cp's OEIS Frontend

A118742 Numbers n for which the expression n!/(n+1) is an integer.

%I A118742 #17 Jan 08 2024 07:02:27
%S A118742 0,5,7,8,9,11,13,14,15,17,19,20,21,23,24,25,26,27,29,31,32,33,34,35,
%T A118742 37,38,39,41,43,44,45,47,48,49,50,51,53,54,55,56,57,59,61,62,63,64,65,
%U A118742 67,68,69,71,73,74,75,76,77,79,80,81,83,84,85,86,87,89,90,91,92,93,94,95,97
%N A118742 Numbers n for which the expression n!/(n+1) is an integer.
%C A118742 Also set of all n>=0, excluding 3, for which n+1 is composite. [Proof: (i) If n+1 is prime, there cannot be any factor in n! to cancel the n+1 in the denominator of the expression. (ii) If n+1=composite=a*b, a<b, consider the equivalent expression (n+1)!/(n+1)^2=1*2*..*a*..*b*..(a*b)/(a^2*b^2) in which factors obviously cancel. (iii) If n+1=square=a^2, a>2, (n+1)!/(n+1)^2 = 1*2*..*a*...*(2a)*..*a^2/a^4 in which factors also cancel.] - _R. J. Mathar_, Nov 22 2006
%F A118742 a(n) = A002808(n+1)-1 for n>=1. - _R. J. Mathar_, Nov 22 2006
%e A118742 n=5 5!/(5+1)= 5*4*3*2*1/6 = 20.
%p A118742 P:=proc(n) local i,j; for i from 0 by 1 to n do j:=i!/(i+1); if trunc(j)=j then print(i); fi; od; end: P(200);
%t A118742 Select[Range[0,100],IntegerQ[#!/(#+1)]&] (* _Harvey P. Dale_, Aug 24 2014 *)
%Y A118742 Cf. A060462, A120416, A270441.
%Y A118742 Essentially the same as A072668.
%K A118742 nonn
%O A118742 0,2
%A A118742 _Paolo P. Lava_ and _Giorgio Balzarotti_, May 22 2006
%E A118742 Corrected (39 inserted) by _Harvey P. Dale_, Aug 24 2014