|
| |
|
|
A191363
|
|
Numbers m such that sigma(m) = 2*m - 2.
|
|
14
|
| |
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Let k be a nonnegative integer such that F(k) = 2^(2^k) + 1 is prime (a Fermat prime A019434), then m = (F(k)-1)*F(k)/2 appears in the sequence.
Conjecture: a(1)=3 is the only odd term of the sequence.
Conjecture: All terms of the sequence are of the above form derived from Fermat primes.
The sequence has 5 (known) terms in common with sequences A055708 (k-1 | sigma(k)) and A056006 (k | sigma(k)+2) since {a(n)} is a subsequence of both.
The first five terms of the sequence are respectively congruent to 3, 4, 4, 4, 4 modulo 6.
After a(5) there are no further terms < 8*10^9.
Up to m = 1312*10^8 there are no further terms in the class congruent to 4 modulo 6.
See A125246 for numbers with deficiency 4, i.e., sigma(m) = 2*m - 4, and A141548 for numbers with deficiency 6. - M. F. Hasler, Jun 29 2016 and Jul 17 2016
A term m of this sequence multiplied by a prime p not dividing it is abundant if and only if p < m-1. For each of a(2..5) there is such a prime near this limit (here: 7, 127, 30197, 2147483647) such that a(k)*p is a primitive weird number, cf. A002975. - M. F. Hasler, Jul 19 2016
Any term m of this sequence can be combined with any term j of A088831 to satisfy the property (sigma(m) + sigma(j))/(m+j) = 2, which is a necessary (but not sufficient) condition for two numbers to be amicable. [Proof: If m = a(n) and j = A088831(k), then sigma(m) = 2m-2 and sigma(j) = 2j+2. Thus, sigma(m) + sigma(j) = (2m-2) + (2j+2) = 2m + 2j = 2(m+j), which implies that (sigma(m) + sigma(j))/(m+j) = 2(m+j)/(m+j) = 2.] - Timothy L. Tiffin, Sep 13 2016
Conjectures: all terms are second hexagonal numbers (A014105). There are no terms with middle divisors. - Omar E. Pol, Oct 31 2018
The symmetric representation of sigma(m) of each of the 5 numbers in the sequence consists of 2 parts of width 1 that meet at the diagonal (subsequence of A246955). - Hartmut F. W. Hoft, Mar 04 2022
The first five terms coincide with the sum of two successive terms of A058891. The same is not true for a(6), if such exists. - Omar E. Pol, Mar 03 2023
|
|
|
LINKS
|
Gianluca Amato, Maximilian Hasler, Giuseppe Melfi, and Maurizio Parton, Primitive weird numbers having more than three distinct prime factors, Riv. Mat. Univ. Parma, 7(1), (2016), 153-163, arXiv:1803.00324 [math.NT], 2018.
|
|
|
FORMULA
|
|
|
|
EXAMPLE
|
For n=1, a(1) = 3 since sigma(3) = 4 = 2*3 - 2.
|
|
|
MATHEMATICA
|
ok[n_] := DivisorSigma[1, n] == 2*n-2; Select[ Table[ 2^(2^k-1) * (2^(2^k)+1), {k, 0, 5}], ok] (* Jean-François Alcover, Sep 14 2011, after conjecture *)
Select[Range[10^6], DivisorSigma[1, #] == 2 # - 2 &] (* Michael De Vlieger, Sep 14 2016 *)
|
|
|
PROG
|
(PARI) zp(a, b) = {my(c, c1, s); c = a; c1 = 2*c-2;
while(c<b, s = sigma(c); if(s == c1, print(c); ); c1 = c1 + 2; c = c+1); }
(PARI) a(k)=(2^2^k+1)<<(2^k-1) \\ For k<6. - M. F. Hasler, Jul 27 2016
(Magma) [n: n in [1..9*10^6] | (SumOfDivisors(n)-2*n) eq -2]; // Vincenzo Librandi, Sep 15 2016
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,hard,more
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
