cp's OEIS Frontend

A131078 Periodic sequence (1, 1, 1, 1, 0, 0, 0, 0).

%I A131078 #44 Dec 09 2024 06:00:54
%S A131078 1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,
%T A131078 1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,
%U A131078 0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,0,0,0,0,1
%N A131078 Periodic sequence (1, 1, 1, 1, 0, 0, 0, 0).
%H A131078 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,-1,1).
%F A131078 a(1) = a(2) = a(3) = a(4) = 1, a(5) = a(6) = a(7) = a(8) = 0; for n > 8, a(n) = a(n-8).
%F A131078 G.f.: x/((1-x)*(1+x^4)).
%F A131078 a(n) = floor(((n+4) mod 8)/4). [_Gary Detlefs_, May 17 2011]
%F A131078 From _Wesley Ivan Hurt_, May 30 2015: (Start)
%F A131078 a(n) = a(n-1)-a(n-4)+a(n-5), n>5.
%F A131078 a(n) = (1+(-1)^((2*n+11-(-1)^n+2*(-1)^((2*n+5-(-1)^n)/4))/8))/2. (End)
%F A131078 From _Ridouane Oudra_, Nov 17 2019: (Start)
%F A131078 a(n) = binomial(n+3,4) mod 2
%F A131078 a(n) = floor((n+3)/4) - 2*floor((n+3)/8). (End)
%o A131078 (PARI) {m=105; for(n=1, m, print1((n-1)%8<4, ","))}
%o A131078 (Magma) m:=105; [ [1, 1, 1, 1, 0, 0, 0, 0][ (n-1) mod 8 + 1 ]: n in [1..m] ];
%o A131078 (Magma) &cat[[1, 1, 1, 1, 0, 0, 0,0]: n in [0..10]]; // _Vincenzo Librandi_, May 31 2015
%o A131078 (Magma) [Floor((1+(-1)^((2*n+11-(-1)^n+2*(-1)^((2*n+5-(-1)^n)/4))/8))/2): n in [1..60]]; // _Vincenzo Librandi_, May 31 2015
%o A131078 (Python)
%o A131078 def A131078(n): return int(not n-1&4) # _Chai Wah Wu_, Jan 31 2023
%Y A131078 Cf. A131074, A000035.
%Y A131078 Period 2*k: repeat k ones followed by k zeros: A000035(n+1) (k=1), A133872(n) (k=2), A088911 (k=3), this sequence (k=4), and A112713(n-1) (k=5).
%K A131078 nonn,easy
%O A131078 1,1
%A A131078 _Klaus Brockhaus_, following a suggestion of _Paul Curtz_, Jun 14 2007