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Search: id:A049998
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| A049998 |
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a(n)=b(n)-b(n-1), where b=A049997 (differences of products of Fibonacci numbers). |
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+0
2
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| 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 5, 3, 1, 1, 8, 5, 1, 2, 13, 8, 1, 1, 3, 21, 13, 2, 1, 5, 34, 21, 3, 1, 1, 8, 55, 34, 5, 1, 2, 13, 89, 55, 8, 1, 1, 3, 21, 144, 89, 13, 2, 1, 5, 34, 233, 144, 21, 3, 1, 1, 8, 55, 377, 233, 34, 5, 1, 2, 13, 89, 610, 377, 55, 8, 1, 1, 3, 21, 144, 987, 610, 89
(list; graph; listen)
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OFFSET
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1,7
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COMMENT
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David W. Wilson conjectured (Dec 14 2005) that this sequence consists only of Fibonacci numbers. Proofs were found by Franklin T. Adams-Watters and Don Reble, Dec 14 2005. The following is Reble's proof:
Rearrange A049997, as suggested by Bernardo Boncompagni (redgolpe(AT)redgolpe.com):
1
2
3 4
5 6
8 9 10
13 15 16
21 24 25 26
34 39 40 42
55 63 64 65 68
89 102 104 105 110
144 165 168 169 170 178
233 267 272 273 275 288
377 432 440 441 442 445 466
Then we know that
F(a+1) * F(a-1) - F(a) * F(a) = (-1)^a
F(a+1) * F(b-1) - F(a-1) * F(b+1)
= + (-1)^b F(a-b), if a>b
= - (-1)^a F(b-a), if a<b
Use these to show that from F(x) to F(x+1), the representable numbers are
F(x) = F(x) * F(2)
< F(x-2) * F(4)
< F(x-4) * F(6)
< ...
< F(x-3) * F(5)
< F(x-1) * F(3)
< F(x+1) * F(1) = F(x+1)
(If x is even, the first identity is needed when the parity changes in the middle.)
Each Fibonacci-product is in one of those subsequences and the identities show that each difference is a Fibonacci number.
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MATHEMATICA
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t = Take[ Union@Flatten@Table[ Fibonacci[i]Fibonacci[j], {i, 0, 20}, {j, 0, i}], 85]; Drop[t, 1] - Drop[t, -1] (* Robert G. Wilson v *)
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CROSSREFS
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A049997 gives numbers of the form F(i)*F(j), when these Fibonacci-products are arranged in order without duplicates.
Sequence in context: A137569 A089177 A023996 this_sequence A029253 A016441 A131333
Adjacent sequences: A049995 A049996 A049997 this_sequence A049999 A050000 A050001
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KEYWORD
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nonn
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AUTHOR
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Clark Kimberling (ck6(AT)evansville.edu)
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EXTENSIONS
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More terms from Robert G. Wilson v (rgwv(at)rgwv.com), Dec 14 2005
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