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%I #74 Sep 15 2023 18:35:01
%S 5,65,325,1025,2501,5185,9605,16385,26245,40001,58565,82945,114245,
%T 153665,202501,262145,334085,419905,521285,640001,777925,937025,
%U 1119365,1327105,1562501,1827905,2125765,2458625,2829125,3240001,3694085,4194305,4743685,5345345,6002501,6718465,7496645
%N a(n) = 4*n^4 + 1.
%C Except for the first term, all terms are composite. a(n) is divisible by 5 if n is not.
%C Long before Aurifeuille, Euler discovered that 4n^4 + 1 = (2n^2 + 2n + 1)*(2n^2 - 2n + 1). For example, 325 = 4 * 3^4 + 1 = (2 * 3^2 + 2 * 3 + 1)*(2 * 3^2 - 2 * 3 + 1) = 25 * 13. Euler shared this discovery with Goldbach in a letter dated August 28, 1742. [Euler identity corrected by _Graham Holmes_, Jun 02 2023]
%C The terms of the sequence are the arithmetic mean of eight numbers located on concentric circles (see Avilov link). - _Nicolay Avilov_, Jan 22 2021
%D Don Knuth, The Art of Computer Programming: Seminumerical Algorithms, 3rd ed., New York: Addison-Wesley Professional (1997), p. 392.
%D David Wells, Prime Numbers: The Most Mysterious Figures in Math. Hoboken, New Jersey: John Wiley & Sons (2005), p. 15.
%H Vincenzo Librandi, <a href="/A211412/b211412.txt">Table of n, a(n) for n = 1..1000</a>
%H Nicolay Avilov, <a href="/A211412/a211412.jpg">Drawing with circles</a>
%H P. H. Fuss, <a href="http://eulerarchive.maa.org/correspondence/fuss/">Correspondance mathématique et physique de quelques célèbres géomètres du XVIIIème siècle</a>, Saint-Pétersbourg, 1843, p. 145; <a href="https://archive.org/details/correspondancem01goldgoog/page/n295/mode/2up">alternative link</a>. See in particular <a href="http://eulerarchive.maa.org/correspondence/fuss/goldbach110-169">Lettre XLVI (Euler to Goldbach)</a>, Aug 28 1742
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F G.f.: -x*(x^4+50*x^2+40*x+5) / (x-1)^5. - _Colin Barker_, Feb 11 2013
%F a(n) = A053755(n^2). - _Michel Marcus_, Sep 18 2015
%F a(n) = (2*n^2)^2 + 1^2 = (2*n^2-1)^2 + (2*n)^2. - _Thomas Ordowski_, Sep 18 2015
%F a(n) = A001844(n) * A001844(n+1) = A141046(n) + 1 = (A000583(n) * 4 ) + 1 = A016742(n) + A173121(n) + 1. - _Bruce J. Nicholson_, Jun 06 2017
%F From _Amiram Eldar_, Jul 26 2022: (Start)
%F Sum_{n>=1} 1/a(n) = tanh(Pi/2)*Pi/4 - 1/2.
%F Sum_{n>=1} (-1)^n/a(n) = 1/2 - sech(Pi/2)*Pi/4. (End)
%t 4 Range[44]^4 + 1
%t Table[4 n^4 + 1, {n, 50}] (* _Vincenzo Librandi_, Jun 11 2017 *)
%t LinearRecurrence[{5,-10,10,-5,1},{5,65,325,1025,2501},40] (* _Harvey P. Dale_, Sep 15 2023 *)
%o (PARI) a(n) = 4*n^4+1 \\ _Felix Fröhlich_, Jun 07 2017
%o (Magma) [4*n^4 + 1: n in [1..50]]; // _Vincenzo Librandi_, Jun 11 2017
%Y Cf. A207262 (subset).
%Y After the first term, subsequence of A121944.
%Y Cf. A053755.
%K nonn,easy
%O 1,1
%A _Alonso del Arte_, Feb 10 2013
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