A268037 Numbers k such that the number of divisors of k+2 divides k and the number of divisors of k divides k+2.

4, 30, 48, 110, 208, 270, 320, 368, 510, 590, 688, 750, 1070, 1216, 1328, 1566, 1808, 2030, 2190, 2510, 2670, 2768, 3008, 3088, 3728, 4110, 4208, 4430, 4528, 4688, 4698, 4910, 5008, 5696, 5870, 5886, 5968, 6128, 6592, 6846, 7088, 7310, 7790, 8384, 9008, 9230, 9390, 9488, 9534, 9710

Offset: 1

Author

Waldemar Puszkarz, Apr 27 2016

Keywords

• nonn

Comments

One can call such pairs {n, n+2} mutually (or amicably) step 2 refactorable numbers.

All terms are even.

Proof. Let us suppose n is odd. Then so is n+2 and the number of their divisors has to be odd, which implies they are squares (see the proof in A268066). Since the separation between closest squares is 2n+1, it is always greater than 2, except for n=0, when it is 1.

Contains 48 + 160*k if 3 + 10*k and 5 + 16*k are prime. Dickson's conjecture implies that there are infinitely many of these. - Robert Israel, May 09 2016

Examples

			4 is a term because its number of divisors (3) divides 6=4+2 and the number of divisors of 6 (4) divides 4.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005 (number of divisors), A033950 (refactorable numbers), A272353, A269781 (related sequences).

Programs

Maple

select(n -> n mod numtheory:-tau(n+2)=0 and (n+2) mod numtheory:-tau(n) = 0,
2*[$1..10000]); # Robert Israel, May 09 2016

Mathematica

lst={}; Do[ If[ Divisible[n, DivisorSigma[0, n+2]]&&Divisible[n+2, DivisorSigma[0, n]], AppendTo[lst, n]], {n, 12000}]; lst
Select[Range[12000], Divisible[#, DivisorSigma[0, # + 2]] && Divisible[# + 2, DivisorSigma[0, #]] &]

PARI

for(n=1, 12000, (n%numdiv(n+2)==0)&&((n+2)%numdiv(n)==0)&&print1(n ", "))

Python

from sympy import divisors
def ok(n): return n%len(divisors(n+2)) == 0 and (n+2)%len(divisors(n)) == 0
print(list(filter(ok, range(1, 9711)))) # Michael S. Branicky, Apr 30 2021