[seqfan] Re: Nice conjecture from Enrique Navarrete re divisors of triangular numbers A111273

[Peter Munn]

Does this attempt at a proof stand up?:

For odd k, k appears by A111273(k). Proof: choose m such that k-1 <= m <=
k and T_m is odd. k is a divisor of T_m and (by induction) all smaller odd
divisors have occurred earlier, so a(m) = k if k has not occurred earlier.

For even k, k appears by A111273(2k-1), as k divides T_(2k-1) and by
induction all smaller divisors have occurred earlier.

So, for odd prime p, the first triangular number it divides is T_(p-1) =
p*(p-1)/2. But (p-1)/2 and any smaller divisors have occurred by (p-1)-1,
so A111273(p-1) = p.

In respect of the conjecture in A309196 that after A111273(1), the
smallest missing value is always even, note that even numbers can only
appear in A111273 at indices congruent to 0 and 3 modulo 4. So as
A111273(4) is not even, for n >= 4 there is always an even number less
than or equal to n that has not appeared by A111273(n), whereas all such
odd numbers have.

Best Regards,

Peter

On Wed, July 24, 2019 8:40 pm, israel at math.ubc.ca wrote:
> No, q is not a(q-1) in general, just for odd primes. For example,
> if p = 127, the divisors of t_126 less than p are 1 = a(1), 3 = a(2),
> 7 = a(6), 9 = a(9), 21 = a(14), 63 = a(62).
> Cheers,
> Robert

> On Jul 24 2019, Fred Lunnon wrote:
>>  By induction: if q|(p-1) then q = a(q-1) has appeared earlier;
>> the only
>> divisor of t_p remaining is p itself, so a(p-1) = p . QED WFL

>>On 7/24/19, Neil Sloane <njasloane at gmail.com> wrote:
>>> E.N. noticed that in A111273 (definition: a(n) = smallest divisor of
t_n = n(n+1)/2 that has not yet appeared in the sequence), it seems
that
>>> a(p-1)=p for all odd primes p. Surely this can't be a hard problem? It
would follow if we knew that the smallest missing number (smn) in
A111273 (which is now A309195) is always >n/2.