

A125246


Numbers m whose abundance sigma(m)  2m = 4. Numbers whose deficiency is 4.


16



5, 14, 44, 110, 152, 884, 2144, 8384, 18632, 116624, 8394752, 15370304, 73995392, 536920064, 2147581952, 34360131584, 27034175140420610, 36028797421617152, 576460753914036224
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OFFSET

1,1


COMMENTS

a(17) <= b(28) = 36028797421617152 ~ 3.6*10^16, since b(k) := 2^(k1)*(2^k+3) is in this sequence for all k in A057732, i.e., whenever 2^k+3 is prime, and 28 = A057732(11). Further terms of this form are b(30), b(55), b(67), b(84), ... The only terms not of the form b(k), below 10^13, are {110, 884, 18632, 116624, 15370304, 73995392}.  M. F. Hasler, Apr 27 2015, edited on Jul 17 2016
A term of this sequence multiplied with a prime p not dividing it is abundant if and only if p < sigma(a(n))/4. For each of a(2..16) there is such a prime, near this limit, such that a(n)*p is a primitive weird number, cf. A002975.  M. F. Hasler, Jul 17 2016
Any term x of this sequence can be combined with any term y of A088832 to satisfy the property (sigma(x)+sigma(y))/(x+y) = 2, which is a necessary (but not sufficient) condition for two numbers to be amicable.  Timothy L. Tiffin, Sep 13 2016
Is 5 the only odd number in this sequence? Is it possible to prove this?  M. F. Hasler, Feb 22 2017


LINKS



EXAMPLE

The abundance of 5 = (1+5)10 = 4.
More generally, whenever p = 2^k + 3 is prime (as p = 5 for k = 1), then A(2^(k1)*p) = (2^k1)*(p+1)  2^k*p = 2^k  p  1 = 4.


MATHEMATICA

Select[Range[10^7], DivisorSigma[1, #]  2 # == 4 &] (* Michael De Vlieger, Jul 18 2016 *)


PROG

(PARI) for(n=1, 1000000, if(((sigma(n)2*n)==4), print1(n, ", ")))
(Magma) [n: n in [1..9*10^6]  (SumOfDivisors(n)2*n) eq 4]; // Vincenzo Librandi, Sep 15 2016


CROSSREFS



KEYWORD

nonn,more


AUTHOR



EXTENSIONS



STATUS

approved



