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%I #51 Feb 25 2024 05:58:10
%S 5,8,9,12,16,24,36,64,576,4096,65536,262144,1073741824,39582418599936,
%T 1152921504606846976,41505174165846491136,
%U 85070591730234615865843651857942052864,14809541015890854379394722643016154544844622790702218770137481216
%N Numbers having just two anti-divisors.
%C See A066272 for definition of anti-divisor.
%C 2^42*3^2, 2^62*3^2, 2^210*3^2, 2^60 and 2^126 are terms. If 2*k-1 and 2*k+1 are both prime and k has exactly three odd divisors, then k is a term. Also if 2^p-1 is a Mersenne prime and 2^p+1 is the product of two distinct primes, then 2^(p-1) is a term. - _Donovan Johnson_, Jan 21 2013
%C From _David A. Corneth_, Oct 23 2019: (Start)
%C Terms greater than 5 are of the form 2^i * 3^j where i >= 0 and 0 <= j <= 2.
%C From the number of antidivisors of n (A066272) we have:
%C A066272(n) = A000005(2*n-1) + A000005(2*n+1) + A001227(n) - 5. [See formula in A066272 by _Max Alekseyev_, Apr 27 2010]
%C Let tau(k) be the number of divisors of k (A000005(k)).
%C Let odd(k) be the odd part of k (A001227(k)).
%C So tau(2*m + 1) + tau(2*m - 1) + tau(odd(m)) = 7.
%C 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}.
%C If tau(odd(m)) = 1 then m = 2^k which confirms our claim.
%C If tau(odd(m)) = 2 then m = 2^k * p for some odd prime p.
%C If p > 3 then p = 6*t +- 1 for some t > 0.
%C 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.
%C Similar reasoning holds for tau(odd(m)) = 3, i.e., m = 2^k * p^2. (End)
%H Jon Perry, <a href="http://www.users.globalnet.co.uk/~perry/maths/antidivisor.htm">The Anti-Divisor</a>
%H Jon Perry, <a href="/A066272/a066272a.html">The Anti-divisor</a> [Cached copy]
%H Jon Perry, <a href="/A066272/a066272.html">The Anti-divisor: Even More Anti-Divisors</a> [Cached copy]
%e For m = 12: 2m-1, 2m, 2m+1 are 23, 24, 25 with odd divisors > 1 {23}, {8}, {5} and quotients 1, 3, 5 so the anti-divisors of 12 are 3 and 5. Therefore 12 is a term of this sequence. - _Bernard Schott_, Oct 23 2019
%t 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 & ]
%o (Python)
%o from sympy.ntheory.factor_ import antidivisor_count
%o A066467_list = [n for n in range(1,10**5) if antidivisor_count(n) == 2]
%o # _Chai Wah Wu_, Jul 17 2015
%o (PARI) nb(n) = if(n>1, numdiv(2*n+1) + numdiv(2*n-1) + numdiv(n/2^valuation(n, 2)) - 5, 0); \\ A066272
%o isok(m) = nb(m) == 2; \\ _Michel Marcus_, Oct 28 2019
%Y Cf. A066272.
%K nonn
%O 1,1
%A _Robert G. Wilson v_, Jan 02 2002
%E a(12)-a(13) from _Donovan Johnson_, Jun 19 2010
%E a(14) from _Jud McCranie_, Oct 22 2019
%E Four terms found by _Donovan Johnson_, Jan 21 2013 confirmed as next terms by _David A. Corneth_, Oct 23 2019
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