Sequence

Warning: Don't try unless you're bored.

"The problem is:

What number is missing?
1 -> 5 -> 28 -> 260 -> ??


Hint: The base of the sequence is a simple but famous number sequence related to the golden ratio."

Oh yeah, I've not figured it out yet.

Warning: Don't try unless you're bored.

"The problem is:

What number is missing?
1 -> 5 -> 28 -> 260 -> ??


Hint: The base of the sequence is a simple but famous number sequence related to the golden ratio."

Oh yeah, I've not figured it out yet.

Well, that's the Fibonacci sequence. Now to figure out what it means by "the base."

1 1 2 3 5 8 13 21 34 55 89 144 i assume is the sequence they are referring to...
Alternate seeds
1 3 4 7 11 18 29 47 76 123
1 4 5 9 14 23 37 60 97 157
1 5 6 11 17 28 45 73 118 191
Moving to prev. 3
1 1 1 3 5 9 17 31 57 105
1 1 2 4 7 13 24 44 81 149
1 2 2 5 9 16 30 55 101


factors..
1
5
-1 = 4: 2.2
2.2.7 | 4.7 | 2.14 | 28
-5 = 23 (prime)
2.2.5.13 | 4.5.13 | 2.10.13 | 2.2.65 | 13.20 | 10.26 | 5.52 | 4.65 | 2.130 | 260
- 28 = 232: 2.2.2.29

If we assume the first 2 are "seeding" like in Fibonacci, and thus not any particular pattern until the 3rd element, this is not much info to go on.

I dunno :x

I thought about tribonacci as well.

say, we got the fib:
1, 1, 2, 3, 5, etc.

1 = 1
1 + 2^2 = 5
1 + 2^2 + 3^2 = 14 (which is half of 28.)
or
1 + 3^3 = 28, though doesn't make sense or a pattern, really. Besides, 4^5 or 5^4 is way higher than 260.

now, trib:
1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193


57 + 193 = 260, though, that may just be a coincidence. Most likely.

1^1 = 1
0^0 + 2^2 = 5
1^1 + 3^3 = 28
2^2 + 4^4 = 4+256 = 260
3^3 + 5^5 = 27+3125 = 3152

The base of the sequence? Made me think of fibonacci-base numbers (where each digit place represents a number in the fib sequence), but those consist of 0's and 1's, with an added restriction.

Above: what does that have to do with the golden ratio, though?
Hm..

Above: what does that have to do with the golden ratio, though?
Hm..

Well, it partly works.
Fib(n) may be defined as fib (n - 1) + fib (n - 2). However, for any n which is not a positive integer, it is defined as 0, and for 1, it is defined as 1.

for f(1) = 1, then f(n) = (n-2)^(n-2) + n^n