A085151 Numbers generated by the Fibonacci polynomial x^4 + 3x^2 + 1.

5, 29, 109, 305, 701, 1405, 2549, 4289, 6805, 10301, 15005, 21169, 29069, 39005, 51301, 66305, 84389, 105949, 131405, 161201, 195805, 235709, 281429, 333505, 392501, 459005, 533629, 617009, 709805, 812701, 926405, 1051649, 1189189

Offset: 1

Author

Gary W. Adamson, Jun 21 2003

Keywords

• nonn

Comments

Start with the Fibonacci polynomials of A011973 (see "examples") and put in appropriate exponents, e.g. {1,1} = x^2 + 1, the generator of A002522; {1,2} = x^3 + 2x, the generator of A054602; and to get the next polynomial, multiply by x and add the previous polynomial, such that the generator for A085151 = x^4 + 3x^2 + 1 = (x)(x^3+2x) + (x^2+1).

Examples

			a(2) = f(2) of x^4 + 3x^2 + 1 = 29.
a(2) = 29 = (2)*A054602(2) + A002522(2) = (2)(12) + 5.
[2,2,2,2] = 12/29; a(2) = 29, & 12 = A054602(2). Thus [n,n,n,n] = A054602(n)/A085151(n).
		

Links

• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A002522, A011973, A054602, A057721, A085151.

Programs

Mathematica

a[n_] := n^4 + 3n^2 + 1; Array[a, 33]

Python

def A085151(n): return (m:=n**2)*(m+3)+1 # Chai Wah Wu, May 20 2025

Formula

a(n) = n^4 + 3*n^2 + 1.

a(n) = n*A054602(n) + A002522(n).

a(n) = denominator of [n, n, n, n]; with numerator = A054602(n).

a(n) = A057721(n). - R. J. Mathar, Sep 12 2008

From Chai Wah Wu, May 20 2025: (Start)

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5.

G.f.: x*(-x^4 - 14*x^2 - 4*x - 5)/(x - 1)^5. (End)

Sum_{n>=1} 1/a(n) = - 1/2 - (Pi/10)*((5*sinh(Pi)+sqrt(5)*sinh(sqrt(5)*Pi))/(cosh(Pi)-cosh(sqrt(5)*Pi))). - Amiram Eldar, Jun 04 2025

Extensions

More terms from Robert G. Wilson v, Aug 06 2006