Conjecture: A103974(n)^2 - A011922(n)^2 = ( 4*A007655(n) )^2

Sat Apr 16 02:23:56 CEST 2005

Hello again, 

The new symseq "lestes" (which has just been defined) gives A103974
(along with  sequences in context) at 
http://www.crowdog.de/SeqContext/A103974lestes.html 
It also returns the relation A103974(n+1) = A098301(n+1) + A055793(n+2)
(among others). I will check how many of the symseq relationships on the
above page look important and will submit them as comments, tomorrow. 

http://www.research.att.com/projects/OEIS?Anum=A098301
http://www.research.att.com/projects/OEIS?Anum=A103974
http://www.research.att.com/projects/OEIS?Anum=A055793

Sincerely, 
Creighton 

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> Date: Mon,  7 Mar 2005 19:48:53 +0100
> Subject: Re: Conjecture:  A103974(n)^2 - A011922(n)^2  =  (
> 4*A007655(n)  )^2
> From: David C Terr <David_C_Terr at raytheon.com>
> To: "Paul D. Hanna" <pauldhanna at juno.com>

> I proved that A103974(n) = 4*A001353(n)^2 + 1, or at least the latter
> is a subsequence.
> 
> Dave
> 
> 
> 
> 
> 
> 
> "Paul D. Hanna" <pauldhanna at juno.com>
> 03/04/2005 08:44 PM
> 
> 
> To:     seqfan at ext.jussieu.fr
> cc:     zakseidov at yahoo.com, mathchess at velucchi.it
> Subject:        Conjecture:  A103974(n)^2 - A011922(n)^2  =  (
> 4*A007655(n)  )^2 
> Hello Seqfans,
> Just noticed Zak Seidov's nice A103974:
> "Smaller sides (a) in (a,a,a+1)-integer triangle with integer area." 
> Can anyone prove the following conjectures?
> 
> (1) A103974(n)^2 - A011922(n)^2  =  ( 4*A007655(n)  )^2
> 
> (2) all triangles meeting the criteria in A103974 are given by (1).
> 
> 
> If true, then we have a formulas for the triangles of A103974 from: 
> (3) 1/2 base of triangles = A011922(n) = ( A103974(n) + 1 )/2
> 
> (4) 1/4 height of triangles = A007655(n)
> 
> and by employing the nice formulas for A011922 and A007655.
> 
> 
> The connections seem rather significant.
> 
> See below for sequences ...
> Paul
> 
> =============================================================== ID
> Number: A103974
> URL:       http://www.research.att.com/projects/OEIS?Anum=A103974
> Sequence:  1,5,65,901,12545,174725,2433601
> Name:      Smaller sides (a) in (a,a,a+1)-integer triangle with
> integer area.
> Comments:  Corresponding areas are:
> 0,12,1848,351780,68149872,13219419708,2564481115560. What
> is
> the next term? Is the sequence finite? The possible last
> two digits of
> "a" are (it may help in searching for more terms):
> 
>
{01,05,09,15,19,25,29,33,35,39,45,49,51,55,59,65,69,75,79,83,85,89,95,99}
> .
> See also:  Cf. A102341, A103975 - A103977.
> Keywords:  more,nonn,new
> Offset:    1
> Author(s): Zak Seidov (zakseidov(AT)yahoo.com), Feb 23 2005
> 
> =============================================================== ID
> Number: A011922
> URL:       http://www.research.att.com/projects/OEIS?Anum=A011922
> Sequence: 
> 1,3,33,451,6273,87363,1216801,16947843,236052993,3287794051,
> 45793063713,637815097923,8883618307201,123732841202883,
> 1723376158533153,24003533378261251,334326091137124353
> Name:     
> (2+sqrt(1+((((2+sqrt(3))^(2*n)-(2-sqrt(3))^(2*n))^2)/4)))/3.
> References Mario Velucchi, Seeing couples, in Recreational and
> Educational
> Computing, to appear 1997.
> Formula:   sqrt 3 = 1 + Sum(1 through infinity) 2/a(n) = 2/2 + 2/3 +
> 2/33 + 2/451 +
> 2/6273 + 2/87363 + 2/1216801... - Gary W. Adamson
> (qntmpkt(AT)yahoo.com), Jun 12 2003
> See also:  Cf. A011916, A011918, A011920.
> Keywords:  nonn,easy
> Offset:    0
> Author(s): Mario Velucchi (mathchess(AT)velucchi.it)
> 
> =============================================================== ID
> Number: A007655 (Formerly M4948)
> URL:       http://www.research.att.com/projects/OEIS?Anum=A007655
> Sequence:  0,1,14,195,2716,37829,526890,7338631,102213944,1423656585,
> 19828978246,276182038859,3846719565780,53577891882061,
> 746243766783074,10393834843080975,144767444036350576
> Name:      Standard deviation of A007654.
> Comments:  a(n)=A001353(2n)/4. a(n) corresponds also to one-sixth the
> area of
> Fleenor-Heronian triangle with middle side A003500(n). -
> Lekraj
> Beedassy (boodhiman(AT)yahoo.com), Jul 15 2002
> a(n) give all (nontrivial, integer) solutions of Pell equation
> b(n+1)^2 - 48*a(n+1)^2 = +1 with b(n+1)=A011943(n), n>=0.
> References D. A. Benaron, personal communication.
> E. K. Lloyd (E.K.Lloyd(AT)maths.soton.ac.uk), "The standard
> deviation of 1, 2, .., n, Pell's equation and rational
> triangles",
> preprint.
> Links:     Index entries for sequences related to Chebyshev
> polynomials.
> C. Dement, The Floretions.
> Formula:   a(n) = 14*a(n-1) - a(n-2). G.f.: (x^2)/(1-14*x+x^2).
> a(n+1) ~ 1/24*sqrt(3)*(2 + sqrt(3))^(2*n) - Joe Keane
> (jgk(AT)jgk.org), May 15 2002
> a(n+1) = S(n-1,14), n>=0, with S(n,x) := U(n,x/2) Chebyshev's
> polynomials of the second kind. S(-1,x) := 0. See A049310.
> a(n+1) = ((7+4*sqrt(3))^n - (7-4*sqrt(3))^n)/(8*sqrt(3)).
> a(n+1) = sqrt((A011943(n)^2 - 1)/48), n>=0.
> Chebyshev's polynomials U(n-2,x) evaluated at x=7.
> 4*a(n+1) + A046184(n) = A055793(n+2) + A098301(n+1) 4*a(n+1) +
> A098301(n+1) + A055793(n+2) = A046184(n+1) (4*a(n+1))^2 =
> A098301(2n+1) (conjectures) - Creighton Dement
> (crowdog(AT)crowdog.de), Nov 02 2004
> See also:  Cf. A001353, A003500.
> Cf. A011945, A067900.
> Keywords:  nonn,easy
> Offset:    1
> Author(s): njas
> Extension: Chebyshev comments from W. Lang
> (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Nov 08 2002
> =============================================================== 
> 
> 
> 
> 