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

2,1

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.

Links

Table of n, a(n) for n=2..73.

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

Sequence in context: A331590 A326496 A058740 * A110868 A110869 A110876

Adjacent sequences: A160639 A160640 A160641 * A160643 A160644 A160645

Keyword

nonn

Author

Hagen von Eitzen, May 21 2009

Status

approved