cp's OEIS Frontend

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

%I A160642 #2 Mar 31 2012 14:12:22
%S A160642 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 A160642 27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,
%U A160642 50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72
%N A160642 Minimal number k such that n! can be written as product of k (>= 2) consecutive integers.
%C A160642 Sequence starts at n=2 because 1! cannot be written as product of 2 (or more) consecutive integers.
%C A160642 For suitable m >= n, we have n! = m!/(m-a(n))!
%C A160642 For n >= 3, we have a(n) <= n-1 because n! = 2*...*n.
%C A160642 For n = m! - 1, we have a(n) <= m!-m because n! = (m+1)*(m+2)*...*(m!-1)*m! = (n+1)!/m!.
%C A160642 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 A160642 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 A160642 (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 A160642 csmin(N) = local(k=2); while(csfac(N,k)==0,k=k+1);k
%o A160642 \p 200; for(n=2,200, print(csmin(n!)))
%K A160642 nonn
%O A160642 2,1
%A A160642 _Hagen von Eitzen_, May 21 2009