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A125246
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Numbers m whose abundance sigma(m) - 2m = -4. Numbers whose deficiency is 4.
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16
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5, 14, 44, 110, 152, 884, 2144, 8384, 18632, 116624, 8394752, 15370304, 73995392, 536920064, 2147581952, 34360131584, 27034175140420610, 36028797421617152, 576460753914036224
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OFFSET
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1,1
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COMMENTS
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a(17) <= b(28) = 36028797421617152 ~ 3.6*10^16, since b(k) := 2^(k-1)*(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
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LINKS
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EXAMPLE
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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^(k-1)*p) = (2^k-1)*(p+1) - 2^k*p = 2^k - p - 1 = -4.
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MATHEMATICA
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Select[Range[10^7], DivisorSigma[1, #] - 2 # == -4 &] (* Michael De Vlieger, Jul 18 2016 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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