A001533 a(n) = (8*n+1)*(8*n+7).

7, 135, 391, 775, 1287, 1927, 2695, 3591, 4615, 5767, 7047, 8455, 9991, 11655, 13447, 15367, 17415, 19591, 21895, 24327, 26887, 29575, 32391, 35335, 38407, 41607, 44935, 48391, 51975, 55687, 59527, 63495, 67591, 71815, 76167, 80647, 85255, 89991, 94855, 99847, 104967

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

From Klaus Purath, Aug 18 2022: (Start)

This is A028560(8*n+1), and thus a(n) + 9 is a square. (See formulas)

7 is the only prime number of this sequence in which all odd prime factors occur.

Each prime factor p appears exactly twice in any interval of p consecutive terms. If a(m) and a(n) are within such an interval containing p, then m + n == -1 (mod p). (End)

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A001539, A004771, A017077, A017113, A028560, A250129.

Programs

Mathematica

a[n_] := (8 n + 1)*(8 n + 7); Array[a, 40, 0] (* Amiram Eldar, Sep 08 2022 *)
LinearRecurrence[{3,-3,1},{7,135,391},40] (* Harvey P. Dale, Jan 07 2023 *)

PARI

a(n)=(8*n+1)*(8*n+7) \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 4*A001539(n) - 5.

a(n) = 128*n + a(n-1) with a(0)=7. - Vincenzo Librandi, Nov 12 2010

Sum_{n>=0} 1/a(n) = (Psi(7/8)-Psi(1/8))/48 = 0.1580099..., see A250129. - R. J. Mathar, May 30 2022 [ = (sqrt(2)+1)*Pi/48. - Amiram Eldar, Sep 08 2022]

From Klaus Purath, Aug 18 2022: (Start)

a(n) = A028560(8*n+1).

a(n) + 9 = ((a(n+1) - a(n-1))/32)^2 = A017113(n)^2.

a(2*n) = (a(n+1) - a(n-1))*n + 7. (End)

From Amiram Eldar, Feb 19 2023: (Start)

a(n) = A017077(n)*A004771(n).

Sum_{n>=0} (-1)^n/a(n) = (cos(Pi/8) * log(cot(Pi/16)) + sin(Pi/8) * log(cot(3*Pi/16)))/12.

Product_{n>=0} (1 - 1/a(n)) = cosec(Pi/8)*cos(sqrt(5/2)*Pi/4).

Product_{n>=0} (1 + 1/a(n)) = cosec(Pi/8)*cos(sqrt(2)*Pi/4). (End)

G.f.: -(7+114*x+7*x^2)/(x-1)^3. - R. J. Mathar, Apr 23 2024

From Elmo R. Oliveira, Oct 25 2024: (Start)

E.g.f.: exp(x)*(7 + 64*x*(2 + x)).

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