A156640 a(n) = 169*n^2 + 140*n + 29.

29, 338, 985, 1970, 3293, 4954, 6953, 9290, 11965, 14978, 18329, 22018, 26045, 30410, 35113, 40154, 45533, 51250, 57305, 63698, 70429, 77498, 84905, 92650, 100733, 109154, 117913, 127010, 136445, 146218, 156329, 166778

Offset: 0

Author

Vincenzo Librandi, Feb 15 2009

Keywords

• nonn
• easy

Comments

The identity (57122*n^2 +47320*n +9801)^2 - (169*n^2 +140*n +29)*(4394*n +1820)^2 = 1 can be written as A156735(n)^2 - a(n)*A156636(n)^2 = 1.

The continued fraction expansion of sqrt(a(n)) is [13n+5; {2, 1, 1, 2, 26n+10}]. - Magus K. Chu, Sep 15 2022

From Klaus Purath, Apr 06 2025: (Start)

a(n)*13^2-1 is a square, and a(n) is the sum of two squares (see FORMULA). There are no squares in this sequence. The odd prime factors of these terms are always of the form 4*k + 1.

All a(n) = D satisfy the Pell equation (k*x)^2 - D*(13*y)^2 = -1 for any integer n where a(1-n) = A156639(n). The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*13^2 - 1), x(0) = 1, x(1) = 4*D*13^2 - 1, y(0) = 1, y(1) = 4*D*13^2 - 3. The two recurrences are of the form (4*D*13^2 - 2, -1).

It follows from the above that this sequence and A156639 belong to A031396. (End)

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A156636, A156718, A156735.

Cf. A154609 (13n+5).

Subsequence of A031396.

Programs

Magma

I:=[29, 338, 985]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]];

Maple

A156640:= n-> 169*n^2 + 140*n + 29; seq(A156640(n), n=0..50); # G. C. Greubel, Feb 28 2021

Mathematica

LinearRecurrence[{3,-3,1},{29,338,985},50]
CoefficientList[Series[(29 +251x +58x^2)/(1-x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, May 03 2014 *)

PARI

a(n)=169*n^2+140*n+29 \\ Charles R Greathouse IV, Dec 23 2011

Sage

[169*n^2 + 140*n + 29 for n in (0..50)] # G. C. Greubel, Feb 28 2021

Formula

a(n) = 3*a(n-1) -3*a(n-2) +a(n-3) for n>2.

G.f.: (29 + 251*x + 58*x^2)/(1-x)^3. - Vincenzo Librandi, May 03 2014

E.g.f.: (29 +309*x +169*x^2)*exp(x). - G. C. Greubel, Feb 28 2021

From Klaus Purath, Apr 06 2025: (Start)

a(n) = (5*n + 2)^2 + (12*n + 5)^2 for any integer n.

169*a(n) - 1 = (169*n + 70)^2 for any integer n. (End)

Extensions

Edited by Charles R Greathouse IV, Jul 25 2010