A127931 Numbers k such that 13 divides 11*k + 2^k.

1, 2, 6, 9, 23, 29, 70, 72, 103, 112, 128, 147, 157, 158, 162, 165, 179, 185, 226, 228, 259, 268, 284, 303, 313, 314, 318, 321, 335, 341, 382, 384, 415, 424, 440, 459, 469, 470, 474, 477, 491, 497, 538, 540, 571, 580, 596, 615, 625, 626, 630, 633, 647, 653

Offset

1,2

Comments

Sequence is infinite: starting with the 13th term, a(13)=157, a(i)=a(i-12)+156. In general, for p and p-2 both prime, starting with p-th term, a(i-(p-1))+p(p-1). This particular sequence corresponds to the case p=13.

First differences have period 12. - Charles R Greathouse IV, Oct 11 2013

Links

G. C. Greubel, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,1,-1).

Formula

a(n) = a(n-1) + a(n-12) - a(n-13). - Wesley Ivan Hurt, Feb 22 2022

Mathematica

Select[Range[700], Divisible[11#+2^#, 13]&] (* or *) LinearRecurrence[ {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {1, 2, 6, 9, 23, 29, 70, 72, 103, 112, 128, 147, 157}, 60] (* Harvey P. Dale, Sep 03 2016 *)

Prog

(PARI) isok(n) = ((11*n + 2^n) % 13) == 0; \\ Michel Marcus, Oct 11 2013

Crossrefs

Cf. A125957.

Sequence in context: A047161 A318193 A048083 * A346673 A077237 A056879

Adjacent sequences: A127928 A127929 A127930 * A127932 A127933 A127934

Keyword

nonn,easy

Author

Zak Seidov, Feb 07 2007, Feb 09 2007

Status

approved