A242215 a(n) = 18*n + 5.

5, 23, 41, 59, 77, 95, 113, 131, 149, 167, 185, 203, 221, 239, 257, 275, 293, 311, 329, 347, 365, 383, 401, 419, 437, 455, 473, 491, 509, 527, 545, 563, 581, 599, 617, 635, 653, 671, 689, 707, 725, 743, 761, 779, 797, 815, 833, 851, 869, 887, 905, 923, 941, 959

Offset: 0

Author

Arkadiusz Wesolowski, May 07 2014

Keywords

• nonn
• easy

Comments

Conjecture: there are infinitely many composite Fermat numbers such that no one of them has a divisor that belongs to this sequence.

Links

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

Crossrefs

Supersequence of A061240.

Cf. A229855.

Programs

Magma

[18*n+5: n in [0..53]];

Maple

seq(18*n+5, n=0..53);

Mathematica

Table[18*n + 5, {n, 0, 53}]
LinearRecurrence[{2,-1},{5,23},60] (* Harvey P. Dale, Aug 25 2017 *)

PARI

for(n=0, 53, print1(18*n+5, ", "));

Formula

G.f.: (5 + 13*x)/(1 - x)^2.

From Elmo R. Oliveira, Dec 08 2024: (Start)

E.g.f.: exp(x)*(5 + 18*x).

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