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

0,23

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)

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

Example

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

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 *)

Prog

(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
(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

Crossrefs

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

Sequence in context: A237347 A075017 A060586 * A076662 A164359 A178307

Adjacent sequences: A385455 A385456 A385457 * A385459 A385460 A385461

Keyword

nonn,tabl

Author

David Radcliffe, Jun 29 2025

Status

approved