A141545 Numbers k whose abundance is 12: sigma(k) - 2*k = 12.

24, 30, 42, 54, 66, 78, 102, 114, 138, 174, 186, 222, 246, 258, 282, 304, 318, 354, 366, 402, 426, 438, 474, 498, 534, 582, 606, 618, 642, 654, 678, 762, 786, 822, 834, 894, 906, 942, 978, 1002, 1038, 1074, 1086, 1146, 1158, 1182, 1194, 1266, 1338, 1362

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Aug 16 2008

Keywords

• nonn

Comments

Numbers k such that sigma(k) = 2k + 12. - Wesley Ivan Hurt, Jul 11 2013

Any term x = a(m) can be combined with any term y = A141549(n) to satisfy the property (sigma(x)+sigma(y))/(x+y) = 2. Although this property is a necessary condition for two numbers to be amicable, it is not a sufficient one. So far, these two sequences have not produced an amicable pair. However, if one is ever found, then it will exhibit y-x = 12. - Timothy L. Tiffin, Sep 13 2016

From Tomohiro Yamada, Jan 01 2023: (Start)

6p belongs to this sequence if p > 3 is prime since sigma(6p) = 12(p + 1) = 12p + 12. Moreover, 2^m * (2^(m+1) - 13) is also a term of this sequence if 2^(m+1) - 13 is prime (m+1 is a term of A096818) since sigma(2^m * (2^(m+1) - 13)) = (2^(m+1) + 1) * (2^(m+1) - 13) = 2^(m+1) * (2^(m+1) - 13) + 12. So 24, 304, 127744, 33501184, and 8589082624 also belong to this sequence.

Problem: is 54 the only term of this sequence which is of neither type given above? (End)

Examples

			30 is in the sequence since sigma(30) = sigma(2*3*5) = sigma(2)*sigma(3)*sigma(5) = 3*4*6 = 72 = 2(30)+12.  Since this is the second such number whose abundance is 12, a(2) = 30. - _Wesley Ivan Hurt_, Jul 11 2013
		

Links

• Donovan Johnson, Table of n, a(n) for n = 1..10000
• Farideh Firoozbakht and M. F. Hasler, Variations on Euclid's formula for Perfect Numbers, JIS 13 (2010) #10.3.1.

Crossrefs

Cf. A000203, A005101, A141549 (deficiency 12).

Cf. A076496 (sigma(k) - a*k = 12).

Programs

Magma

[n: n in [1..1400] | (SumOfDivisors(n)-2*n) eq 12]; // Vincenzo Librandi, Sep 14 2016

Mathematica

lst={};Do[If[n==Plus@@Divisors[n]-n-12,AppendTo[lst,n]],{n,10^4}];Print[lst];
Select[Range[1, 10^4], DivisorSigma[1, #] - 2 # == 12 &] (* Vincenzo Librandi, Sep 14 2016 *)

PARI

is(n)=sigma(n)==2*n+12 \\ Charles R Greathouse IV, Feb 21 2017