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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).
2
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
1,1
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.
LINKS
R. Mestrovic, On a Congruence Modulo n^3 Involving Two Consecutive Sums of Powers, Journal of Integer Sequences, Vol. 17 (2014), 14.8.4.
EXAMPLE
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.
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);
CROSSREFS
KEYWORD
nonn
AUTHOR
Paolo P. Lava, Jun 12 2013
STATUS
approved