

A066467


Numbers having just two antidivisors.


1



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
From David A. Corneth, Oct 23 2019: (Start)
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:
A066272(n) = A000005(2*n1) + A000005(2*n+1) + A001227(n)  5. [See formula in A066272 by Max Alekseyev, Apr 27 2010]
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

Table of n, a(n) for n=1..18.
Jon Perry, The AntiDivisor
Jon Perry, The Antidivisor [Cached copy]
Jon Perry, The Antidivisor: Even More AntiDivisors [Cached copy]


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


MAPLE

A066467:= proc(q)
local k, n, t;
for n from 1 to q do
t:=0; for k from 2 to n1 do if abs((n mod k)k/2)<1 then t:=t+1; fi; od;
if t=2 then print(n); fi; od; end:
A066467 (10^10); # Paolo P. Lava, Feb 22 2013


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]
# Chai Wah Wu, Jul 17 2015
(PARI) nb(n) = if(n>1, numdiv(2*n+1) + numdiv(2*n1) + numdiv(n/2^valuation(n, 2))  5, 0); \\ A066272
isok(m) = nb(m) == 2; \\ Michel Marcus, Oct 28 2019


CROSSREFS

Cf. A066272.
Sequence in context: A034812 A260256 A314573 * A180244 A072833 A229469
Adjacent sequences: A066464 A066465 A066466 * A066468 A066469 A066470


KEYWORD

nonn


AUTHOR

Robert G. Wilson v, Jan 02 2002


EXTENSIONS

a(12)a(13) from Donovan Johnson, Jun 19 2010
a(14) from Jud McCranie, Oct 22 2019
Four terms found by Donovan Johnson, Jan 21 2013 confirmed as next terms by David A. Corneth, Oct 23 2019


STATUS

approved



