A226365 Composite numbers such that Sum_{i=1..k} (1 + 1/p_i) - Product_{i=1..k} (1 + 1/p_i) is an integer, where p_i are the k prime factors of n (with multiplicity).

152, 432, 1620, 1728, 3456, 4752, 22464, 46656, 80892, 139968, 186624, 237168, 326592, 746496, 1651968, 2052864, 2426112, 2985984, 5971968, 10257408, 12177216, 12690432, 14048240, 14183424, 20155392, 20901888, 26127360, 38817792

Offset: 1

Author

Paolo P. Lava, Jun 12 2013

Keywords

• nonn

Comments

The numbers of the sequence are the solution of the differential equation n' = (a-k)*n + b, which can also be written as A003415(n) = (a-k)*n + A003958(n), where k is the number of prime factors of n, and a is the integer Sum_{i=1..k} (1 + 1/p_i) - Product_{1=1..k} (1 + 1/p_i).

The numbers of the sequence satisfy also Sum_{i=1..k} (1 - 1/p_i) + Product_{i=1..k} (1 + 1/p_i) = some integer.

Examples

			237168 has prime factors 2, 2, 2, 2, 3, 3, 3, 3, 3, 61. 4*(1 + 1/2) + 5*(1 + 1/3) + (1 + 1/61) = 2504/183 is the sum over the 1 + 1/p_i. (1 + 1/2)^4 * (1 + 1/3)^5 * (1 + 1/61) = 3968/183 is the product of the 1 + 1/p_i. The difference over sum and product is 2504/183 - 3968/183 = -8, an integer.
		

Links

• J. M. Borwein and E. Wong, A survey of results relating to Giuga's conjecture on primality, May 8, 1995.
• R. Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.

Crossrefs

Cf. A003415, A003958, A198391, A199767, A224912.

Programs

Maple

with(numtheory); ListA226365:=proc(q) local a,d,n,p;
for n from 1 to q do if not isprime(n) then p:=ifactors(n)[2];
a:=add(op(2,d)+op(2,d)/op(1,d),d=p)-mul((1+1/op(1,d))^op(2,d),d=p);
if type(a,integer) then print(n); fi; fi;
od; end: ListA226365(10^9);