A082285 a(n) = 16*n + 13.

13, 29, 45, 61, 77, 93, 109, 125, 141, 157, 173, 189, 205, 221, 237, 253, 269, 285, 301, 317, 333, 349, 365, 381, 397, 413, 429, 445, 461, 477, 493, 509, 525, 541, 557, 573, 589, 605, 621, 637, 653, 669, 685, 701, 717, 733, 749, 765, 781, 797, 813, 829, 845

Offset: 0

Author

Cino Hilliard, May 10 2003

Keywords

• easy
• nonn

Comments

Solutions to (7^x + 11^x) mod 17 = 13.

a(n-2), n>=2, gives the second column in triangle A238476 related to the Collatz problem. - Wolfdieter Lang, Mar 12 2014

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Tanya Khovanova, Recursive Sequences.
• Leo Tavares, Illustration: Bounded Star Crosses.
• Index entries for linear recurrences with constant coefficients, signature (2,-1).

Crossrefs

Cf. A004770, A008594, A008598, A016813, A098502, A106839, A119413.

Programs

Magma

[[ n : n in [1..1000] | n mod 16 eq 13]]; // Vincenzo Librandi, Oct 10 2011

Mathematica

Range[13, 1000, 16] (* Vladimir Joseph Stephan Orlovsky, May 31 2011 *)
LinearRecurrence[{2,-1},{13,29},60] (* Harvey P. Dale, Jan 28 2023 *)

PARI

\\ solutions to 7^x+11^x == 13 mod 17
anpbn(n) = { for(x=1,n, if((7^x+11^x-13)%17==0,print1(x" "))) }

Formula

a(n) = 16*n + 13.

a(n) = 32*n - a(n-1) + 10; a(0)=13. - Vincenzo Librandi, Oct 10 2011

From Stefano Spezia, Dec 27 2019: (Start)

O.g.f.: (13 + 3*x)/(1 - x)^2.

E.g.f.: exp(x)*(13 + 16*x). (End)

a(n) = A008594(n+1) + A016813(n+1) - 4. - Leo Tavares, Sep 22 2022

From Elmo R. Oliveira, Apr 12 2025: (Start)

a(n) = 2*a(n-1) - a(n-2).

a(n) = A004770(2*n+2). (End)