OFFSET
2,1
COMMENTS
Sequence starts at n=2 because 1! cannot be written as product of 2 (or more) consecutive integers.
For suitable m >= n, we have n! = m!/(m-a(n))!
For n >= 3, we have a(n) <= n-1 because n! = 2*...*n.
For n = m! - 1, we have a(n) <= m!-m because n! = (m+1)*(m+2)*...*(m!-1)*m! = (n+1)!/m!.
For n>=8, it appears that the preceding two inequalities completely describe a(n), i.e. a(n) = m!-m if n=m!-1 and a(n)=n-1 otherwise.
EXAMPLE
a(2) = 2 because 2! = 1*2. a(3) = 2 because 3! = 2*3. a(4) = 3 because 4! = 2*3*4. a(5) = 3 because 5! = 4*5*6. a(6) = 3 because 6! = 8*9*10. a(7) = 4 because 7! = 7*8*9*10.
PROG
(PARI) csfac(N, k) = local(d, w=floor(N^(1/k))); while((d=prod(i=1, k, w+i))>N, w=w-1); if(d==N, 1, 0)
csmin(N) = local(k=2); while(csfac(N, k)==0, k=k+1); k
\p 200; for(n=2, 200, print(csmin(n!)))
CROSSREFS
KEYWORD
nonn
AUTHOR
Hagen von Eitzen, May 21 2009
STATUS
approved