A078370 a(n) = 4*(n+1)*n + 5.

5, 13, 29, 53, 85, 125, 173, 229, 293, 365, 445, 533, 629, 733, 845, 965, 1093, 1229, 1373, 1525, 1685, 1853, 2029, 2213, 2405, 2605, 2813, 3029, 3253, 3485, 3725, 3973, 4229, 4493, 4765, 5045, 5333, 5629, 5933, 6245, 6565, 6893, 7229, 7573, 7925, 8285, 8653, 9029

Offset: 0

Author

Wolfdieter Lang, Nov 29 2002

Keywords

• nonn
• easy

Comments

This is the generic form of D in the (nontrivially) solvable Pell equation x^2 - D*y^2 = -4. See A078356, A078357.

1/5 + 1/13 + 1/29 + ... = (Pi/8)*tanh Pi [Jolley]. - Gary W. Adamson, Dec 21 2006

Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n+1)^2 + 4), n = 1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010

(2*n + 1 + sqrt(a(n)))/2 = [2*n + 1; 2*n + 1, 2*n + 1, ...], n >= 0, with the regular continued fraction with period length 1. This is the odd case. See A087475 for the general case with the Schroeder reference and comments. For the even case see A002522.

Primes in the sequence are in A005473. - Russ Cox, Aug 26 2019

The continued fraction expansion of sqrt(a(n)) is [2n+1; {n, 1, 1, n, 4n+2}]. For n=0, this collapses to [2; {4}]. - Magus K. Chu, Aug 27 2022

Discriminant of the binary quadratic forms y^2 - x*y - A002061(n+1)*x^2. - Klaus Purath, Nov 10 2022

From Klaus Purath, Apr 08 2025: (Start)

There are no squares in this sequence. The 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*(m*y)^2 = -1 for any integer n where m = (D - 3)/2. The values for k and the solutions x, y can be calculated using the following algorithm: k = sqrt(D*m^2 - 1), x(0) = 1, x(1) = 4*D*m^2 - 1, y(0) = 1, y(1) = 4*D*m^2 - 3. The two recurrences are of the form (4*D*m^2 - 2, -1).

It follows from the above that this sequence belongs to A031396. (End)

References

• L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 176.

Links

• Indranil Ghosh, Table of n, a(n) for n = 0..1000
• Leo Tavares, Square illustration.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Subsequence of A077426 (D values (not a square) for which Pell x^2 - D*y^2 = -4 is solvable in positive integers).

Cf. A002061, A002522, A004770, A005473, A016754.

Cf. A054413, A078356, A078357, A086902, A087475.

Subsequence of A031396.

Programs

Magma

[4*n^2+4*n+5 : n in [0..80]]; // Wesley Ivan Hurt, Aug 29 2022

Mathematica

Table[4 n (n + 1) + 5, {n, 0, 45}] (* or *)
Table[8 Binomial[n + 1, 2] + 5, {n, 0, 45}] (* or *)
CoefficientList[Series[(5 - 2 x + 5 x^2)/(1 - x)^3, {x, 0, 45}], x] (* Michael De Vlieger, Jan 04 2017 *)

PARI

a(n)=4*n^2+4*n+5 \\ Charles R Greathouse IV, Sep 24 2015

Python

a= lambda n: 4*n**2+4*n+5 # Indranil Ghosh, Jan 04 2017

Scala

(1 to 99 by 2).map(n => n * n + 4) // Alonso del Arte, May 29 2019

Formula

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

a(n) = 4*(n+1)*n + 5 = 8*binomial(n+1, 2) + 5, hence subsequence of A004770 (5 (mod 8) numbers). [Typo fixed by Zak Seidov, Feb 26 2012]

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

a(n) = 8*n + a(n-1), with a(0) = 5. - Vincenzo Librandi, Aug 08 2010

a(n) = A016754(n) + 4. - Leo Tavares, Feb 22 2023

From Elmo R. Oliveira, Oct 31 2024: (Start)

E.g.f.: (5 + 8*x + 4*x^2)*exp(x).

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

Extensions

More terms from Max Alekseyev, Mar 03 2010