A376583 Parity of A002260.

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, 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, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1

Offset

1

Comments

This is also the triangle:

1;

1, 0;

1, 0, 1;

1, 0, 1, 0;

1, 0, 1, 0, 1;

...

This is also the array:

1 1 1 1 1 1 1 1 ...

0 0 0 0 0 0 0 0 ...

1 1 1 1 1 1 1 1 ...

0 0 0 0 0 0 0 0 ...

1 1 1 1 1 1 1 1 ...

0 0 0 0 0 0 0 0 ...

1 1 1 1 1 1 1 1 ...

0 0 0 0 0 0 0 0 ...

...

read by antidiagonals.

As this sequence is "the opposite" of A060510, most things mentioned there also apply here.

Essentially the same as A177990. - R. J. Mathar, Oct 24 2024

The choice of offset 0 together with signed terms might be advantageous as this corresponds to the sign pattern of various kinds of Chebyshev polynomials. Under these assumptions T(n, k) = cos(k*Pi/2) and T(n, k) = A369736(n, n-k). - Peter Luschny, Jan 03 2025

Links

Table of n, a(n) for n=1..87.

Formula

a(n) = A000035(A002260(n)).

a(n) = 1 - A060510(n-1).

Mathematica

fp[i_]:=If[OddQ[i], 1, 0]; fp/@Flatten[Table[Range[n], {n, 14}]] (* James C. McMahon, Oct 23 2024 *)

Prog

(PARI) a(n) = (n - binomial(floor(sqrt(2*n)+1/2), 2)) % 2;
(Python)
from math import isqrt
def A376583(n): return n&1^((m:=isqrt(n<<3)+1>>1)*(m-1)>>1&1) # Chai Wah Wu, Oct 23 2024

Crossrefs

Cf. A000035, A002260, A060510 (parity of A002262), A128174 (mirror triangle), A369736.

Sequence in context: A189163 A189129 A267269 * A187972 A248396 A285952

Adjacent sequences: A376580 A376581 A376582 * A376584 A376585 A376586

Keyword

nonn,tabl,easy

Author

Ruud H.G. van Tol, Sep 29 2024

Status

approved