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A251606
Let {d(i)}, i=1..q, be the set of the q prime distinct divisors of a number n, and consider the alternating sum s = Sum_{i=1..q} i*((-1)^(i+1))/d(i) and the product p = Product_{i=1..q} d(i) respectively. The sequence lists the composite numbers n such that s = 1/p.
0
4, 8, 9, 10, 16, 20, 21, 25, 27, 32, 40, 49, 50, 55, 63, 64, 80, 81, 100, 102, 121, 125, 128, 147, 160, 169, 189, 200, 204, 243, 250, 253, 256, 275, 289, 306, 320, 343, 361, 400, 408, 441, 500, 512, 529, 567, 605, 612, 625, 640, 729, 800, 816, 841, 918, 961
OFFSET
1,1
COMMENTS
The semiprimes p*q, p and q prime with q = 2*p + 1 (A156592) are in the sequence.
The trivial values of prime powers p^m (A025475) are in the sequence.
The primitive values of the sequence are 4, 9, 10, 21, 25, 49, 55, ... .
EXAMPLE
21 is in the sequence because the prime factors of 21 are {3,7} => s = 1/3 - 2/7, 1/p = 1/21 and s = p = 1/21.
MAPLE
with(numtheory):nn:=1000:
for n from 1 to nn do:
if type(n, prime)=false
then
x:=factorset(n):n0:=nops(x):
s:=sum('i*((-1)^(i+1))/x[i]', 'i'=1..n0):s0:=product('x[i]', 'i'=1..n0):
s2:=s-1/s0:
if s2=0
then
printf(`%d, `, n):
else
fi:
fi:
od:
PROG
(PARI) lista(nn) = {forcomposite(n=2, nn, my(f = factor(n)[, 1]); if (sum(i=1, #f, i*((-1)^(i+1))/f[i]) == prod(i=1, #f, 1/f[i]), print1(n, ", ")); ); } \\ Michel Marcus, Sep 13 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Dec 13 2014
STATUS
approved