 1, 3;
6, 2, 4, 12;
24, 6, 10, 20, 60;
120, 24, 36, 60, 120, 360;
MAPLE
T:=proc(n, k) if k=1 then factorial(n-1) elif k <= n then factorial(n)/((n-k+1)*(n-k+2)) else 0 end if end proc: for n to 10 do seq(T(n, k), k=1..n) end do; # yields sequence in triangular form
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Nov 10 2008
STATUS
approved