OFFSET
1,1
COMMENTS
Equivalently, binomial(n!+n,n). Proof: (n!+1)(n!+2)...(n!+k) == k! mod n! == 0 mod n! if and only if k >= n (for n >= 2). - _Paul D Hanna_ and Robert Israel, Aug 31 2010.
Note that k <= n. Subsidiary sequence to be investigated: n such that k < n.
This is just a coincidence, but k=2,6,84 are also such that floor(exp(1)*10^k) is a prime, cf. A064118. - M. F. Hasler, Aug 31 2013
EXAMPLE
a(4)=25*26*27*28/24=20475.
MAPLE
A109892 := proc(n) local k, fn; k := 1; fn := n! ; while mul(fn+i, i=1..k) mod fn <> 0 do k := k+1; od ; RETURN(mul(fn+i, i=1..k)/fn) ; end: seq(A109892(n), n=1..10) ; # R. J. Mathar, Aug 15 2007
MATHEMATICA
Table[(n+n!)!/(n!*(n!)!), {n, 1, 9}] (* Jean-François Alcover, Mar 04 2014, after first comment *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Amarnath Murthy, Jul 13 2005
EXTENSIONS
Corrected and extended by R. J. Mathar, Aug 15 2007
STATUS
approved