A379048 Irregular triangular array: row n is the linear recurrence signature of F(i)^n - F(i-1)^n, where F = A000045 (Fibonacci numbers).

1, 1, 2, 2, -1, 3, 6, -3, -1, 4, 19, 4, -1, 8, 40, -60, -40, 8, 1, 13, 104, -260, -260, 104, 13, -1, 21, 273, -1092, -1820, 1092, 273, -21, -1, 33, 747, -3894, -16270, -3894, 747, 33, -1, 55, 1870, -19635, -85085, 136136, 85085, -19635, -1870, 55, 1, 89

Offset

1,3

Comments

(column 1) = (1, 2, 3, 4, 8, 13, 21, 33, 55, 89, 144, 232, ...) is linearly recurrent with signature (1,1,0,1,-1,-1).

(column 2) = (1, 2, 6, 19, 40, 104, 273, 747, 1870, 4895, 12816, 33784, ...) is linearly recurrent with signature (3, -1, 0, 8, -24, 8, 0, -8, 24, -8, 0, 1, -3, 1).

Row n is also the linear recurrence signature of L(i)^n - L(i-1)(^n, where L = A000032 (Lucas numbers).

Links

Table of n, a(n) for n=1..53.

Formula

First 10 rows:

1 1

2 2 -1

3 6 -3 -1

4 19 4 -1

8 40 -60 -40 8 1

13 104 -260 -260 104 13 -1

21 273 -1092 -1820 1092 273 -21 -1

33 747 -3894 -16270 -3894 747 33 -1

55 1870 -19635 -85085 136136 85085 -19635 -1870 55 1

89 4895 -83215 -582505 1514513 1514513 -582505 -83215 4895 89 -1

Mathematica

z = 25; w[i_] := Fibonacci[i];
t[p_] := Table[w[i]^p - w[i - 1]^p, {i, 1, z}];
Column[Table[FindLinearRecurrence[t[p]], {p, 1, 12}]]  (* array *)
Flatten[Table[FindLinearRecurrence[t[p]], {p, 1, 12}]]    (* sequence *)

Crossrefs

Cf. A000032, A000045, A226205, A346513.

Sequence in context: A094436 A373436 A286012 * A094441 A107230 A159830

Adjacent sequences: A379045 A379046 A379047 * A379049 A379050 A379051

Keyword

sign

Author

Clark Kimberling, Dec 16 2024

Status

approved