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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

Let m and k be distinct integers and numdiv(n) be the number of divisors of n (A000005(n)). We call m and k amicably refactorable if numdiv(m) divides k and numdiv(k) divides m.

For any n with no amicably refactorable partner, a(n) = 0.

Conjecture: the sequence contains no zeros.

1 does not have an amicable partner as all other numbers have more than one divisor and 2 does not have an amicable partner as all other numbers with two divisors are odd primes and cannot be divided by the number of divisors of 2, also 2. All other numbers may have an amicably refactorable partner, though for some, primes, semiprimes and squares of primes in particular, this number can be quite large.

For primes and semiprimes, a(n) = 2^(f(n) - 1), (see A061286), where f(n) is their largest prime factor. For squares of primes, a(n) = 3^(|sqrt(n)| - 1), except for n = 9 for which this formula yields 9; this forces us to choose the next best candidate: 36.