A211412 a(n) = 4*n^4 + 1.

5, 65, 325, 1025, 2501, 5185, 9605, 16385, 26245, 40001, 58565, 82945, 114245, 153665, 202501, 262145, 334085, 419905, 521285, 640001, 777925, 937025, 1119365, 1327105, 1562501, 1827905, 2125765, 2458625, 2829125, 3240001, 3694085, 4194305, 4743685, 5345345, 6002501, 6718465, 7496645

Offset: 1

Author

Alonso del Arte, Feb 10 2013

Keywords

• nonn
• easy

Comments

Except for the first term, all terms are composite. a(n) is divisible by 5 if n is not.

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]

The terms of the sequence are the arithmetic mean of eight numbers located on concentric circles (see Avilov link). - Nicolay Avilov, Jan 22 2021

References

• Don Knuth, The Art of Computer Programming: Seminumerical Algorithms, 3rd ed., New York: Addison-Wesley Professional (1997), p. 392.
• David Wells, Prime Numbers: The Most Mysterious Figures in Math. Hoboken, New Jersey: John Wiley & Sons (2005), p. 15.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Nicolay Avilov, Drawing with circles
• P. H. Fuss, Correspondance mathématique et physique de quelques célèbres géomètres du XVIIIème siècle, Saint-Pétersbourg, 1843, p. 145; alternative link. See in particular Lettre XLVI (Euler to Goldbach), Aug 28 1742
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A207262 (subset).

After the first term, subsequence of A121944.

Cf. A053755.

Programs

Magma

[4*n^4 + 1: n in [1..50]]; // Vincenzo Librandi, Jun 11 2017

Mathematica

4 Range[44]^4 + 1
Table[4 n^4 + 1, {n, 50}] (* Vincenzo Librandi, Jun 11 2017 *)
LinearRecurrence[{5,-10,10,-5,1},{5,65,325,1025,2501},40] (* Harvey P. Dale, Sep 15 2023 *)

PARI

a(n) = 4*n^4+1 \\ Felix Fröhlich, Jun 07 2017

Formula

G.f.: -x*(x^4+50*x^2+40*x+5) / (x-1)^5. - Colin Barker, Feb 11 2013

a(n) = A053755(n^2). - Michel Marcus, Sep 18 2015

a(n) = (2*n^2)^2 + 1^2 = (2*n^2-1)^2 + (2*n)^2. - Thomas Ordowski, Sep 18 2015

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

From Amiram Eldar, Jul 26 2022: (Start)

Sum_{n>=1} 1/a(n) = tanh(Pi/2)*Pi/4 - 1/2.

Sum_{n>=1} (-1)^n/a(n) = 1/2 - sech(Pi/2)*Pi/4. (End)