- #1
ramsey2879
- 841
- 3
Conjecture, For p>2, the [tex]2^(p-1)[/tex] th square triangular number is divisible by [tex]M_{p}[/tex] if and only if [tex]M_{p}[/tex] is prime. I checked this for 2<p<27. For instance the first four square triangular numbers are 0,1,36 and 1225 and the fourth is divisible by .
PS In fact it appears that if is prime then for any starting integers [tex]S_{1}[/tex] and [tex]S_2[/tex] having the recursive relation ,
[tex]S_{n} = 6*S_{n-1} - S_{n-2}[/tex] the following congruence holds:
[tex]S_{2^{p-1}} = S_{1} \mod M_{p}. There is prize money lurking here for those who are interested.
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PS In fact it appears that if is prime then for any starting integers [tex]S_{1}[/tex] and [tex]S_2[/tex] having the recursive relation ,
[tex]S_{n} = 6*S_{n-1} - S_{n-2}[/tex] the following congruence holds:
[tex]S_{2^{p-1}} = S_{1} \mod M_{p}. There is prize money lurking here for those who are interested.
_________________
Have a useful day