Location via proxy:   [ UP ]  
[Report a bug]   [Manage cookies]                
a(n) = P(n)/(L(n)*P(n/2)*P(n/3)*P(n/7)*P(n/43)*...) with P(n) = floor(n)!, L(n) the LCM of the first n integers and where the sequence 2, 3, 7, 43, ... is A000058.
1

%I #22 Jul 16 2019 19:51:02

%S 1,1,1,1,1,1,1,1,1,2,2,1,1,1,3,3,3,1,1,2,2,4,4,1,5,10,10,5,5,1,1,1,3,

%T 6,42,7,7,14,42,84,84,2,2,4,12,24,24,3,3,6,18,36,36,4,220,55,165,330,

%U 330,33,33,66,22,22,1430,130,130,260,780,156,156,13

%N a(n) = P(n)/(L(n)*P(n/2)*P(n/3)*P(n/7)*P(n/43)*...) with P(n) = floor(n)!, L(n) the LCM of the first n integers and where the sequence 2, 3, 7, 43, ... is A000058.

%C G. Myerson actually proved that P(n)/(P(n/2)*P(n/3)*P(n/7)*P(n/43)*...) is divisible by L(n) in a more general case. That is when n in the above expression is replaced by the terms of a sequence u(n) that satisfies GCD(u(n),u(m))=u(GCD(m,n)). And also when the sequence of quotients q(n)=2,3,7,43,... is replaced by a sequence q(n) such that sum(1/q(n))<=1.

%C The behavior of a(n) is quite erratic for small values of n, for instance a(26)=10, a(32)=1, a(65)=1430, a(84)=2, a(95)=542640, a(114)=3 (cf. Myerson 1994).

%H G. Bachman, <a href="http://dx.doi.org/10.1006/jnth.1997.2098">On divisibility properties of certain multinomial coefficients</a>, Journal of Number Theory, Volume 63, Issue 2, April 1997, Pages 244-255.

%H G. Bachman and T. Kessler, <a href="http://dx.doi.org/10.1016/j.jnt.2003.12.001">On divisibility properties of certain multinomial coefficients—II</a>, Journal of Number Theory, Volume 106, Issue 1, May 2004, Pages 1-12.

%H G. Myerson, <a href="http://dx.doi.org/10.1006/jnth.1994.1054">What the Least Common Multiple Divides</a>, Journal of Number Theory, Volume 48, Issue 1, July 1994, Pages 80-87.

%H G. Myerson and J. W. Sander, <a href="http://dx.doi.org/10.1006/jnth.1996.0138">What the Least Common Multiple Divides, II</a>, Journal of Number Theory, Volume 61, Issue 1, November 1996, Pages 67-84.

%o (PARI) a(n)=my(t=n!/lcm(vector(n,i,i))/(n\2)!,a1=2,a2=3);while(a2<n, t /= (n\a2)!; [a1,a2]=[a2,a2^2-a2+1]); t \\ _Charles R Greathouse IV_, Nov 19 2012

%Y Cf. A000058.

%K nonn

%O 1,10

%A _Michel Marcus_, Nov 19 2012