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