A348845 Part two of the trisection of A017101: a(n) = 11 + 24*n.

11, 35, 59, 83, 107, 131, 155, 179, 203, 227, 251, 275, 299, 323, 347, 371, 395, 419, 443, 467, 491, 515, 539, 563, 587, 611, 635, 659, 683, 707, 731, 755, 779, 803, 827, 851, 875, 899, 923, 947, 971, 995, 1019, 1043, 1067

Offset: 0

Author

Wolfdieter Lang, Dec 11 2021

Keywords

• nonn
• easy

Comments

The trisection of A017101 = {3 + 8*k}A017077%20=%20%7B3*(1%20+%2012*n)%7D">{k>=0} gives 3*A017077 = {3*(1 + 12*n)}{n>=0}, {a(n)}A350051%20=%20%7B19%20+%2024*n%7D">{n >= 0} and A350051 = {19 + 24*n}{n>=0}. These three sequences are congruent to 3 modulo 8 and to 3, 5, and 1 modulo 6, respectively.

Links

• Index entries for linear recurrences with constant coefficients, signature (2,-1).

Crossrefs

Cf. A008606, 3*A017077, A017101, A350051.

Programs

Mathematica

24 * Range[0, 44] + 11 (* Amiram Eldar, Dec 18 2021 *)

Formula

a(n) = 11 + 24*n = 11 + A008606(n), for n >= 0

a(n) = 2*a(n-1) - a(n-2), for n >= 1, with a(-1) = -13, a(0) = 11.

G.f.: (11 + 13*x)/(1-x)^2.

E.g.f.: (11 + 24*x)*exp(x).