

A066542


Nonnegative integers all of whose antidivisors are either 2 or odd.


2



3, 4, 5, 7, 8, 11, 13, 16, 17, 19, 23, 29, 31, 32, 37, 41, 43, 47, 53, 59, 61, 64, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 128, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251
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OFFSET

1,1


COMMENTS

See A066272 for definition of antidivisor.
The following conjectures have been proved by Bob Selcoe.  Michael Somos, Feb 28 2014
Additional conjectures suggested by computational experiments:
1) Numbers all of whose antidivisors (AD's) are odd => {2^k} (A000079).
2) Numbers with AD 2, all other AD's odd => primes (A000040).
3) Numbers none of whose AD's are multiples of 3 => 3*2^k (A007283).
4) Numbers all of whose AD's are even => 3*A002822 = A040040 (except for a(0)=1), both related to twin prime pairs.
Calculations suggest the following conjecture. This sequence consists of all odd primes and nonnegative powers of 2 and no other terms. This has been verified for to n=100000. Robert G. Wilson v extended the conjecture out to 2^20.
The sequence consists of all odd primes and powers of two (>=2^2) and no other terms.
Proof: Denote the even antidivisors of n as ADe(n). ADe(n) is defined as the set of numbers x satisfying the equation n(mod x)=x/2. Substitute x = 2n/y, since it can be shown that ADe(n) => 2n divided by the odd divisors of n when n>1 (This is because 2j antidivides only numbers of the form 3j+2j*k; j>=1, k>=0. For example: j=7; 14 antidivides only 21,35,49,63.... So in other words, even numbers antidivide only odd multiples (>=3) of themselves, divided by 2). Therefore, ADe(n) is n(mod [2n/y])=n/y, and y must be an odd divisor of n and 2n, y>1. Since y is the only odd divisor of n when y>1 iff n is prime, then ADe(n) => 2 when n is prime. Since 2n has no odd divisors when n=2^k, then ADe(n) is null when n=2^k. Therefore, the only numbers whose antidivisors are either 2 or odd must be primes and powers of 2.
Similarly, for odd antidivisors (ADo(n)): Given 2j+1 (odd numbers) antidivide only numbers of the forms [(3j+1)+(2j+1)*k] and [(3j+2)+(2j+1)*k]; j>=1, k>=0. (For example: j=6; 13 antidivides only 19,20, 32,33, 45,46...). Since odd n divided by its odd divisors ARE its odd divisors, then ADo(n) => the divisors of 2n1 and 2n+1 (except 1, 2n1 and 2n+1).
By extension:
1) Numbers all of whose antidivisors (AD's) are odd => {2^k} (A000079).
2) Numbers with ADe(n)=2, all other AD's odd => primes (A000040).
3) Numbers none of whose AD's are multiples of j => j*2^k.
4) When 2n1 and 2n+1 are twin primes, (A040040, except for a(0)=1) then n has only even AD's.
(End)
If 1 and 2 are included, this sequence contains all positive integers not contained in A111774.  Bob Selcoe, Sep 09 2014 [corrected by Wolfdieter Lang, Nov 06 2020]


LINKS



EXAMPLE

ADe(420): Odd divisors of 420 are: 3,5,7,15,21,35, 105. ADe(420) => 840/{3,5,7,15,21,35,105} = 8,24,40,56,120,168 and 280.
ADo(420) => the divisors of 839 and 841, which are (a) for 839: null (839 is prime); and (b) for 841: 29 (841 is 29^2).
All AD's (AD(420)) => 8,24,29,40,56,120,168 and 280 (End)


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 & ]; f[n_] := Select[ antid[n], EvenQ[ # ] && # > 2 & ]; Select[ Range[3, 300], f[ # ] == {} & ]


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



