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%I #14 Feb 22 2022 14:34:22
%S 1,2,6,9,23,29,70,72,103,112,128,147,157,158,162,165,179,185,226,228,
%T 259,268,284,303,313,314,318,321,335,341,382,384,415,424,440,459,469,
%U 470,474,477,491,497,538,540,571,580,596,615,625,626,630,633,647,653
%N Numbers k such that 13 divides 11*k + 2^k.
%C 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.
%C First differences have period 12. - _Charles R Greathouse IV_, Oct 11 2013
%H G. C. Greubel, <a href="/A127931/b127931.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,0,0,0,0,0,1,-1).
%F a(n) = a(n-1) + a(n-12) - a(n-13). - _Wesley Ivan Hurt_, Feb 22 2022
%t 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 *)
%o (PARI) isok(n) = ((11*n + 2^n) % 13) == 0; \\ _Michel Marcus_, Oct 11 2013
%Y Cf. A125957.
%K nonn,easy
%O 1,2
%A _Zak Seidov_, Feb 07 2007, Feb 09 2007
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