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

= 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 and. 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, ... ]. 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 $$p\in\R,q_i\in\C,$$ 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 ατn/2cos(nβ), where α and β are determined by qi and ξi. (It's easy to see that |ξi| = |τ|-1/2, since τξ1ξ2=1. ln(ξ1/|ξ1|)=-ln(ξ2/|ξ2|)=iβ.) You can bound the ατn/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)

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)