A385434 Triangle of Gaussian binomial coefficients (or q-binomial coefficients) [n,k] for q = 2, reduced mod 3.

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 1, 2, 2, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1

Offset: 0

Author

David Radcliffe, Jun 28 2025

Keywords

• nonn
• tabl

Comments

Row sums give A385435.

Examples

			Triangle begins:
  [ 0] [1]
  [ 1] [1, 1]
  [ 2] [1, 0, 1]
  [ 3] [1, 1, 1, 1]
  [ 4] [1, 0, 2, 0, 1]
  [ 5] [1, 1, 2, 2, 1, 1]
  [ 6] [1, 0, 0, 0, 0, 0, 1]
  [ 7] [1, 1, 0, 0, 0, 0, 1, 1]
  [ 8] [1, 0, 1, 0, 0, 0, 1, 0, 1]
  [ 9] [1, 1, 1, 1, 0, 0, 1, 1, 1, 1]
  [10] [1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 1]
  [11] [1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1]
		

Links

• Donald E. Knuth and Herbert S. Wilf, The power of a prime that divides a generalized binomial coefficient, J. Reine Angew. Math., 396:212-219, 1989.
• Romeo Meštrović, Lucas' theorem: its generalizations, extensions and applications (1878--2014), arXiv preprint arXiv:1409.3820 [math.NT], 2014. (See Equation (80) on p. 31.)

Crossrefs

Cf. A007318, A022166, A385435.

Programs

Mathematica

Table[Mod[QBinomial[n, k, 2],3], {n, 0, 11}, {k, 0, n}] // Flatten  (* James C. McMahon, Jun 29 2025 *)

SageMath

def T(n, k): return mod(gaussian_binomial(n, k).subs(q=2), 3)
for n in range(12): print([T(n, k) for k in range(n+1)])  # Peter Luschny, Jun 29 2025

Formula

a(n) = A022166(n) mod 3.

T(2n, 2k) = T(2n+1, 2k) = T(2n, 2k+1) = binomial(n, k) mod 3; T(2n, 2k+1) = 0.