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A274426
Numbers that are a product of two distinct Fibonacci numbers >1 or two distinct Lucas numbers > 1.
5
6, 10, 12, 15, 16, 21, 24, 26, 28, 33, 39, 40, 42, 44, 54, 63, 65, 68, 72, 77, 87, 102, 104, 105, 110, 116, 126, 141, 165, 168, 170, 178, 188, 198, 203, 228, 267, 272, 273, 275, 288, 304, 319, 329, 369, 432, 440, 442, 445, 466, 492, 517, 522, 532, 597, 699
OFFSET
1,1
COMMENTS
Let U = {F(i)F(j), 2 < i < j}, where F = A000045 (Fibonacci numbers), and V = {L(i)L(j), 1 < i < j}, where L = A000032 (Lucas numbers). The sets U and V are disjoint, and their union, arranged as a sequence in increasing order, is A274426.
Writing u for a Fibonacci product and v for a Lucas product, the numbers in A274426 are represented by the infinite word uuvuuvuuvvuuuvvuuuvvv... This is the concatenation of uuv and the words (u^k)(v^(k-1))(u^k)(v^k) for k >= 2. Thus, there are runs of Lucas products of every finite length and runs of Fibonacci products of every finite length except 1.
Guide to related sequences:
A274426 = union of (U = {F(i)F(j), 2 < i < j} and V = {L(i)L(j), 1 < i < j})
A274429 = union of (U = {F(i)F(j), 2 < i < j} and V = {L(i)L(j), 0 < i < j})
A274374 = union of (U = {F(i)F(j), 1 < i < j} and V = {L(i)L(j), 0 < i < j})
LINKS
EXAMPLE
U = {6,10,15,16,...}, V = {12,21,28,...}, so that A274426 = (6,10,12,15,16,21,...).
MATHEMATICA
z = 200; f[n_] := Fibonacci[n];
u = Take[Sort[Flatten[Table[f[m] f[n], {n, 3, z}, {m, 3, n - 1}]]], z]
g[n_] := LucasL[n];
v = Take[Sort[Flatten[Table[g[u] g[v], {u, 2, z}, {v, 2, u - 1}]]], z]
Intersection[u, v] (* empty *)
w = Union[u, v] (* A274426 *)
f1 = Select[Range[300], MemberQ[u, w[[#]]] &] (* A274427 *)
g1 = Select[Range[300], MemberQ[v, w[[#]]] &] (* A274428 *)
CROSSREFS
Cf. A274427 (positions of numbers in U), A274428 (positions of numbers in V), A000032, A000045, A274429, A274432.
Sequence in context: A362180 A081390 A210578 * A345995 A324456 A324857
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 22 2016
STATUS
approved