A225028 - OEIS

%I #12 Jan 28 2020 04:04:53

%S 36,42,50,1316,12540240,29559057,131080256

%N Numbers m such that A206773(m) = k and A206773(k) = m.

%C 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.

%C We introduce in a single sequence the “pseudo-amicable numbers” and the “pseudo-perfect numbers” in the case m = k. In this sequence 42 and 1316 are pseudo-perfect 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.

%C The pseudo-amicable numbers are (36, 50), (12540240, 29559057). (see the example).

%C Is the sequence finite?

%C a(8) > 5*10^9. - _Donovan Johnson_, Apr 25 2013

%C a(8) <= 37778715690312487141376. It is easy to verify that if p = (2^k-1) and q = 4^k-2*2^k-1 are two primes, then n = 2^(k-1)*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

%C a(8) <= 72872313094554244192. - _Giovanni Resta_, Jan 28 2020

%e The divisors of m = 50 are {1, 2, 5, 10, 25, 50} and (1 + 2 + 5 + 10 + 25) - (2 + 5) = 43 - 7 = 36;

%e 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.

%p 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)-n-s: y:=factorset(s1):n2:=nops(y): ss:=sum('y[i] ', 'i'=1..n2):s2:=sigma(s1)-s1-ss:if s2=n then printf(`%d, `,n):else fi:od:

%Y Cf. A023890, A001065, A000203, A000396, A063990, A206773.

%K nonn,hard

%O 1,1

%A _Michel Lagneau_, Apr 24 2013

%E a(7) from _Donovan Johnson_, Apr 25 2013