A095122 a(n) = Fibonacci(n)*(2*Fibonacci(n)-1).

0, 1, 1, 6, 15, 45, 120, 325, 861, 2278, 5995, 15753, 41328, 108345, 283881, 743590, 1947351, 5099221, 13351528, 34957341, 91523685, 239618886, 627341331, 1642418641, 4299936480, 11257426225, 29472399505, 77159865030, 202007345631, 528862414653, 1384580291160

Offset

0,4

Comments

a(n) mod 2 = Fibonacci(n) mod 2 = A011655(n).

The unique primitive Pythagorean triple (a,b,c) such that (a-b+c)/2 = A000045(n) and its long leg and hypotenuse are consecutive natural numbers is (2*A000045(n)-1, 2*A000045(n)*(A000045(n)-1), 2*A000045(n)*(A000045(n)-1)+1) and its semiperimeter is a(n). - Miguel-Ángel Pérez García-Ortega, Apr 13 2025

References

Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2025.

Links

Table of n, a(n) for n=0..30.

Index entries for linear recurrences with constant coefficients, signature (3,1,-5,-1,1).

Formula

G.f.: x*(1-2*x+2*x^2+x^3)/((1+x)*(1-x-x^2)*(1-3*x+x^2)).

a(n) = 2*(Fibonacci(2n-1)+Fibonacci(2n+1))/5-Fibonacci(n)+4*(-1)^n/5.

a(n) = 2*Lucas(2n)/5-Fibonacci(n)+4*(-1)^n/5.

a(n) = 2*A000032(2n)/5-A000045(n)+4*(-1)^n/5.

a(n) = 3*a(n-1)+a(n-2)-5*a(n-3)-a(n-4)+a(n-5), with a(0)=0, a(1)=1, a(2)=1, a(3)=6, a(4)=15. - Harvey P. Dale, Jan 14 2012

Mathematica

#(2#-1)&/@Fibonacci[Range[0, 30]] (* or *) LinearRecurrence[{3, 1, -5, -1, 1}, {0, 1, 1, 6, 15}, 30] (* Harvey P. Dale, Jan 14 2012 *)

Crossrefs

Cf. A000032, A000045, A011655, A382843, A382844, A382845.

Sequence in context: A117961 A327769 A318482 * A215917 A082637 A244023

Adjacent sequences: A095119 A095120 A095121 * A095123 A095124 A095125

Keyword

easy,nonn

Author

Paul Barry, May 29 2004

Status

approved