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

David C Terr David_C_Terr at raytheon.com
Mon Mar 7 19:29:15 CET 2005

I don't know about your conjecture, but I just found a formula for an
infinite family of (probably all) solutions to A103974. The solution is
$a_k = 4 U_k^2 + 1$, where $(U_k, V_k)$ is the $kth$ solution to Pell's
equation $3 U_k^2 + 1 = V_k^2$, namely $U_0 = 0$, $U_1 = 1$,
$V_0 = 1$, $V_1 = 2$, and $U_{k+1} = 4 U_k - U_{k-1}$ and $V_{k+1} = 4V_k - V_{k-1}$ for $k\geq 1$. Specifically we have
$U_k = (\alpha^k - \beta^k)/(2\sqrt{3})$, where $\alpha = 2+\sqrt{3}$ and
$\beta = 2-\sqrt{3}$. The first few solutions are $a=1,5,65, 901,12545,174725,2433601, 33895685$ corresponding to
$U=0,1,4,15,56,209,780,2911$.

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.
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}
.
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
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
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
===============================================================

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