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

[hv at crypt.org]

Earlier I wrote:
:"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.

Let f(n) = (n - d(n)) / d^2(n), then since x=43243200 is in A002182, I
_think_ it is enough to show that f(px) > 102 for p prime, px in A002182:
that should be enough to guarantee that f(A002182(n)) > 102 for all
A002182(n) > x, and therefore that f(n) > 102 for n > x.

Hugo