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%I #14 Sep 08 2022 08:46:20
%S 1,2,4,5,7,9,11,12,14,17,17,19,23,22,24,29,27,29,35,32,34,41,37,39,47,
%T 42,44,53,47,49,59,52,54,65,57,59,71,62,64,77,67,69,83,72,74,89,77,79,
%U 95,82,84,101,87,89,107,92,94,113,97,99,119,102,104,125
%N a(0)=1; thereafter, a(n) = 2n-1 if n == 0 (mod 3), (5n+1)/3 if n == 1 (mod 3), (5n+2)/3 if n == 2 (mod 3).
%D A. V. Shutov, On the number of words of a given length in plane crystallographic groups (Russian), Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 302 (2003), Anal. Teor. Chisel i Teor. Funkts. 19, 188--197, 203; translation in J. Math. Sci. (N.Y.) 129 (2005), no. 3, 3922-3926 [MR2023041]. See Table 1, line "p4".
%H Vincenzo Librandi, <a href="/A301728/b301728.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).
%F G.f.: -(-2*x^6-x^5-3*x^4-3*x^3-4*x^2-2*x-1)/(x^6-2*x^3+1).
%F Recurrence (there are several equivalent versions):
%F -n a(n) - 2 a(n + 1) - a(n + 2) + (n - 3) a(n + 3) + 2 a(n + 4) + a(n + 5) = 0, with a(0) = 1, a(1) = 2, a(2) = 4, a(3) = 5, a(4) = 7.
%F a(n) = 2*a(n-3) - a(n-6) for n > 6. - _Chai Wah Wu_, Mar 30 2018
%p f:= proc(n) if n=0 then 1
%p elif (n mod 3) = 0 then 2*n-1
%p elif (n mod 3) = 1 then (5*n+1)/3
%p else (5*n+2)/3; fi;
%p end;
%p [seq(f(n),n=0..70)];
%t Join[{1}, LinearRecurrence[{0, 0, 2, 0, 0, -1}, {2, 4, 5, 7, 9, 11}, 100]] (* _Vincenzo Librandi_, Mar 31 2018 *)
%t nxt[{n_,a_}]:={n+1,Which[Divisible[n+1,3],2n+1,Mod[n+1,3]==1,(5n+6)/3,True,(5n+7)/3]}; NestList[nxt,{0,1},70][[All,2]] (* _Harvey P. Dale_, Sep 08 2021 *)
%o (Magma) I:=[1,2,4,5,7,9,11]; [n le 7 select I[n] else 2*Self(n-3)-Self(n-6): n in [1..70]]; // _Vincenzo Librandi_, Mar 31 2018
%K nonn
%O 0,2
%A _N. J. A. Sloane_, Mar 30 2018
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