A049998

[Don Reble]

> Is A049998 composed entirely of Fibonacci numbers?

> Someone willing to find a proof?

Rearranging A049997:
      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
(Thanks for the idea, Berny.)

First prove these identities:

    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 those 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.

-- 
Don Reble  djr at nk.ca

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