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%I #91 Sep 08 2022 08:45:17
%S 1,17,241,3361,46817,652081,9082321,126500417,1761923521,24540428881,
%T 341804080817,4760716702561,66308229755041,923554499868017,
%U 12863454768397201,179164812257692801,2495443916839302017,34757050023492535441,484103256412056194161
%N Larger of two sides in a (k,k,k-1)-integer-sided triangle with integer area.
%C Corresponding areas are 0, 120, 25080, 4890480, 949077360, 184120982760, ...
%C Values of (x^2 + y^2)/2, where the pair (x, y) satisfies x^2 - 3*y^2 = -2, i.e., a(n) = {(A001834(n))^2 + (A001835(n))^2}/2 = {(A001834(n))^2 + A046184(n)}/2. - _Lekraj Beedassy_, Jul 13 2006
%C The heights of these triangles are given in A028230. (A028230(n), A045899(n), A103772(n)) forms a primitive Pythagorean triple.
%C Shortest side of (k,k+2,k+3) triangle such that median to longest side is integral. Sequence of such medians is A028230. - _James R. Buddenhagen_, Nov 22 2013
%C Numbers n such that (n+1)*(3n-1) is a square. - _James R. Buddenhagen_, Nov 22 2013
%H Colin Barker, <a href="/A103772/b103772.txt">Table of n, a(n) for n = 1..850</a>
%H J. B. Cosgrave, <a href="/A103772/a103772.txt">The Gauss-Factorial Motzkin connection</a> (Maple worksheet, change suffix to .mw)
%H J. B. Cosgrave and K. Dilcher, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.118.09.812">An Introduction to Gauss Factorials</a>, The American Mathematical Monthly, 118 (Nov. 2011), 812-829.
%H Project Euler, <a href="https://projecteuler.net/problem=94">Problem 94: Almost Equilateral Triangles</a>.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).
%F a(n) = (4*A001570(n+1) - 1)/3, n > 0. - _Ralf Stephan_, May 20 2007
%F a(n) = A052530(n-1)*A052530(n) + 1. - _Johannes Boot_, May 21 2011
%F G.f.: x*(1+x)^2/((1-x)*(1-14*x+x^2)). - _Colin Barker_, Apr 09 2012
%F a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3); a(1)=1, a(2)=17, a(3)=241. - _Harvey P. Dale_, Jan 02 2016
%F a(n) = (-1+(7-4*sqrt(3))^n*(2+sqrt(3))-(-2+sqrt(3))*(7+4*sqrt(3))^n)/3. - _Colin Barker_, Mar 05 2016
%F a(n) = 14*a(n-1) - a(n-2) + 4. - _Vincenzo Librandi_, Mar 05 2016
%F a(n) = A001353(n)^2 + A001353(n-1)^2. - _Antonio Alberto Olivares_, Apr 06 2020
%t a[1] = 1; a[2] = 17; a[3] = 241; a[n_] := a[n] = 15a[n - 1] - 15a[n - 2] + a[n - 3]; Table[ a[n] - 1, {n, 17}] (* _Robert G. Wilson v_, Mar 24 2005 *)
%t LinearRecurrence[{15,-15,1},{1,17,241},20] (* _Harvey P. Dale_, Jan 02 2016 *)
%t RecurrenceTable[{a[1] == 1, a[2] == 17, a[n] == 14 a[n-1] - a[n-2] + 4}, a, {n, 20}] (* _Vincenzo Librandi_, Mar 05 2016 *)
%o (PARI) Vec(x*(1+x)^2/((1-x)*(1-14*x+x^2)) + O(x^25)) \\ _Colin Barker_, Mar 05 2016
%o (Magma) I:=[1,17]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+4: n in [1..20]]; // _Vincenzo Librandi_, Mar 05 2016
%Y Cf. A102341, A103974, A016064, A011945, A028230.
%K nonn,easy
%O 1,2
%A _Zak Seidov_, Feb 23 2005
%E More terms from _Robert G. Wilson v_, Mar 24 2005
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