If a tribonacci sequence has 20 as its second seed and 17 as its third seed, find all positive integers that can be its first seed so that 2017 appears as a term somewhere in the sequence. Note: Tribonicca is the sequence of number the fourth onwards is the sum of the previous three terms. For example: the first 3 numbers called seeds are 1, 2, 3. The next number is 6 then next one is 11 and so on.
Let (ₙ) be the tribonacci sequence defined by:
{ ₁ =
{ ₂ = 20
{ ₃ = 17
{ ₙ₊₃ = ₙ + ₙ₊₁ + ₙ₊₂ where is an unknown positive integer.Thus: ₄ = + 37 → If it is equal to 2017, then =1980 ◄◄◄ VALID
₅ = + 74 → If it is equal to 2017, then =1943 ◄◄◄ VALID
₆ = 2 + 128 → If it is equal to 2017, then =944.5 ◄◄◄ INVALID (not integral)
₇ = 4 + 239 → If it is equal to 2017, then =444.5 ◄◄◄ INVALID
(not integral)₈ = 7 + 441 → If it is equal to 2017, then ≈225.143 ◄◄◄ INVALID
(not integral)₉ = 13 + 808 → If it is equal to 2017, then =93 ◄◄◄ VALID
₁₀ = 24 + 1488 → If it is equal to 2017, then ≈22.0417 ◄◄◄ INVALID (not integral)
₁₁ = 44 + 2737 → If it is equal to 2017, then ≈-16.364 ◄◄◄ INVALID (negative)
... and any higher term will obviously yield a negative .
So if: = 93 → ₉ = 2017 = 1943 → ₅ = 2017 = 1980 → ₄ = 2017 and obviously, if: = 2017 → ₁ = 2017
Thus all the positive integral values of ₁ so that the value 2017 appears in the sequence are: ₁ ∈ { 93; 1943; 1980; 2017 }
I was searching up the same question, and I came up across this answer. Can somebody explain it to me, as I'm having a bit of trouble understanding it. Also, is it a correct proof? Yahoo has lots of wrong answers, according to my experience. I'm particularly worried about the second step, where the person has added a coefficient to $x$ for reasons I don't understand. Also, I'm having a bit of trouble with understanding what the variable $n$ is meant to be.