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.

1 Answer
May 6, 2017

#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#