Question

Tribonacci sequence question?

If a tribonacci sequence has 20 as its second seed and 17 as its third sedd, find all positive integers that can be its first seed so that 2017 appears as a term somewhere along the sequence.

Answer 1

`1980`, `1943` and `93`

Explanation:

Let us see what the sequence looks like:

If the first term is `a` then we get the sequence:

`a, 20, 17, a+37, a+74, 2a+128, 4a+239, 7a+441, 13a+808, 24a+1488, color(red)(cancel(color(black)(44a+2737)))`

So for `a` to be a positive integer, we require one of the following:

`a+37 = 2017" "rarr" "a=1980`

`a+74 = 2017" "rarr" "a=1943`

`color(red)(cancel(color(black)(2a+128 = 2017)))`

`color(red)(cancel(color(black)(4a+239 = 2017)))`

`color(red)(cancel(color(black)(7a+441 = 2017)))`

`13a+808 = 2017" "rarr" "a=93`

`color(red)(cancel(color(black)(24a+1488 = 2017)))`

The cancelled attempts above result in fractional values of `a`.

The only positive integer solutions are `1980`, `1943` and `93`