# [seqfan] Re: A351913: numerator(1/d - 1/n) [was: Re: Upper bound for A091895 and A091896]

hv at crypt.org hv at crypt.org
Thu Apr 14 16:31:54 CEST 2022

```"M. F. Hasler" <oeis at hasler.fr> wrote:
:> On Tue, Apr 5, 2022 at 10:51 AM Michel Marcus wrote:
:>
:>> A bit different, did you see A351913 <http://oeis.org/A351913> ?
:>> Any idea when can we say that Unknown can be set to -1 ?
:>
:>
:This is an interesting question, I'd like to make it more explicit to
:encourage people on this list who might be able to establish some relevant
:results.
:The problem is to show that there is no  n  such that numerator( 1/d(n) -
:1/n ) = 102,
:for example (first Unknown), where d(n) is the number of divisors of n.
:Using  d(p^e) = e+1  one can search solutions of the form  n = 2^k m  where
:m is odd.
:To get an even numerator  x, one must have
:(k+1) d(m) = 2^k (1+2E)  for some integer E >= 0, which then leads to
:m = 1+2E + 2^k g x,  where g = gcd( m(1+2E), m-1-2E ).
:
:Maybe one can scan the space of solutions by increasing values of k and E,
:(k+1 must divide 2^k (2E+1), i.e., k = t*2^K - 1 with  t | 2E+1 ;
:K=0,1,2,...)
:for each of which one can make the list of possible prime signatures of m,
:(cf. oeis.org/A353248) and show that there is no solution  m  for any of
:these.
:But there are still some missing pieces to be filled in here, be it just
:for x=102...

I think it may be easier to do numerically - what's the largest n
such that (n - d) / d^2 <= 102? A quick check suggests it is probably
43243200 = 2^6 3^3 5^2 7 11 13, and there is no solution up to that
value.

Hugo

```