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%I #65 Nov 13 2017 11:07:36
%S 1,3,9,23,61,159,417,1091,2857,7479,19581,51263,134209,351363,919881,
%T 2408279,6304957,16506591,43214817,113137859,296198761,775458423,
%U 2030176509,5315071103,13915036801,36430039299,95375081097,249695203991,653710530877,1711436388639
%N Beginning at the well for the topograph of a positive definite quadratic form with values 1, 1, 1 at a superbase (i.e., 1, 1 and 1 are the vonorms of the superbase), these numbers indicate the labels of the edges of the topograph on a path of greatest ascent.
%C Form the matrix A=[1,1,1;2,1,0;1,0,0]. a(n) is the sum of the second row elements of A^n. - _Paul Barry_, Sep 22 2004
%D J. H. Conway, The Sensual (Quadratic) Form, MAA.
%H Colin Barker, <a href="/A061647/b061647.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-1).
%F a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3) for n > 3, with a(1)=1, a(2)=3, a(3)=9.
%F a(n) = Fibonacci(n)^2 + Fibonacci(n-1)^2 + 2*Fibonacci(n)*Fibonacci(n+1) + 3*Fibonacci(n-1)*Fibonacci(n), with offset 0. - _Paul Barry_, Sep 22 2004
%F a(n) = Fibonacci(n+1)^2 - Fibonacci(n-2)^2 - (-1)^n. - _Thomas Baruchel_, Jul 29 2005
%F From _J. M. Bergot_, Aug 26 2013: (Start)
%F a(n) = F(n-2)*F(n-1) + F(n-1)*F(n) + F(n)*F(n+1), where F=A000045.
%F a(n) = (L(2*n+1) + L(2*n-1) + L(2*n-3) - (-1)^n)/5, where L=A000032. [Corrected by _Ehren Metcalfe_, Mar 26 2016]
%F Starting at n=2 create Pythagorean triangles by using side x = b^2 - a^2, side y = 2*a*b, and side y = a^2 + b^2. For three successive triangles, let a=F(n) and b=F(n+1), a=F(n+1) and b=F(n+2), and a=F(n+2) and b=F(n+3); then a(n+3)=one-half the sum of the three perimeters.
%F (End)
%F G.f.: x*(1+x+x^2) / ( (1+x)*(x^2-3*x+1) ). - _R. J. Mathar_, Aug 28 2013
%F a(n) = A001654(n) + A001654(n-1) + A001654(n-2). - _R. J. Mathar_, Aug 28 2013
%F a(n) = 4*Fibonacci(n-1)*Fibonacci(n) - (-1)^n. - _Bruno Berselli_, Oct 30 2015
%F a(n) = Lucas(2*n-1) - Fibonacci(n-1)*Fibonacci(n). See similar identity for A264080. - _Bruno Berselli_, Nov 04 2015
%F a(n) = (1/5)*(4*Lucas(2*n-1) - (-1)^n). - _Ehren Metcalfe_, Mar 26 2016
%F a(n) = 2^(-n)*(-(-2)^n - 2*(3-sqrt(5))^n*(1+sqrt(5)) + 2*(-1+sqrt(5))*(3+sqrt(5))^n)/5. - _Colin Barker_, Sep 30 2016
%F a(n) = (-1)^(n-1) + 3*a(n-1) - a(n-2) with a(1) = 1 and a(2) = 3. - _Peter Bala_, Nov 12 2017
%e a(7) = 417 since a(7) = 2*a(6) + 2*a(5) - a(4) = 2*159 + 2*6 - 23.
%t LinearRecurrence[{2,2,-1},{1,3,9},40] (* _Harvey P. Dale_, May 31 2015 *)
%o (PARI) x='x+O('x^99); Vec(x*(1+x+x^2)/((1+x)*(x^2-3*x+1))) \\ _Altug Alkan_, Mar 26 2016
%o (PARI) a(n) = round(2^(-n)*(-(-2)^n-2*(3-sqrt(5))^n*(1+sqrt(5))+2*(-1+sqrt(5))*(3+sqrt(5))^n)/5) \\ _Colin Barker_, Sep 30 2016
%Y Cf. A000032, A000045, A001654.
%Y Cf. similar sequences of the type k*F(n)*F(n+1) + (-1)^n listed in A264080.
%K nonn,easy
%O 1,2
%A Darrin Frey (freyd(AT)cedarville.edu), Jun 14 2001
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