Wikipedia:Reference desk/Archives/Mathematics/2023 July 17

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July 17

Zeros in the tribonacci sequence

The tribonacci sequence is defined by a(0) = a(1) = 0, a(2) = 1, a(n) = a(n-1) + a(n-2) + a(n-3). This defines a(n) not only for for n >= 0, but for n < 0 if you rearrange the recursion formula; see OEIS: A000073 and OEIS: A057597. We have a(0) = a(1) = 0, but also a(-3) = 0, and somewhat surprisingly a(-16) = 0. Are there any other values of n with a(n) = 0? Of course n would have to be negative. I strongly doubt there any other values, but I strongly suspect it would be very difficult to prove this. Not coincidentally, the tribonacci constant, τ=1.839286755214161132551852..., has an unexpectedly good rational approximation 103/56, with 56τ - 103 = -τ ^-16. This is reflected in the large entry in its continued fraction expansion [1, 1, 5, 4, 2, 305, ... ] (OEIS: A019712). I checked values up to a(-100000) but no other 0's appeared. --RDBury (talk) 21:50, 17 July 2023 (UTC)

$\tau$ is the sole real root of the polynomial $X^{3}-X^{2}-X-1,$ the other two being formed by a complex conjugate pair with the approximate values of $-0.41964\pm 0.60629i\,.$ Denoting these by $\xi _{1}$ and $\xi _{2},$ we have the identity $a(n)=p\tau ^{n}+q_{1}\xi _{1}^{n}+q_{2}\xi _{2}^{n}$ for some $p\in \mathbb {R} ,q_{i}\in \mathbb {C} ,$ in which the $q_{i}$ are also a conjugate pair. The values for $p$ and $q_{i}$ can be solved using three values for $n$ and the corresponding $a(n).$ Note that $|\xi _{i}^{{-}1}|=1.3...>1$ while $|\tau ^{{-}1}|=0.54...<1,$ so when going left on the integer line, the powers of the $\xi _{i}$ are going to dominate. I expect the explicit formula for $a(n)$ can be used to set a bound on $n$ below which no more zeros can appear.  --Lambiam 07:34, 18 July 2023 (UTC)

I see what you're saying, but ignoring the τ ^-n term gives an expression of the form ατ^n/2cos(nβ), where α and β are determined by q_i and ξ_i. (It's easy to see that |ξ_i| = |τ|^-1/2, since τξ₁ξ₂=1. ln(ξ₁/|ξ₁|)=-ln(ξ₂/|ξ₂|)=iβ.) You can bound the ατ^n/2 below, but the cos(nβ) part can and will get as close to 0 as you want. You could also argue informally that if these bounds were going to work at all then they would work for -16. --RDBury (talk) 14:18, 18 July 2023 (UTC)

Correction: I should have included a sine term as well as a cosine term. The full expression is:

$a(n)=\zeta \tau ^{n}+\xi ^{n}(\alpha _{1}\cos(n\beta )+\alpha _{2}\sin(n\beta )),$

where τ≈1.83928675521416, ζ≈0.182803532968296, ξ=τ^−1/2≈0.737352705760328, α₁≈-0.182803532968296, α₂≈0.681093061654159, β≈2.17623354549187. In any case, it still holds that the trigonometric part can be arbitrarily close to 0 based on β/π being irrational. (There are proofs and references for this in this Stack Exchange post. --RDBury (talk) 01:41, 19 July 2023 (UTC)

@Lambian: I hope you're not writing things like $1.3...>1$ instead of $1.3\ldots >1$ in Wikipedia articles. Michael Hardy (talk) 21:08, 21 July 2023 (UTC)