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A225028 Numbers m such that A206773(m) = k and A206773(k) = m. 1
36, 42, 50, 1316, 12540240, 29559057, 131080256 (list; graph; refs; listen; history; text; internal format)
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 “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.
The pseudo-amicable numbers are (36, 50), (12540240, 29559057). (see the example).
Is the sequence finite?
a(8) > 5*10^9. - Donovan Johnson, Apr 25 2013
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
a(8) <= 72872313094554244192. - Giovanni Resta, Jan 28 2020
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)-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:
CROSSREFS
Sequence in context: A044862 A162526 A078299 * A239802 A266242 A274241
KEYWORD
nonn,hard
AUTHOR
Michel Lagneau, Apr 24 2013
EXTENSIONS
a(7) from Donovan Johnson, Apr 25 2013
STATUS
approved

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Last modified June 16 23:34 EDT 2024. Contains 373432 sequences. (Running on oeis4.)