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(PARI) isok(n) = (Mod(3, 17)^n- == Mod(n) % , 17) == 0; \\ Michel Marcus, Dec 02 2013
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<a href="/index/Rec#order_16">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-1).
Numbers n k such that 17 divides 3^nk-nk.
a(n+1)-a(n) is the 16-periodic sequence (24, 13, 33, 57, 2, 6, 12, 7, 3, 1, 38, 5, 9, 26, 15, 21).
Select[Range[1000], PowerMod[3, #, 17] == Mod[#, 17] &] (* Amiram Eldar, May 14 2022 *)
Amiram Eldar, <a href="/A081116/b081116.txt">Table of n, a(n) for n = 1..10000</a>
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