A269781 -id:A269781 - cp's OEIS frontend

A272353 Numbers n such that the number of divisors of n+1 divides n and the number of divisors of n divides n+1.

3, 15, 1023, 6399, 10815, 15375, 26895, 53823, 55695, 65535, 80655, 107583, 118335, 262143, 309135, 440895, 614655, 633615, 817215, 891135, 1236543, 1784895, 2676495, 2715903, 2849343, 2985983, 3182655, 3225615, 3268863, 4194303, 4326399, 4343055, 4596735, 5053503

Offset: 1

Waldemar Puszkarz, Apr 26 2016

nonn

One may call such pairs {n, n+1} mutually (or amicably) one step refactorable numbers.
While most terms appear to be divisible by 3, some are not, the first being 2985983=7*11*13*19*157.
From Robert Israel, May 09 2016: (Start)
2^k-1 is a term if k+1 is an odd prime and 2^k-1 is squarefree.
If n is an odd term, then n+1 is a square, and n == 3 mod 4. (End)
There is no term such that last digit of it 1 or 7. Proof: If n is an odd term, then n+1 is a square. For any odd number k, last digit can be trivially 1, 3, 5, 7, 9, that is, the last digit of k+1 is 2, 4, 6, 8, 0 for corresponding odd k values. There cannot be square such that last digit of it 2 or 8. So in this sequence, there is no term such that the last digit of it 1 or 7. - Altug Alkan, May 11 2016

			15 is a term because the number of divisors of 16=15+1, which is 5, divides 15, and the number of divisors of 15, which is 4, divides 16.

Giovanni Resta, Table of n, a(n) for n = 1..2500

Cf. A000005 (number of divisors), A033950 (refactorable numbers), A268037, A269781 (related sequences).

select(t -> (t+1) mod numtheory:-tau(t) = 0 and t mod numtheory:-tau(t+1) = 0, [$1..10^6]); # Robert Israel, May 09 2016

lst={}; Do[ If[ Divisible[n, DivisorSigma[0, n+1]]&&Divisible[n+1, DivisorSigma[0, n]], AppendTo[lst, n]], {n, 7000000}]; lst
Select[Range[7000000], Divisible[#, DivisorSigma[0, # + 1]] && Divisible[# + 1, DivisorSigma[0, #]] &]

for(n=1, 7000000, (n%numdiv(n+1)==0) && ((n+1)%numdiv(n)==0)&& print1(n ", "))

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

Original entry on oeis.org

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

Waldemar Puszkarz, Apr 27 2016

nonn

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

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

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

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

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

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, #]] &]

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

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

cp's OEIS Frontend

A272353 Numbers n such that the number of divisors of n+1 divides n and the number of divisors of n divides n+1.

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