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A326137 Numbers with at least five distinct prime factors that satisfy Euler's criterion (A228058) for odd perfect numbers.

%I A326137 #19 Feb 16 2025 08:33:58
%S A326137 17342325,22678425,31674825,38686725,41420925,45090045,49358925,
%T A326137 51740325,54033525,54695025,67660425,68939325,70703325,75818925,
%U A326137 76392225,77106645,78217425,81375525,92400525,96316605,97383825,98750925,99147825,102284325,107694405,113656725,115420725,117890325,118728225,120536325,127766925
%N A326137 Numbers with at least five distinct prime factors that satisfy Euler's criterion (A228058) for odd perfect numbers.
%C A326137 P. P. Nielsen's 2006 paper shows that any odd perfect number must have at least nine distinct prime factors, thus if such numbers exist at all, they must occur in this sequence.
%C A326137 I conjecture that it is eventually possible to find an easy proof that this sequence has no common terms with A325981, and/or several other sequences (A326064, A326074, A326141, A326148, etc.) listed under index entry "sequences where odd perfect numbers must occur", thus settling the question about the existence of such numbers.
%H A326137 Antti Karttunen, <a href="/A326137/b326137.txt">Table of n, a(n) for n = 1..1032; all terms < 2^31</a>
%H A326137 Charles Greathouse and Eric W. Weisstein, <a href="https://mathworld.wolfram.com/OddPerfectNumber.html">MathWorld: Odd perfect number</a>
%H A326137 Oliver Knill, <a href="http://www.math.harvard.edu/~knill/seminars/perfect/handout.pdf">The oldest open problem in mathematics</a>, Handout for NEU Math Circle, December 2, 2007
%H A326137 P. P. Nielsen, <a href="http://arxiv.org/abs/math/0602485">Odd Perfect Numbers Have At Least Nine Distinct Prime Factors</a>, arXiv:math/0602485 [math.NT], 2006.
%H A326137 Wikipedia, <a href="https://en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers">Perfect number: Odd perfect numbers</a>
%H A326137 <a href="/index/O#opnseqs">Index entries for sequences where any odd perfect numbers must occur</a>
%o A326137 (PARI)
%o A326137 isA228058(n) = if(!(n%2)||(omega(n)<2),0,my(f=factor(n),y=0); for(i=1,#f~,if(1==(f[i,2]%4), if((1==y)||(1!=(f[i,1]%4)),return(0),y=1), if(f[i,2]%2, return(0)))); (y));
%o A326137 isA326137(n) = ((omega(n)>=5)&&isA228058(n));
%Y A326137 Subsequence of A228058.
%Y A326137 Cf. A001221, A325981, A326064, A326074, A326141, A326148.
%K A326137 nonn
%O A326137 1,1
%A A326137 _Antti Karttunen_, Jun 12 2019