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%I #2 Mar 31 2012 14:12:22
%S 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,
%T 27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,
%U 50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72
%N Minimal number k such that n! can be written as product of k (>= 2) consecutive integers.
%C Sequence starts at n=2 because 1! cannot be written as product of 2 (or more) consecutive integers.
%C For suitable m >= n, we have n! = m!/(m-a(n))!
%C For n >= 3, we have a(n) <= n-1 because n! = 2*...*n.
%C For n = m! - 1, we have a(n) <= m!-m because n! = (m+1)*(m+2)*...*(m!-1)*m! = (n+1)!/m!.
%C 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.
%e 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.
%o (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)
%o csmin(N) = local(k=2); while(csfac(N,k)==0,k=k+1);k
%o \p 200; for(n=2,200, print(csmin(n!)))
%K nonn
%O 2,1
%A _Hagen von Eitzen_, May 21 2009