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

0,1

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).

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

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 *)

Prog

(Magma) [n^4+4*n^2+3: n in [0..40]];
(GAP) List([0..40], n -> (n^2+2)^2-1); # Muniru A Asiru, Jun 03 2018

Crossrefs

Cf. A058071, A059100, A217554.

Subsequence of A005563.

Sequence in context: A077291 A192212 A148918 * A226679 A216541 A347896

Adjacent sequences: A304723 A304724 A304725 * A304727 A304728 A304729

Keyword

nonn,easy

Author

Vincenzo Librandi, May 31 2018

Status

approved