A385458 Triangle read by rows: T(n,k) = exponent of the highest power of 2 dividing each Fibonomial coefficient fibonomial(n, k).

0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 3, 2, 3, 3, 0, 0, 0, 3, 2, 2, 3, 0, 0, 0, 0, 0, 2, 2, 2, 0, 0, 0, 0, 1, 1, 0, 3, 3, 0, 1, 1, 0, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 3, 4, 4, 1, 4, 4, 3, 4, 4, 0

Offset: 0

Author

David Radcliffe, Jun 29 2025

Keywords

• nonn
• tabl

Examples

			Triangle begins:
   n\k  0  1  2  3  4  5  6  7  8  9 10 11 12
   0:   0
   1:   0  0
   2:   0  0  0
   3:   0  1  1  0
   4:   0  0  1  0  0
   5:   0  0  0  0  0  0
   6:   0  3  3  2  3  3  0
   7:   0  0  3  2  2  3  0  0
   8:   0  0  0  2  2  2  0  0  0
   9:   0  1  1  0  3  3  0  1  1  0
  10:   0  0  1  0  0  3  0  0  1  0  0
  11:   0  0  0  0  0  0  0  0  0  0  0  0
  12:   0  4  4  3  4  4  1  4  4  3  4  4  0
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).
• 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.
• David Radcliffe, Matrix plot of the first 768 rows
• Diana L. Wells, The Fibonacci and Lucas triangles modulo 2, Fibonacci Quart. 32, no. 2 (1994), 111-123. (Theorem 2)

Crossrefs

Cf. A000120, A010048, A007814, A065040, A337923, A385456, A385457, A385608.

Programs

Julia

function T_row(n)
    function T(n, k)
        c(a, b) = 2 * a + b ÷ 6 - count_ones(a)
        (nd, nm) = divrem(n, 3)
        (kd, km) = divrem(k, 3)
        !(nm < km || (kd & (nd - kd)) != 0) && return 0
        c(nd, n) - c(kd, k) - c((n - k) ÷ 3, n - k)
    end
    [T(n, k) for k in 0:n]
end
for n in 0:12 println(T_row(n)) end  # Peter Luschny, Jul 02 2025

Mathematica

A385608[n_] := A385608[n] = 2*# + Quotient[n, 6] - DigitSum[#, 2] & [Quotient[n, 3]];
A385458[n_, k_] := A385608[n] - A385608[k] - A385608[n-k];
Table[A385458[n, k], {n, 0, 15}, {k, 0, n}] (* Paolo Xausa, Jul 04 2025 *)

Python

def b(n): return 2*(n//3) + n//6 - (n//3).bit_count()
def T(n, k): return b(n) - b(k) - b(n-k) # David Radcliffe, Jul 01 2025

Formula

T(n, k) = A007814(A010048(n, k)).

T(n, k) = Sum_{i=1..k} (A337923(n+1-i) - A337923(i)).

T(n, k) = b(n) - b(k) - b(n - k), where b(n) = 2*floor(n/3) + floor(n/6) - A000120(floor(n/3)) = A385608(n) is the 2-adic valuation of the product of the first n Fibonacci numbers.

sign(T(n, k)) = 1 - A385456(n, k). - Peter Luschny, Jul 03 2025