From Rolfe Petschek comes a puzzle that’s good fun in snow, rain, heat or gloom of night. (Okay, maybe not in gloom of night.)
A postal worker and his customer joke about the various ways the customer could mathematically encode her post office box number.
The customer realizes that every integer greater than 1 can be encoded via at least one Fibonacci-like sequence using an ordered triple (m, n, q). The encoded number is the qth member of the sequence after the first two positive integers m and n, where each term is the sum of the previous two terms. For example, 7 has the encodings (3, 4, 1) and (1, 3, 2).
In an attempt to stump the postal worker, the customer prefers encodings with a maximal value of q. What encoding should she use for the number 81?
Extra credit: What encoding should she use for the number 179?
The number in the sequence prior to 81 should be $\dfrac{81}{\phi} \approx 50$ to make the sequence be as long as possible.
This yields 2, 5, 7, 12, 19, 31, 50, 81.
Note that using 49 yields 2, 15, 17, 32, 49, 81.
Note that using 51 yields 9, 21, 30, 51, 81.
An additional 3 can be appended to the beginning. The answer is $(3, 2, 7)$.
Similarly, the number prior to 179 should be $\dfrac{179}{\phi} \approx 111$ to make the sequence be as long as possible.
This yields 11, 7, 18, 25, 43, 68, 111, 179. The answer is $(11, 7, 6)$.