OFFSET
1,1
COMMENTS
Run through all natural numbers i = 1, 2, 3, ... in order, and record for each the abundancy index sigma(i)/i. When we reach an abundancy that has been seen before, output first the "old" number which had that abundancy (unless that number has already been output earlier), and output secondly the current i.
By construction, no number can occur more than once in the sequence.
Friendly numbers that are not smallest in their club, appear in increasing order. Friendly numbers that are smallest in their club, appear just before the second-smallest member.
If we were to "forget" to output the smallest member in each club, we would get instead A095301.
Oppositely, if we output the smallest members only, we get instead A259918.
It is not known whether the number 10 belongs to this sequence.
LINKS
Jeppe Stig Nielsen, Table of n, a(n) for n = 1..6000
Doyon Kim, Friends of 12, arXiv:1608.06834 [math.HO], 2016.
Jeffrey Ward, Does Ten Have a Friend?, arXiv:0806.1001 [math.NT], 2008.
Eric Weisstein's World of Mathematics, Friendly Pair
Wikipedia, Friendly number.
PROG
(PARI) known=List(); for(i=1, 10^5, a=sigma(i)/i; match=0; for(j=1, #known, if(known[j][1]==a, match=j; break())); if(match, old=known[match][2]; if(old, print1(old, ", "); known[match]=[a, 0]); print(i, ", "), listput(known, [a, i])))
CROSSREFS
KEYWORD
nonn
AUTHOR
Jeppe Stig Nielsen, Jul 08 2015
STATUS
approved