
COMMENTS

a(n) is the smallest number m such that tau(m) = tau(m+1)  1 = n, where tau(m) = the number of divisors of m (A000005).
Other terms: a(32) = 1476224, a(48) = 7529535, a(80) = 27709695. There are no other positive terms <= 10^8.
Next terms: a(44) = 358913024, a(63) = 288422289, a(64) = 116985855, a(224) = 10702937024.  Vaclav Kotesovec, Apr 07 2021
The number m = 27285093123 is the first start of run of 3 consecutive integers m, m+1 and m+2 with triplet [tau(m), tau(m+1), tau(m+2)] = [tau(m), tau(m) + 1, tau(m) + 2]: [tau(27285093123), tau(27285093124), tau(27285093125)] = [8, 9, 10].
a(5) = 0. Proof: If tau(m) = 5 then m = p^4, where p is an odd prime. Then m+1 is even and tau(m+1) = 6.
There are 3 possible forms of m+1 = p^4+1 that we have to consider:
1) m+1 = 2^5 which is not a solution since tau(31) = 2.
2) m+1 = 4*q, where q is an odd prime, which is impossible since p^4 + 1 = 4*q has no solution as p^4 == 1 (mod 4) for an odd prime p.
3) m+1 = 2*q^2, where q is an odd prime, which is impossible since p^4 + 1 = 2*q^2 has no solution with p and q primes (see Cohn, 1997). (End)
a(10) = 0. Proof: if m has 10 divisors and m + 1 has 11 divisors then m + 1 = p^10 for some prime p. Then m = p^10  1 = (p  1)*(p + 1)*(p^4  p^3 + p^2  p + 1)*(p^4 + p^3 + p^2 + p + 1) which has as least 16 divisors for p > 2. Case p = 2 does not give a solution.
a(12) = 0. Proof: if m has 12 divisors and m + 1 has 13 divisors then m + 1 = p^12 for some prime p. Then m = p^12  1 = (p  1)*(p + 1)*(p^2  p + 1)*(p^2 + 1)*(p^2 + p + 1)*(p^4  p^2 + 1) which has at least 24 divisors for p >= 2. So m cannot have 12 divisors. (End)
For each n (except where a(n) = 0), since m and m+1 have n and n+1 divisors, respectively, and either n or n+1 is odd, it follows that either m or m+1 is a square; i.e., each term is either a square or one less than a square.
The only terms <= 10^16 not listed above are a(39) = 5633825050624, a(56) = 8601122276288, a(95) = 281354730471424, a(104) = 9274308890624, a(128) = 2326566604374015, a(144) = 5409275354546175, a(255) = 1778655385595664, a(384) = 357507737015624, and a(512) = 14765267353599. (Two additional known terms and one known upper bound are far larger; see below.)
For each prime p, a(p) = q^(p1) where q is the smallest prime such that q^(p1) + 1 has p+1 divisors (or, if no such prime q exists, a(p) = 0). At present, the only primes p for which such a prime q is known to exist are p = 2, 3, 7, 11, 23, 31, and 47, yielding the terms a(2) = 3^1 = 3, a(3) = 3^2 = 9, a(7) = 3^6 = 729, a(11) = 3^10 = 59049, a(23) = 41^22 = 302862043149743582494593171234930481, a(31) = 2183431^30 (a 191digit number), and the upper bound a(47) <= 1483^46, respectively. (a(47) is 1459^46 or 1483^46, depending on whether 1459^46+1 has 48 divisors.) (Observation: for each prime p in {2, 3, 7, 11, 23, 31, 47}, p+1 is 3smooth.)
Conjecture: a(n) = 0 for more than 95% of the indices n in 1..1000. (End)
