[seqfan] Re: A327265 and A309981

[M. F. Hasler]

Not sure I understand correctly but in any case, as already said in a
comment by Jon Schoenfeld in A309981, we do know that any n has a unique
tau signature not longer than (r+1)² - n where r = ceiling(sqrt(n)) is the
root of the next larger square : the squares s=k² are exactly the numbers
for which tau is odd (in view of the central divisor k) and (k+1)² - k² =
2k+1 uniquely identifies the position.
(My idea with the primes was unnecessarily complicated.)
- Maximilian

On Mon, Apr 10, 2023, 07:44 Peter Munn <techsubs at pearceneptune.co.uk> wrote:

> On Sat, April 1, 2023 2:13 pm, M. F. Hasler wrote:
> > call "tau signature" sequences of the form
> > t(n,k) = (tau(n), tau(n+1), ..., tau(n+k)), [tau = sigma_0 = numdiv]
> > either for any k (in that case /a/ tau signature doesn't always uniquely
> > identify n)
> > or for the least k* = A309981(n) such that t(n,k*) uniquely identifies n
> > (and then let t(n) = t(n,k*))
> > and we should call it *the* tau signature of n .
> >
> [...]
> > Do we have a formal proof that any n has a tau signature?
> > Yes, because if this weren't the case, it would mean that there's m >
> n such that
> > tau(m+k) = tau(n+k) for all k,
>
> This does not quite follow, because there could be a sequence of such m,
> each requiring a longer signature to distinguish m from n.


> Nevertheless, the idea of equivalent sequences that use the characteristic
> function of primes instead of tau seems worth pursuing. The proof of the
> existence of the equivalent signature for n would come from considering
> the a prime p > n and the residues of subsequent primes modulo p.
>
> Best regards,
>
> Peter
>