A super-Fibonacci sequence is a list of whole numbers with the property that, from the third term onwards, every term is the sum of all the previous terms. For example,

Question

1, 4, 5, 10, ...
How many super-Fibonacci sequences with 1 involve the number 2016?

A.1
B.3
C.5
D.7
E.9

Answer 1

If I read this correctly, we would get:

1, 4, 5, 10, 20, 40, 80, ...
in effect, each term is twice its previous term from the 4th term on

Where does the number 2016 come in?
clearly not in this sequence.

suppose my term start with 1, 2 , then we get
1,2, 3, 6, 12, 24, , again we double each time
we will hit 2016

starting with 1, 3, we get
1,3,4,8,16, ... same doubling pattern

notice in all cases, the terms after the third are multiples of the third term.
So the third term establishes the multiples
2016 = 63(2^5)
= (9)(7)(2^5)

so as long as the third terms starts with
63, 63(2), 63(4), 63(8), 63(16) or 63(32)
or
63, 126, 252, 504, 1008, 2016 it will work

1, 62, 63, 126, 252, 504, 1008, 2016, ... YUP
or
1, 125, 126, 252, ..., 2016
1, 251, 252, ... , 2016
...
1, 2015, 2016

I count 6 of those.

But what about 1, 2016, 2017, .... ?
wouldn't that also involve 2016?

So I would go with 7

Very interesting question.

Answer 2

Thanks so much for giving a great explanation

Answer 3

1 or 0, 1 part = i
2016
1016
06
03
01
0(1)
Modulus division by prime numbers baby

Answer 4

2i dimensional analysis is what I'm trying to do, please tell me im wrong