[seqfan] Simple sequence based on Pythagorean triples

[Jack Brennen]

This sequence seems simple enough and yet is not in the OEIS:

Begin with a(0) = 3.

Let a(n) for n > 0 be the smallest positive integer not yet
in the sequence which forms part of a Pythagorean triple
when paired with a(n-1).

I believe that the sequence begins:

   3,4,5,12,9,15,8,6,10,24,7,25,20,16,30,18,80,39,36,27,
   45,28,21,29,420,65,33,44,55,48,14,50,40,32,60,11,61,
   1860,341,541,146340,15447,20596,25745,32208,2540,
   1524,635,381,508,16125,4515,936,75,72,54,90,56,42,
   58,840,41,841,580,68,51,85,13,84,35,37,684,285,152,
   114,190,336,52,165,88,66,110,96,100,105,63,87,116,
   145,17,144,108,81,135,153,104,78,130,112,113,6384,
   640,312,91,109,5940,567,540,57,76,95,168,26,170,102,
   136,64,120,22,122,3720,682,1082,292680,30894,41192,
   51490,64416,2513,8616,3590,2154,2872,5385,3231,4308,
   1795,1077,1436,128877,12920,663,180,19,...

(Any typos are part of my cut-and-paste...)

Two questions:

Is the sequence infinite?  Can it "paint itself into
a corner" at any point?  Note that picking any starting
point >= 5 seems to lead to a finite sequence ending in
5,3,4:

    6,8,10,24,7,25,15,9,12,5,3,4 stop

By beginning with 3 or 4, you make sure that the 5,3,4
dead-end is never available.


If infinite, is it a permutation of the integers >= 3?
It seems likely.  Proving it doesn't seem easy though.