

A066467


Numbers having just two antidivisors.


0



5, 8, 9, 12, 16, 24, 36, 64, 576, 4096, 65536, 262144, 1073741824, 39582418599936, 1152921504606846976, 41505174165846491136, 85070591730234615865843651857942052864, 14809541015890854379394722643016154544844622790702218770137481216
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OFFSET

1,1


COMMENTS

See A066272 for definition of antidivisor.
2^42*3^2, 2^62*3^2, 2^210*3^2, 2^60 and 2^126 are terms. If 2*k1 and 2*k+1 are both prime and k has exactly three odd divisors, then k is a term. Also if 2^p1 is a Mersenne prime and 2^p+1 is the product of two distinct primes, then 2^(p1) is a term.  Donovan Johnson, Jan 21 2013
Terms greater than 5 are of the form 2^i * 3^j where i >= 0 and 0 <= j <= 2.
From the number of antidivisors of n (A066272) we have:
Let tau(k) be the number of divisors of k (A000005(k)).
Let odd(k) be the odd part of k (A001227(k)).
So tau(2*m + 1) + tau(2*m  1) + tau(odd(m)) = 7.
As m > 1, we have tau(2*m + 1) >= 2 and tau(2*m  1) >= 2, i.e., tau(odd(m)) in {1, 2, 3}.
If tau(odd(m)) = 1 then m = 2^k which confirms our claim.
If tau(odd(m)) = 2 then m = 2^k * p for some odd prime p.
If p > 3 then p = 6*t + 1 for some t > 0.
Then 2*m + 1 or 2*m  1 is divisible by 3, so they can only be 3^2, which gives m = 5. Otherwise, it is a semiprime and one of tau(2*m + 1) or tau(2*m  1) = 4 and we have too many antidivisors.
Similar reasoning holds for tau(odd(m)) = 3, i.e., m = 2^k * p^2. (End)


LINKS



EXAMPLE

For m = 12: 2m1, 2m, 2m+1 are 23, 24, 25 with odd divisors > 1 {23}, {8}, {5} and quotients 1, 3, 5 so the antidivisors of 12 are 3 and 5. Therefore 12 is a term of this sequence.  Bernard Schott, Oct 23 2019


MATHEMATICA

antid[n_] := Select[ Union[ Join[ Select[ Divisors[2n  1], OddQ[ # ] && # != 1 & ], Select[ Divisors[2n + 1], OddQ[ # ] && # != 1 & ], 2n/Select[ Divisors[ 2*n], OddQ[ # ] && # != 1 &]]] }, # < n & ]]; Select[ Range[10^5], Length[ antid[ # ]] == 2 & ]


PROG

(Python)
from sympy.ntheory.factor_ import antidivisor_count
A066467_list = [n for n in range(1, 10**5) if antidivisor_count(n) == 2]
(PARI) nb(n) = if(n>1, numdiv(2*n+1) + numdiv(2*n1) + numdiv(n/2^valuation(n, 2))  5, 0); \\ A066272


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



