# [seqfan] Re: Regarding Pairs Of Number-Of-Divisors

hv at crypt.org hv at crypt.org
Fri May 14 11:00:58 CEST 2010

```Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:
:Are these sequences in the EIS?
:
:a) a(n) is the minimum positive integer k such that d(m) = n and d(m+1)=k for exactly one positive integer m. a(n) = 0 if there is no m for any k or always more than one m for all k. d(m) is the number of divisors of m.
:
:For example, let n = 4. Checking k = 1,2,3. d(m)=4 and d(m+1)=1 for no m's.
:d(m) = 4 and d(m+1) = 2 for more than one m. (d(6)=4, a(7)=2. d(10)=4, d(11)=2, for example.)
:Now we have, since d(m) = 4, m = p*q, p and q = distinct primes, or m = p^3.
:So, letting k = 3, then m+1 must be of the form r^2, r = prime.
:So, if m = p*q, then either r is even (equals 2) and p and q are both odd, or (WLOG) p is odd and q = 2 and r is odd. m+1 = r^2 doesn't equal 4, because 4-1 doesn't have 4 divisors. So, we have r^2-1 = 2*p.
:(r-1)*(r+1) = 2p. So, r and r+1 are twin primes, and one of them equals 2. But if r-1 = 2, then r+1 = 4, a composite.
:So, m = p^3. Since p^3+1 = r^2, then p and r have opposite parity. So, p must be 2 and r must be 3. (Again, r^2 isn't 4.)
:Therefore a(4) = 3.
:
:Also consider the related sequence:
:b) b(n) is the minimum positive integer k such that d(m) = n and d(m-1)=k for exactly one positive integer m. b(n) = 0 if there is no m for any k or always more than one m for all k.
:
:I think the sequences c and d would be interesting too, if there are an infinite number of 0's in both sequence a and b:
:
:c) Those positive integers n where a(n) = 0.
:
:d) Those positive integers n where b(n) = 0.
:
:Are these in the EIS, if they are interesting?

It is worth checking the OEIS for relevant keywords, and searching through
the recent archives - there was certainly discussion here quite recently
about a, b such that there is a unique n with d(n)=a, d(n+1)=b. Unaccountably
I haven't kept any of the messages from that discussion in my own archive,
though I did look into the question.

My gut feel is that both c() and d() would be infinite.

Hugo

```