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%I #59 Dec 12 2023 07:45:22
%S 0,1,2,0,2,1,0,1,2,0,2,1,0,1,2,0,2,1,0,1,2,0,2,1,0,1,2,0,2,1,0,1,2,0,
%T 2,1,0,1,2,0,2,1,0,1,2,0,2,1,0,1,2,0,2,1,0,1,2,0,2,1,0,1,2,0,2,1,0,1,
%U 2,0,2,1,0,1,2,0,2,1,0,1,2,0,2,1,0,1,2,0,2,1
%N Period 6 sequence 0,1,2,0,2,1.
%C This sequence can be extended periodically to negative values of n.
%C See a comment on A203571 where a k-family of 2k-periodic sequences P_k has been defined. The present sequence is P_3. - _Wolfdieter Lang_, Feb 02 2012
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,1).
%F a(n) = n (mod 3) if (-1)^floor(n/3)=+1 else (3 - n)(mod 3), n>=0. (-1)^floor(n/3) is the parity of the quotient floor(n/3), sometimes denoted by n\3.
%F O.g.f.: x*(1+2*x+2*x^3+x^4)/(1-x^6).
%F a(n) = 1-( (-1)^(n+1)+cos(Pi*n/3)+3*cos(2*Pi*n/3))/3. - _R. J. Mathar_, Oct 07 2011, corrected by _Vaclav Kotesovec_, Feb 19 2023
%F a(n) = floor((71/364)*3^(n+1)) mod 3. - _Hieronymus Fischer_, Jan 04 2013
%F a(n) = floor((4007/333333)*10^(n+1)) mod 10. - _Hieronymus Fischer_, Jan 04 2013
%F From _Amiram Eldar_, Jan 01 2023: (Start)
%F Multiplicative with a(2^e) = 2, a(3^e) = 0, and a(p^e) = 1 for p >= 5.
%F Dirichlet g.f.: zeta(s)*(1+1/2^s-1/3^s-1/6^s). (End)
%e a(8) = 8(mod 3) = 2 because (-1)^floor(8/3)= +1; 8\3 = 2 is even.
%e a(4) = (3-1)(mod 3) = 2, because (-1)^floor(4/3) is -1; 4\3 = 1 is odd.
%t Table[{0, 1, 2, 0, 2, 1}, {15}] // Flatten (* _Jean-François Alcover_, Jun 21 2013 *)
%t PadRight[{},120,{0,1,2,0,2,1}] (* _Harvey P. Dale_, Jul 25 2020 *)
%o (PARI) a(n)=[0,1,2,0,2,1][n%6+1] \\ _Charles R Greathouse IV_, Oct 07 2011
%Y Cf. signed versions A112300, A186809.
%K nonn,easy,mult
%O 0,3
%A _Wolfdieter Lang_, Sep 30 2011
%E Keyword:mult added by _Andrew Howroyd_, Jul 31 2018
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