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A160009
Numbers that are the product of distinct Fibonacci numbers.
27
0, 1, 2, 3, 5, 6, 8, 10, 13, 15, 16, 21, 24, 26, 30, 34, 39, 40, 42, 48, 55, 63, 65, 68, 78, 80, 89, 102, 104, 105, 110, 120, 126, 130, 144, 165, 168, 170, 178, 195, 204, 208, 210, 233, 240, 267, 272, 273, 275, 288, 312, 315, 330, 336, 340, 377, 390, 432, 440, 442, 445
OFFSET
1,3
COMMENTS
Starts the same as A049862, the product of two distinct Fibonacci numbers. This sequence has an infinite number of consecutive terms that are consecutive numbers (such as 15 and 16) because fib(k)*fib(k+3) and fib(k+1)*fib(k+2) differ by one for all k >= 0.
It follows from Carmichael's theorem that if u and v are finite sets of Fibonacci numbers such that (product of all the numbers in u) = (product of all the numbers in v), then u = v. The same holds for many other 2nd order linear recurrence sequences with constant coefficients. In the following guide to related "distinct product sequences", W = Wythoff array, A035513:
base sequence distinct-product sequence
A000045 (Fibonacci) A160009
A000032 (Lucas, without 2) A274280
A000032 (Lucas, with 2) A274281
A000285 (1,4,5,...) A274282
A022095 (1,5,6,...) A274283
A006355 (2,4,6,...) A274284
A013655 (2,5,7,...) A274285
A022086 (3,6,9,...) A274191
row 2 of W: (4,7,11,...) A274286
row 3 of W: (6,10,16,...) A274287
row 4 of W: (9,15,24,...) A274288
- Clark Kimberling, Jun 17 2016
MATHEMATICA
s={1}; nn=30; f=Fibonacci[2+Range[nn]]; Do[s=Union[s, Select[s*f[[i]], #<=f[[nn]]&]], {i, nn}]; s=Prepend[s, 0]
CROSSREFS
KEYWORD
nonn
AUTHOR
T. D. Noe, Apr 29 2009
STATUS
approved