A385627 Table read by rows: T(n, k) = (binomial(n, k) * fibonomial(n, k)) mod 2.

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

Offset

0

Links

Table of n, a(n) for n=0..90.

Peter Luschny, Illustarting the triangle. The red pixels in this plot of the Sierpiński triangle correspond to the value 1 of this triangle.

Formula

T(n, k) = [(n & k = k) and ((n mod 3 >= k mod 3) and (~floor(n/3) & floor(k/3) = 0))], where '&' denotes the bitwise 'and', '~' the bitwise complement, and [.] is the Iverson bracket.

T(n, k) = A385626(n, k) mod 2.

T(n, k) = (A007318(n, k) * A010048(n, k)) mod 2.

T(n, k) = (A385630(n) / (A385630(n - k) * A385630(k))) mod 2.

Example

  [ 0] 1;
  [ 1] 1, 1;
  [ 2] 1, 0, 1;
  [ 3] 1, 0, 0, 1;
  [ 4] 1, 0, 0, 0, 1;
  [ 5] 1, 1, 0, 0, 1, 1;
  [ 6] 1, 0, 0, 0, 0, 0, 1;
  [ 7] 1, 1, 0, 0, 0, 0, 1, 1;
  [ 8] 1, 0, 0, 0, 0, 0, 0, 0, 1;
  [ 9] 1, 0, 0, 0, 0, 0, 0, 0, 0, 1;
  [10] 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;
  [11] 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1;
  [12] 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1;

Mathematica

isT[n_, k_] :=
  BitAnd[n, k] === k &&
  Mod[n, 3] >= Mod[k, 3] &&
  BitAnd[BitNot[Quotient[n, 3]], Quotient[k, 3]] === 0;
Table[Boole[isT[n, k]], {n, 0, 12}, {k, 0, n}] // Flatten

Prog

(Python)
def isT(n: int, k: int) -> bool:
    return (n & k) == k and (n % 3 >= k % 3 and ((~(n // 3)) & (k // 3)) == 0)
for n in range(13): print([int(isT(n, k)) for k in range(n + 1)])

Crossrefs

Cf. A385626, A385395, A007318, A010048, A385630.

Cf. A047999, A385456.

Sequence in context: A103451 A103452 A131219 * A127970 A158856 A154957

Adjacent sequences: A385624 A385625 A385626 * A385628 A385629 A385630

Keyword

nonn,tabl

Author

Peter Luschny, Jul 06 2025

Status

approved