# d(m) = d(m+1) = 2n

Hugo van der Sanden human at google.com
Sat Aug 16 02:54:33 CEST 2008

```I think the proof can be made quite rigorous: it is clear that the least m
with tau(m) = n must be composed of minimal primes. as you mentioned, so for
n > 1, m must be even. So we have m+1 odd, tau(m+1) = n, and m+1 is the
second smallest k with tau(k) = n. It follows (for n > 1) that m+1 = 3^x and
m = 2^x, satisfiable only by x=1.

It is of related but less vital interest to look at the sequence of t(n, 2)
- t(n, 1) where t(a, b) represents the b'th +ve integer with a factors, or
of successive minima in the sequence.

If my (hand-) calculations are correct, the first looks like [ 1, 5, 2, 65,
6, 665, 6, 64, 32, 58025, 12, ... ], not currently in OEIS. The second
probably looks like [ 1, 2, 6, 12, 48?, ... ], but constructing the
appropriate proof for each value is more work.

Hugo

2008/8/14 Leroy Quet <q1qq2qqq3qqqq at yahoo.com>

> Never mind. It seems quite obvious that there are no more m's or n's.
>
> Someone please correct me if I am wrong:
> Let us say that m has 2n divisors. Then there are a number of ways the
> exponents in the prime-factorization of m, independent of the primes in the
> prime-factorization, can be written. (Since 2n = product(1 + e_k), where
> each e_k is an exponent in the prime-factorization.)
>
> Since m is the SMALLEST m with 2n divisors, then the primes dividing m are
> all consecutive primes starting with 2.
>
> But m+1 is coprime to m. So its primes must differ from those dividing m,
> and therefore be larger than the largest prime dividing m.
> Now, it is possible that the exponents in the prime-factorization of m+1
> (which lead to the number of divisors of m+1 being 2n) would allow m+1 to be
> very close to m (one away from m, in fact), because the prime-factorization
> exponents of m+1 are smaller than those of m. But if that is so, then
> replacing the primes that divide m+1 with consecutive primes starting at 2,
> using the same prime-factorization exponents, would produce an integer <= m.
>
> This is all unrigorous, of course. But I am tired and lazy. The makings of
> the start of a rigorous proof are here, hopefully.
>
> Thanks,
> Leroy Quet
>
>
>
> --- On Thu, 8/14/08, Leroy Quet <q1qq2qqq3qqqq at yahoo.com> wrote:
>
> > From: Leroy Quet <q1qq2qqq3qqqq at yahoo.com>
> > Subject: d(m) = d(m+1) = 2n
> > To: seqfan at ext.jussieu.fr
> > Cc: qq-quet at mindspring.com
> > Date: Thursday, August 14, 2008, 9:38 PM
> > Consider the two sequences, one of positive integers m and
> > one of positive integers n:
> > m and n are such that m is the smallest positive integer
> > with 2n divisors, and m+1 also has 2n divisors.
> >
> > For example, 2 has 2 divisors, and 2+1 also has 2 divisors.
> > And 2 is the smallest positive integer with 2 divisors.
> >
> > So the m sequence starts: 2,...
> > And the n sequence starts: 1,...
> >
> > This may be easily answered, I bet, but are there any other
> > m's and n's?
> >
> > Thanks,
> > Leroy Quet
>
>
>
>
>
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