A304726 a(n) = n^4 + 4*n^2 + 3.

3, 8, 35, 120, 323, 728, 1443, 2600, 4355, 6888, 10403, 15128, 21315, 29240, 39203, 51528, 66563, 84680, 106275, 131768, 161603, 196248, 236195, 281960, 334083, 393128, 459683, 534360, 617795, 710648, 813603, 927368, 1052675, 1190280, 1340963, 1505528, 1684803

Offset: 0

Author

Vincenzo Librandi, May 31 2018

Keywords

• nonn
• easy

Comments

Alternating sum of all points on the fourth row of the Hosoya triangle composed of Fibonacci polynomials, where F_{0}(n) = 1 and F_{1}(n) = n, hence a(n) = F_{5}(n)/F_{1}(n) for n>0 (see Florez et al. reference, page 7, Table 4 and following sum).

Apart from 8, all terms belong to A217554 because a(n) = (n^2+1)^2 + (n+1)^2 + (n-1)^2 = (n^2+2)^2 - 1. - Bruno Berselli, Jun 04 2018

Links

• Muniru A Asiru, Table of n, a(n) for n = 0..10000
• Rigoberto Florez, Robinson A. Higuita, and Antara Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Journal of Integer Sequences, Vol. 17 (2014), Article 14.9.5.
• Eric Weisstein's World of Mathematics, Fibonacci Polynomial.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A058071, A059100, A217554.

Subsequence of A005563.

Programs

GAP

List([0..40], n -> (n^2+2)^2-1); # Muniru A Asiru, Jun 03 2018

Magma

[n^4+4*n^2+3: n in [0..40]];

Maple

seq((n^2+2)^2-1,n=0..40); # Muniru A Asiru, Jun 03 2018

Mathematica

Table[n^4 + 4 n^2 + 3, {n, 0, 35}]
LinearRecurrence[{5,-10,10,-5,1},{3,8,35,120,323},40] (* Harvey P. Dale, Mar 04 2021 *)

Formula

G.f.: (3 - 7*x + 25*x^2 - 5*x^3 + 8*x^4)/(1-x)^5.

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

a(n) = A059100(n)^2 - 1.

Sum_{n>=0} 1/a(n) = 1/6 + coth(Pi)*Pi/4 - coth(sqrt(3)*Pi)*Pi/(4*sqrt(3)). - Amiram Eldar, Feb 24 2023