A160372 The 4-tuple (2, ((2*n+1)^2-1)/2, ((2*n+3)^2-1)/2, a(n)), where a(n) = 4*(3 + 20n + 42n^2 + 32n^3 + 8n^4), has Diophantus's property that the product of any two distinct terms plus one is a square.

420, 2380, 7812, 19404, 40612, 75660, 129540, 208012, 317604, 465612, 660100, 909900, 1224612, 1614604, 2091012, 2665740, 3351460, 4161612, 5110404, 6212812, 7484580, 8942220, 10603012, 12485004, 14607012, 16988620, 19650180, 22612812, 25898404, 29529612, 33529860

Offset

1,1

Links

Paolo Xausa, Table of n, a(n) for n = 1..10000

Lenny Jones, A polynomial Approach to a Diophantine Problem, Math. Mag. 72 (1999), 52-55.

Eric Weisstein's World of Mathematics, Diophantus Property.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

G.f.: 4*x*(3*x^4-14*x^3+28*x^2+70*x+105)/(1-x)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009 [corrected by R. J. Mathar, Sep 16 2009]

a(n) = 4 * (2*n + 1) * (2*n + 3) * (2*n^2 + 4*n + 1). - Paolo Xausa, Jan 16 2024

From Amiram Eldar, Jan 22 2024: (Start)

Sum_{n>=1} 1/a(n) = -cot(Pi/sqrt(2))*Pi/(8*sqrt(2)) - 5/24.

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 - cosec(Pi/sqrt(2))*Pi/(8*sqrt(2)) - 1/24. (End)

Example

For n=2, we get (2,12,24,2380), and 2*12+1 = 25 = 5^2, 2*24+1 = 49 = 7^2, 2*2380+1 = 4761 = 69^2, 12*24+1 = 289 = 17^2, 12*2380+1 = 28561 = 169^2, 4*2380+1 = 57121 = 239^2.

Mathematica

A160372[n_] := 4 (2 n + 1) (2 n + 3) (2 n^2 + 4 n + 1);
Array[A160372, 50] (* Paolo Xausa, Jan 16 2024 *)

Crossrefs

Cf. A086302.

Sequence in context: A251084 A250383 A145678 * A171259 A061125 A269016

Adjacent sequences: A160369 A160370 A160371 * A160373 A160374 A160375

Keyword

nonn,easy

Author

John W. Layman, May 11 2009

Status

approved