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Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z-(X+1) values.
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%I #30 Sep 20 2017 02:51:49

%S 1,8,49,288,1681,9800,57121,332928,1940449,11309768,65918161,

%T 384199200,2239277041,13051463048,76069501249,443365544448,

%U 2584123765441,15061377048200,87784138523761,511643454094368,2982076586042449,17380816062160328,101302819786919521

%N Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z-(X+1) values.

%C Old name was A001653(n) - A046090(n).

%H Vincenzo Librandi, <a href="/A115598/b115598.txt">Table of n, a(n) for n = 1..200</a>

%H L. J. Gerstein, <a href="http://www.jstor.org/stable/30044157">Pythagorean triples and inner products</a>, Math. Mag., 78 (2005), 205-213.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).

%F a(n) = (-2+(3-2*sqrt(2))^n+(3+2*sqrt(2))^n)/4.

%F a(n) = 7*a(n-1)-7*a(n-2)+a(n-3).

%F G.f.: -x*(x+1) / ((x-1)*(x^2-6*x+1)).

%F a(n)^2 + (a(n)+1)^2 = A001542(n)^2 + 1^2. - _Hermann Stamm-Wilbrandt_, Jul 27 2014

%t LinearRecurrence[{7,-7,1},{1,8,49},30] (* _Harvey P. Dale_, Oct 27 2013 *)

%t CoefficientList[Series[-(x + 1)/((x - 1) (x^2 - 6 x + 1)), {x, 0, 30}], x] (* _Vincenzo Librandi_, Jul 28 2014 *)

%Y Essentially a duplicate of A001108.

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_, Mar 14 2006

%E Corrected and edited by _Colin Barker_, Jul 31 2013