




OFFSET

1,1


COMMENTS

Members of a pair (m,k) such that m = sum of aliquot parts of k minus the sum of the prime distinct divisors of k and k = sum of aliquot parts of m minus the sum of the prime distinct divisors of m.
We introduce in a single sequence the “pseudoamicable numbers” and the “pseudoperfect numbers” in the case m = k. In this sequence 42 and 1316 are pseudoperfect numbers: for example, the sum of the aliquot parts of 42 is 1 + 2 + 3 + 6 + 7 + 14 + 21 = 54 and 54  (2 + 3 + 7) = 54  12 = 42.
The pseudoamicable numbers are (36, 50), (12540240, 29559057). (see the example).
Is the sequence finite?
a(8) <= 37778715690312487141376. It is easy to verify that if p = (2^k1) and q = 4^k2*2^k1 are two primes, then n = 2^(k1)*p*q is in the sequence, because A206773(n) = n. This happens for k = 2, 3, 7, and 19, which give the terms 42, 1316, 131080256, and 3777871569031248714137, respectively. Up to 3*10^12 there are no other n for which A206773(n) = n.  Giovanni Resta, May 03 2016


LINKS



EXAMPLE

The divisors of m = 50 are {1, 2, 5, 10, 25, 50} and (1 + 2 + 5 + 10 + 25)  (2 + 5) = 43  7 = 36;
The divisors of k = 36 are {1, 2, 3, 4, 6, 9, 12, 18, 36} and (1+2+3+4+6+9+12+18)  (2 + 3) = 55  5 = 50.


MAPLE

with(numtheory):for n from 1 to 10^7 do:x:=factorset(n):n1:=nops(x):s:=sum('x[i] ', 'i'=1..n1):s1:=sigma(n)ns: y:=factorset(s1):n2:=nops(y): ss:=sum('y[i] ', 'i'=1..n2):s2:=sigma(s1)s1ss:if s2=n then printf(`%d, `, n):else fi:od:


CROSSREFS



KEYWORD

nonn,hard


AUTHOR



EXTENSIONS



STATUS

approved



