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a(n) = n^4 + 3*n^2 + 1.
12

%I #36 May 22 2026 10:04:59

%S 1,5,29,109,305,701,1405,2549,4289,6805,10301,15005,21169,29069,39005,

%T 51301,66305,84389,105949,131405,161201,195805,235709,281429,333505,

%U 392501,459005,533629,617009,709805,812701,926405,1051649,1189189,1339805,1504301,1683505

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

%C Longest possible side c of a triangle with integer sides a <= b < c and inradius n. Triangle has sides (n^2+2, n^4+2*n^2+1, n^4+3*n^2+1). Proved by _Joseph Myers_, Jun 11 2006.

%H G. C. Greubel, <a href="/A057721/b057721.txt">Table of n, a(n) for n = 0..1000</a>

%H Michelle Rudolph-Lilith, <a href="https://arxiv.org/abs/1508.07894">On the Product Representation of Number Sequences, with Application to the Fibonacci Family</a>, arXiv:1508.07894 [math.NT], 2015.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F a(n) = denominator of Integral_{x=0..oo} sin(n*x)/exp((n^2+1)*x). - _Francesco Daddi_, Jul 07 2013

%F From _Elmo R. Oliveira_, May 22 2026: (Start)

%F G.f.: (1 + 14*x^2 + 4*x^3 + 5*x^4)/(1 - x)^5.

%F E.g.f.: (1 + 4*x + 10*x^2 + 6*x^3 + x^4)*exp(x).

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

%t a[n_]:=n^4+3n^2+1; Array[a,40,0] (* _Vladimir Joseph Stephan Orlovsky_, Nov 03 2009 *)

%o (PARI) vector(40, n, n--; n^4+3*n^2+1) \\ _G. C. Greubel_, Aug 12 2019

%o (Magma) [n^4+3*n^2+1: n in [0..40]]; // _G. C. Greubel_, Aug 12 2019

%o (GAP) List([0..40], n-> n^4+3*n^2+1); # _G. C. Greubel_, Aug 12 2019

%Y See A120062 for sequences related to integer-sided triangles with integer inradius n.

%Y Cf. A120062 [triangles with integer inradius], A120063 [minimum of their longest sides].

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Oct 27 2000