[seqfan] Simple sequence based on Pythagorean triples
Jack Brennen
jfb at brennen.net
Thu Dec 26 19:48:46 CET 2013
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.
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