

A160642


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


0



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
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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!/(ma(n))!
For n >= 3, we have a(n) <= n1 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)=n1 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=w1); 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



