[seqfan] Re: A269254

[Brad Klee]

Hi Neil & Hans,

Table I. With Integer sqrt(n+2)
====================
m=3, n = 7, A033890,  A001519
m=4, n=14, A028230, A002530
m=5, n=23, A004254, A136211
m=6, n=34, A046176, A041011
m=7, n=47, A049678, A001519 *
m=8, n=62, A258684, A041023
m=9, n=79, A018913,  N.A.N.

The second column of A-Numbers are the more enlightening, as they seem to
generalize your Fibonacci example to other cases with integer sqrt(n+2).
These sequences are linear recurrences with signature (0,m,0,-1). This
probably is the property needed for a proof. All tests immediately show
that a(2n+1)|a(4n+2) for small n, i.e. the odds appear to divide one even
4Z coset.

I did not find a record for case m=9, but the axiom (0,1,1,8) appears to
produce a sequence with the desired divisibility property:

0, 1, 1, 8, 9, 71, 80, 631, 711, 5608, 6319, 49841, 56160, 442961, 499121 .
. .
4Z+2: 1, 80, 6319, 499121 . . .
2Z+1: 1, 8, 71, 631 . . .

As for the odd case n=110, it does overlap with A041351. However, sequence
A041351 does not have the same divisibility properties as seen in the
regular examples. There could be another function from n ---> m aside from
m=sqrt(n+2), but this is just a poor guess with no real support.

Hope this helps,

Brad



On Fri, Oct 20, 2017 at 9:55 PM, Neil Sloane Wrote:

> " But what about the other -1 entries? "
>