About sequences

[creigh at o2online.de]

Please excuse all the postings, below is a IX 'th  property 
(without proof - although, 
Limit as n-> infinity of A001353(n)/A001353(n-1) = 2 + sqrt(3). 
-Gregory V. Richardson (omomom(AT)hotmail.com), 
Oct 06 2002 has been proven- I was informed by Neil that 
"jes(n)" =   [1, -4, 15, -56, ...] is (-1)^(n+1)*A001353(n+1)).
I also fixed a mistake at the end. 

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Just found an additional property (VIII) which perplexes 
me. I also fixed a mistake at the end. 

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Apparently, none of the four sequences 

jes seq: 1(0), -4(1), 15(2), -56(3), 209(4), -780(5), 2911(6), -10864(7), 
40545(8), -151316(9), 564719(10) 

les seq: -2(0), 1(1), -6(2), 16(3), -62(4), 225(5), -842(6), 3136(7), -11706 
(8), 43681(9), -163022(10) 

tes seq: -1(0), 4(1), -13(2), 49(3), -181(4), 676(5), -2521(6), 9409(7), -35113 
(8), 131044(9), -489061(10) 

ves seq: -2(0), 1(1), -4(2), 9(3), -34(4), 121(5), -452(6), 1681(7), -6274(8), 
23409(9), -87364(10) 

are listed at OEIS. However, notice the symmetries (for all n): 

I: jes(n) + les(n) + tes(n) = ves(n) 

II: les(2n+1); tes(2n+1); ves(2n+1); 
ves(2n+1) - jes(2n+1) - 1 = les(2n+1) + tes(2n+1) - 1; 
3*les(2n+1) + 1 = 3*jes(n)^2 + 1 (see IV and VII) are perfect squares 

III: les(2n+1) divides ves(2n+1) - jes(2n+1) - 1 = les(2n+1) + tes(2n+1) -
 
1 

IV: (jes(n))^2 = les(2n+1) 

V: tes(2n) = A001570(n), sqrt( tes(2n+1) ) = A001075(n) [both are "nice" 
sequences] 

VI: ves(2n), ves(2n)/2 do not exist in OEIS, however: sqrt( ves(2n+1) ) 
= A001835(n) [ "nice" ] 

VII: sqrt( les(2n+1) ) = A001353(n) [ Comments: 3*a(n)^2 + 1 is a perfect square. 
] 

VIII: les(n) + tes(n) = ves(2+n) + jes(n) 

IX: lim n |jes(n+1)/jes(n)| = 
     lim n |les(n+1)/les(n)| = 
     lim n |tes(n+1)/tes(n)| =
     lim n |ves(n+1)/ves(n)| = 2 + sqrt(3)

It follows immediately from properties I + II that... 
1^2 (les(1) + 2^2 (tes(1)) = 2^2 (ves(1) - jes(1) - 1) + 1
4^2 (les(3))+ 7^2 (tes(3)) = 8^2 (ves(3) - jes(3) - 1)  + 1 
15^2 (les(5)) + 26^2 (tes(5)) = 30^2 (ves(5) - jes(5) - 1)  + 1 

Sincerely, 
Creighton