login
Nonnegative integers all of whose anti-divisors are either 2 or odd.
2

%I #48 Nov 20 2020 13:45:25

%S 3,4,5,7,8,11,13,16,17,19,23,29,31,32,37,41,43,47,53,59,61,64,67,71,

%T 73,79,83,89,97,101,103,107,109,113,127,128,131,137,139,149,151,157,

%U 163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251

%N Nonnegative integers all of whose anti-divisors are either 2 or odd.

%C See A066272 for definition of anti-divisor.

%C The following conjectures have been proved by Bob Selcoe. - _Michael Somos_, Feb 28 2014

%C Additional conjectures suggested by computational experiments:

%C 1) Numbers all of whose anti-divisors (AD's) are odd => {2^k} (A000079).

%C 2) Numbers with AD 2, all other AD's odd => primes (A000040).

%C 3) Numbers none of whose AD's are multiples of 3 => 3*2^k (A007283).

%C 4) Numbers all of whose AD's are even => 3*A002822 = A040040 (except for a(0)=1), both related to twin prime pairs.

%C 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.

%C From _Bob Selcoe_, Feb 24 2014: (Start)

%C The sequence consists of all odd primes and powers of two (>=2^2) and no other terms.

%C Proof: Denote the even anti-divisors 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 anti-divides only numbers of the form 3j+2j*k; j>=1, k>=0. For example: j=7; 14 anti-divides only 21,35,49,63.... So in other words, even numbers anti-divide 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 anti-divisors are either 2 or odd must be primes and powers of 2.

%C Similarly, for odd anti-divisors (ADo(n)): Given 2j+1 (odd numbers) anti-divide 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 anti-divides 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 2n-1 and 2n+1 (except 1, 2n-1 and 2n+1).

%C By extension:

%C 1) Numbers all of whose anti-divisors (AD's) are odd => {2^k} (A000079).

%C 2) Numbers with ADe(n)=2, all other AD's odd => primes (A000040).

%C 3) Numbers none of whose AD's are multiples of j => j*2^k.

%C 4) When 2n-1 and 2n+1 are twin primes, (A040040, except for a(0)=1) then n has only even AD's.

%C (End)

%C 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]

%e From _Bob Selcoe_, Feb 24 2014: (Start)

%e 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.

%e 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).

%e All AD's (AD(420)) => 8,24,29,40,56,120,168 and 280 (End)

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

%Y Cf. A000040, A000079, A002822, A007283, A040040.

%Y Cf. A014545, A006862, A111774.

%K nonn

%O 1,1

%A _John W. Layman_, Jan 07 2002