A385732 Triangle read by rows: the numerators of the Lucas triangle.

1, 1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 7, 28, 7, 1, 1, 11, 77, 77, 11, 1, 1, 18, 66, 231, 66, 18, 1, 1, 29, 174, 957, 957, 174, 29, 1, 1, 47, 1363, 4089, 44979, 4089, 1363, 47, 1, 1, 76, 3572, 25897, 155382, 155382, 25897, 3572, 76, 1, 1, 123, 3116, 36613, 1061777, 19111986, 1061777, 36613, 3116, 123, 1

Offset: 0

Author

Peter Luschny, Jul 08 2025

Keywords

• nonn
• tabl
• frac

Examples

			Triangle begins:
  [0] 1;
  [1] 1,  1;
  [2] 1,  3,    1;
  [3] 1,  4,    4,     1;
  [4] 1,  7,   28,     7,      1;
  [5] 1, 11,   77,    77,     11,      1;
  [6] 1, 18,   66,   231,     66,     18,     1;
  [7] 1, 29,  174,   957,    957,    174,    29,    1;
  [8] 1, 47, 1363,  4089,  44979,   4089,  1363,   47,  1;
  [9] 1, 76, 3572, 25897, 155382, 155382, 25897, 3572, 76, 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 (1989), 212-219.
• Edouard Lucas, Théorie des Fonctions Numériques Simplement Périodiques, American Journal of Mathematics 1 (1878), 184-240, §9.
• Diana L. Wells, The Fibonacci and Lucas triangles modulo 2, Fibonacci Quart. 32, no. 2 (1994), p. 112.

Crossrefs

Cf. A385733 (denominators), A070825 (Lucanorial), A003266 (Fibonorial), A010048 (Fibonomial).

Programs

Maple

c := arccsch(2) - I*Pi/2:
LT := (n, k) -> mul(I^j*cosh(c*j), j = k + 1..n) / mul(I^j*cosh(c*j), j = 1..n - k):
T := (n, k) -> numer(simplify(LT(n, k))): seq(seq(T(n, k), k = 0..n), n = 0..10);

Mathematica

T[n_, k_] := With[{c = ArcCsch[2] - I Pi/2}, Product[I^j Cosh[c j], {j, k + 1, n}] / Product[I^j Cosh[c j], {j, 1, n - k}]];
Table[Simplify[T[n, k]], {n, 0, 8}, {k, 0, n}] // Flatten // Numerator

Formula

LT(n, k) = Product_{j=k+1..n} i^j*cosh(c*j) / Product_{j=1..n-k} i^j*cosh(c*j) where c = arccsch(2) - i*Pi/2 and i is the imaginary unit. If in this formula cosh is substituted by sinh one gets the Fibonomial triangle A010048.

T(n, k) = numerator(LT(n, k)).