[seqfan] Re: Iterating "smallest odd prime divisor of n^2 + 1"

[Neil Fernandez]

Hi all,

The sequence a(n+1) := smallest odd prime divisor (SOPD) of (a(n)^2)+1
may possibly always end in the loop (5,13), but there exist t such that
the sequence a(n+1) := SOPD of (a(n)^2)+t can fall into more than one
loop depending on a(1).

When t=34 (x^2+t always with a SOPD because it cannot be a power of 2),
a(1)=2 => loop is (5,59);
a(1)=3 => loop is (7,83).

When t=7 (for which x^2+t can be a power of 2 and therefore SOPD-free),
a(1)=4 => loop is (23,67,281,9871);
a(1)=8 => loop is (71,631).

Best regards,

Neil



>>     -----Original Message-----
>>     From: Neil Sloane <njasloane at gmail.com>
>>     To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>>     Sent: Fri, Nov 3, 2017 11:59 pm
>>     Subject: [seqfan] Iterating "smallest odd prime divisor of n^2 + 1"
>> 
>>     Dear Seq Fans, While I was at Hofstra Univ. the other day, Zoran Sunik
>>     asked, if you iterate the map "n -> smallest odd prime divisor of n^2+1",
>>     do you always end in the 2-cycle (5 <-> 13) ?Does anyone know?
>>     See A125256 for the map, and also its bisections A256970and A293958.
>>     If this is true, then there could be a sequence giving the number of
>>     iterations needed to reach the loop, or to reach any loop if there are
>>     others ...


-- 
Neil Fernandez