%I #20 Oct 23 2024 16:05:45
%S 1,1,0,1,0,1,1,0,1,0,1,0,1,0,1,1,0,1,0,1,0,1,0,1,0,1,0,1,1,0,1,0,1,0,
%T 1,0,1,0,1,0,1,0,1,0,1,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1,0,
%U 1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1
%N Parity of A002260.
%C This is also the triangle:
%C 1;
%C 1, 0;
%C 1, 0, 1;
%C 1, 0, 1, 0;
%C 1, 0, 1, 0, 1;
%C ...
%C This is also the array:
%C 1 1 1 1 1 1 1 1 ...
%C 0 0 0 0 0 0 0 0 ...
%C 1 1 1 1 1 1 1 1 ...
%C 0 0 0 0 0 0 0 0 ...
%C 1 1 1 1 1 1 1 1 ...
%C 0 0 0 0 0 0 0 0 ...
%C 1 1 1 1 1 1 1 1 ...
%C 0 0 0 0 0 0 0 0 ...
%C ...
%C read by antidiagonals.
%C As this sequence is "the opposite" of A060510, most things mentioned there also apply here.
%F a(n) = A000035(A002260(n)).
%F a(n) = 1 - A060510(n-1).
%o (PARI) a(n) = (n - binomial(floor(sqrt(2*n)+1/2), 2)) % 2;
%o (Python)
%o from math import isqrt
%o def A376583(n): return n&1^((m:=isqrt(n<<3)+1>>1)*(m-1)>>1&1) # _Chai Wah Wu_, Oct 23 2024
%Y Cf. A000035, A002260, A060510 (parity of A002262), A128174 (mirror triangle).
%K nonn,tabl,new
%O 1
%A _Ruud H.G. van Tol_, Sep 29 2024