next in series - Fibonacci-related function #3

The following function #3 is Fibonacci-related:

f(1) = 3 f(2) = 2 f(3) = 2 f(4) = 2 f(5) = 3 f(6) = 8 f(7) = 55 f(8) = 1597 f(9) = 514229 f(10) = 7778742049 f(11) = ?

Find an explicit function for f(n).

Rules for lawyers & spoilsports:

Rule 0 added: 0) Fib(1) = 1, Fib(2) = 1, For n > 2, Fib(n) = Fib(n - 2) - Fib(n - 1). IMPORTANT: For n < 1, Fib(n) = Fib(n + 2) - Fib(n + 1). (Thus Fib(0) = 0, Fib(-1) = 1, Fib(-2) = -1, etc.)

1) Occam's razor applies. 2) Use the pattern that is already in the terms / one number does not a pattern make! 2) Use Fib(whatever) for the Fibonacci function, and Fib(something not constant) must be part of f(n). 3) No Trunc / Floor / Ceiling / Round functions or the like. 4) No alternating functions (example: f(n) = -1^n * whatever), other than the Fibonacci function itself. 5) No mod functions (example: f(n) = n/3 if n = 0 mod 3 f(n) = 2n if n = 1 mod 3 f(n) = n if n = 2 mod 3). 6) Math functions only

f(1) = 3 f(2) = 2 f(3) = 2 f(4) = 2 f(5) = 3 f(6) = 8 f(7) = 55 f(8) = 1597 f(9) = 514229 f(10) = 7778742049 f(11) = ?

Find an explicit function for f(n).

This is f(n)=Fib(Fib(n)-n+4).

But how did you get that? <grin>

Since up to now nobody has solved my Fib-related function #1, I'll show one way you might have solved this one.

f(1) = Fib(4) = Fib(g(1)) f(2) = Fib(3) = Fib(g(2)) f(3) = Fib(3) = Fib(g(3)) f(4) = Fib(3) = Fib(g(4)) f(5) = Fib(4) = Fib(g(5)) f(6) = Fib(6) = Fib(g(6)) f(7) = Fib(10) = Fib(g(7)) f(8) = Fib(17) = Fib(g(8)) f(9) = Fib(29) = Fib(g(9)) f(10) = Fib(49) = Fib(g(10))

Then g(1) = 4 g(2) = 3 3 - 4 = -1 g(3) = 3 3 - 3 = 0 0 - -1 = 1 g(4) = 3 3 - 3 = 0 0 - 0 = 0 g(5) = 4 4 - 3 = 1 1 - 0 = 1 g(6) = 6 6 - 4 = 2 2 - 1 = 1 g(7) = 10 10 - 6 = 4 4 - 2 = 2 g(8) = 17 17 - 10 = 7 7 - 4 = 3 g(9) = 29 29 - 17 = 12 12 - 7 = 5 g(10) = 49 49 - 29 = 20 20 - 12 = 8

The 2nd differences are successive Fib numbers, starting with Fib(-1).

Without the Fib numbers in the 2nd differences, the 2nd differences would all be 0, and the 1st differences would have been constant.

Hence, g(n) = a*n + b + Fib(something).

For g(n) the 2nd diffs were Fib(n - 2). Going back a diff, you'd get Fib(n - 1). Going back a diff to g(n), you'll get Fib(n).

(I'll leave it to folks to confirm this fact of Fib polynomial functions for themselves.)

Hence, g(n) = a* n + b + Fib(n).

Now to solve for a and b: g(1) = 4 = a * 1 + b + 1, so a + b = 3. g(2) = 3 = a * 2 + b + 1, so 2 * a + b = 2.

a = -1 b = 4

Hence, g(n) = -n + 4 + Fib(n) and f(n) = Fib(g(n)) = Fib(-n + 4 + Fib(n)).

My Fib-related function #1 is similar to g(n)