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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.

bio
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1 Answers1

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This proof looks fine to me.

added a coefficient to x for reasons I don't understand

$x$ is just the name he gives to the unknown first seed. Then he repeatedly applies the formula $$u_{n+3} = u_{n} + u_{n+1} + u_{n+2}$$ to find the $n$th element in terms of $x$.

understanding what the variable n is meant to be

$$u_{n+3} = u_{n} + u_{n+1} + u_{n+2}$$ This defines the Tribonacci sequence, it is supposed to hold for all integer $n$. A more complete expression would be $$\forall n \in \mathbb{N} : u_{n+3} = u_{n} + u_{n+1} + u_{n+2}$$

Wouter
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