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

Sat Mar 5 09:05:10 CET 2005

  1, 5, 65, 901, 12545, 174725, 2433601, 33895685, 472105985,
        6575588101, 91586127425, 1275630195845, 17767236614401,
        247465682405765, 3446752317066305, 48007066756522501,
        668652182274248705, 9313123485082959365,
        129715076608887182401, 1806697949039337594245,
        25164056209941839137025, 350490088990146410324101,
        4881697189652107905400385, 67993270566139364265281285,
        947024090736298991808537601, 13190343999742046521054245125,
        183717791905652352302950894145,
        2558858742679390885720258272901,
        35640304605605820047780664926465,
        496405405735802089783209050697605,
        6914035375695623436917146044840001

Formula: A103974(n)=1/3*((2+sqrt(3))^(2*n)+(2-sqrt(3))^(2*n)+1).

Recurrence: a(n+3)=15*a(n+2)-15*a(n+1)+a(n), a(0)=1, a(1)=5, a(2)=65.

Generating function: (1-10*x+5*x^2)/(1-15*x+15*x^2-x^3).

Exponential generating function: 
1/3*(exp(x)+exp((7+4*sqrt(3))*x)+exp((7-4*sqrt(3))*x)).

Maple program: A:=rsolve({-A(n+3)+15*A(n+2)-15*A(n+1)+A(n), A(0) = 1, A(1) = 
5, A(2)=65}, A(n), makeproc);

Proof. Area is (a+1)/4*sqrt((3*a-1)*(a-1)). If a is even, the numerator is 
odd and the area is not integer. That means a=2*k-1. In this case, 
Area=k*sqrt((3*k-1)*(k-1)). Solving equation (3*k-1)*(k-1)=y^2, we get 
k=(2+sqrt(1+3*y^2))/3. That means that 1+3*y^2=x^2 with integer x and y. It 
is a Pell equation, x^2-3*y^2=1, all solutions of which has form 
x=((2+sqrt(3))^n+(2-sqrt(3))^n)/2, 
y=((2+sqrt(3))^n-(2-sqrt(3))^n)/(2*sqrt(3)). From here, k=(x+2)/3 that is 
integer only for even n. Then a=2*k-1=(2*x+1)/3 with even n. Q.E.D.

Alec Mihailovs
http://math.tntech.edu/alec/

----- Original Message ----- 
From: "Paul D. Hanna" <pauldhanna at juno.com>
To: <seqfan at ext.jussieu.fr>
Cc: <zakseidov at yahoo.com>; <mathchess at velucchi.it>
Sent: Friday, March 04, 2005 10:44 PM
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
> ===============================================================
> 