A160642 Minimal number k such that n! can be written as product of k (>= 2) consecutive integers.

2, 2, 3, 3, 3, 4, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 20, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72

Offset: 2

Author

Hagen von Eitzen, May 21 2009

Keywords

• nonn

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.

Examples

			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.
		

Programs

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!)))