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Abundant numbers n such that n/(sigma(n)-2n) is an integer.
12

%I #31 Feb 01 2023 12:28:33

%S 12,18,20,24,40,56,88,104,120,196,224,234,368,464,650,672,992,1504,

%T 1888,1952,3724,5624,9112,11096,13736,15376,15872,16256,17816,24448,

%U 28544,30592,32128,77744,98048,122624,128768,130304,174592,396896,507392

%N Abundant numbers n such that n/(sigma(n)-2n) is an integer.

%C Sigma(n)-2n is the abundance of n.

%C The only odd term in this sequence < 2*10^12 is 173369889. - _Donovan Johnson_, Feb 15 2012

%C Equivalently, the abundancy of n, ab=sigma(n)/n, satisfies the following relation: numerator(ab) = 2*denominator(ab)+1, that is, ab=(2k+1)/k where k is the integer ratio mentioned in definition. - _Michel Marcus_, Nov 07 2014

%C The tri-perfect numbers (A005820) are in this sequence, since their abundancy is 3n/n = 3 = (2k+1)/k with k=1. - _Michel Marcus_, Nov 07 2014

%H Donovan Johnson, <a href="/A153501/b153501.txt">Table of n, a(n) for n = 1..200</a>

%e The abundance of 174592 = sigma(174592)-2*174592 = 43648. 174592/43648 = 4.

%p filter:= proc(n) local s; s:= numtheory:-sigma(n); (s > 2*n) and (n mod (s-2*n) = 0) end proc:

%p select(filter, [$1..10^5]); # _Robert Israel_, Nov 07 2014

%t filterQ[n_] := Module[{s = DivisorSigma[1, n]}, s > 2n && Mod[n, s - 2n] == 0];

%t Select[Range[10^6], filterQ] (* _Jean-François Alcover_, Feb 01 2023, after _Robert Israel_ *)

%o (PARI) isok(n) = ((ab = (sigma(n)-2*n))>0) && (n % ab == 0) \\ _Michel Marcus_, Jul 16 2013

%o (Sage)

%o def A153501_list(len):

%o def is_A153501(n):

%o t = sigma(n,1) - 2*n

%o return t > 0 and t.divides(n)

%o return filter(is_A153501, range(1,len))

%o A153501_list(1000) # _Peter Luschny_, Nov 07 2014

%Y Intersection of A097498 and A005101.

%Y Disjoint union of A181595 and A005820.

%Y Cf. A000203, A033880.

%K nonn

%O 1,1

%A _Donovan Johnson_, Jan 02 2009