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%I #10 Aug 15 2015 11:15:03
%S 24,36,40,48,60,72,80,84,96,108,112,120,132,140,144,156,160,168,176,
%T 180,192,200,204,208,216,224,228,240,252,264,276,280,288,300,312,320,
%U 324,336,348,352,360,372,384,392,396,400,408,416,420,432,440,444,448,456
%N Numbers n such that the sum of reciprocals of even divisors of n > 1.
%C Numbers n such that the sum of reciprocals of even divisors of n equals m/n for some integer m where the fraction m/n > 1. The corresponding numerators m are given by the sequence A204822(n) = {28, 39, 42, 60, 72, 91, 90, 96,...} (Sum of divisors (A000203) of abundant numbers (A005101).
%H Harvey P. Dale, <a href="/A224907/b224907.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = 2*A005101(n).
%e 40 is in the sequence because the even divisors of 40 are 2, 4, 8 , 10, 20, 40 and 1/2 + 1/4 + 1/8 + 1/10 + 1/20 + 1/40 = 42/40 = A204823(3)/a(3), and 42/40 > 1.
%p ***program 1 where sum of reciprocals even divisors > 1***
%p with(numtheory):for n from 2 by 2 to 500 do:x:=divisors(n):n1:=nops(x): s:=0:for i from 1 to n1 do: if irem(x[i],2)=0 then s:=s+1/x[i]:else fi:od: if s>1 then printf(`%d, `,n):else fi:od:
%p ***program 2 where sum of reciprocals even divisors = m/n***
%p with(numtheory):for n from 2 to 500 do:x:=divisors(n):n1:=nops(x): s:=0:for i from 1 to n1 do: if irem(x[i],2)=0 then s:=s+1/x[i]:else fi:od: for m from n+1 to 2*n do: if s=m/n then printf(`%d, `,n):else fi:od:od:
%t Select[Range[500],Total[1/Select[Divisors[#],EvenQ]]>1&] (* _Harvey P. Dale_, Aug 15 2015 *)
%Y Cf. A000203, A005101, A204822, A224832, A225241.
%K nonn
%O 1,1
%A _Michel Lagneau_, Jul 25 2013
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