A061647 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.

1, 3, 9, 23, 61, 159, 417, 1091, 2857, 7479, 19581, 51263, 134209, 351363, 919881, 2408279, 6304957, 16506591, 43214817, 113137859, 296198761, 775458423, 2030176509, 5315071103, 13915036801, 36430039299, 95375081097, 249695203991, 653710530877, 1711436388639

Offset: 1

Author

Darrin Frey (freyd(AT)cedarville.edu), Jun 14 2001

Keywords

• nonn
• easy

Comments

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

Examples

			a(7) = 417 since a(7) = 2*a(6) + 2*a(5) - a(4) = 2*159 + 2*6 - 23.
		

References

• J. H. Conway, The Sensual (Quadratic) Form, MAA.

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Crossrefs

Cf. A000032, A000045, A001654.

Cf. similar sequences of the type k*F(n)*F(n+1) + (-1)^n listed in A264080.

Programs

Mathematica

LinearRecurrence[{2,2,-1},{1,3,9},40] (* Harvey P. Dale, May 31 2015 *)

PARI

x='x+O('x^99); Vec(x*(1+x+x^2)/((1+x)*(x^2-3*x+1))) \\ Altug Alkan, Mar 26 2016

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

Formula

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.

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

a(n) = Fibonacci(n+1)^2 - Fibonacci(n-2)^2 - (-1)^n. - Thomas Baruchel, Jul 29 2005

From J. M. Bergot, Aug 26 2013: (Start)

a(n) = F(n-2)*F(n-1) + F(n-1)*F(n) + F(n)*F(n+1), where F=A000045.

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]

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.

(End)

G.f.: x*(1+x+x^2) / ( (1+x)*(x^2-3*x+1) ). - R. J. Mathar, Aug 28 2013

a(n) = A001654(n) + A001654(n-1) + A001654(n-2). - R. J. Mathar, Aug 28 2013

a(n) = 4*Fibonacci(n-1)*Fibonacci(n) - (-1)^n. - Bruno Berselli, Oct 30 2015

a(n) = Lucas(2*n-1) - Fibonacci(n-1)*Fibonacci(n). See similar identity for A264080. - Bruno Berselli, Nov 04 2015

a(n) = (1/5)*(4*Lucas(2*n-1) - (-1)^n). - Ehren Metcalfe, Mar 26 2016

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

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