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A304726 a(n) = n^4 + 4*n^2 + 3. 1
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 (list; graph; refs; listen; history; text; internal format)
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
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.
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
Subsequence of A005563.
Sequence in context: A077291 A192212 A148918 * A226679 A216541 A347896
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, May 31 2018
STATUS
approved

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Last modified February 28 03:01 EST 2024. Contains 370379 sequences. (Running on oeis4.)