Let F=X^2-PX+Q be the characteristic polynomial of a linear sequence (T_n). For any r≥ 0, define polynomials f_r,j(P,Q), for j=0,…,2r+2, by F^r+1=∑_j=0^2r+2f_r,j(P,Q)X^j. Recurrence conjecture. The sequence (tau_n,r) satisfies the linear recurrence of order 2r+2 ∑_j=0^2r+2f_r,j(P,Q)tau_n+j,r=0. The conjecture is motivated by the explicitly verified recurrences for (tau_n,1) and (tau_n,2). It proposes a uniform recurrence for every r≥ 0, but the supplied text does not indicate whether the claim has been proved or remains open.

Recurrence conjecture for the Maclaurin-series coefficients of Somos-related sequences

Let $F=X^2-PX+Q$ be the characteristic polynomial of a linear sequence $(T_n)$ . For any $r\geq 0$ , define polynomials $f_{r,j}(P,Q)$ , for $j=0,\ldots,2r+2$ , by

F^{r+1}=\sum_{j=0}^{2r+2}f_{r,j}(P,Q)X^j.

Recurrence conjecture. The sequence $(\tau_{n,r})$ satisfies the linear recurrence of order $2r+2$

\sum_{j=0}^{2r+2}f_{r,j}(P,Q)\tau_{n+j,r}=0.

The conjecture is motivated by the explicitly verified recurrences for $(\tau_{n,1})$ and $(\tau_{n,2})$ . It proposes a uniform recurrence for every $r\geq 0$ , but the supplied text does not indicate whether the claim has been proved or remains open.

Progress summary

Open

No public discussion or published progress on this conjecture was found.

No public discussion or published progress addressing this conjecture was found.

Current status (as of August 2026): The conjecture appears open, with no recorded public activity establishing or refuting it.

Sources & referencesView supporting material

Primary source

Andrei K. Svinin, “Somos-4 equation and related equations”, arXiv:2307.05866 (2023).

Additional references

4 papers in this index state this conjecture (2010–2023). The statement above is taken from the most recent of them; the others are arXiv:2210.10577, arXiv:1910.09576, arXiv:1002.3844.

1 week ago

Let E denote the forward shift and put F(E)=E²−PE+Q, with Q≠0 as in the source. The already proved Theorem 7.2 gives τ_n(x)=(T_n−T_{n−1}B(x))e^{nx}, where F(E)T=0, B(0)=0, and B is independent of n. Write B(x)=Σ_{j≥1}b_jx^j/j!, τ_n(x)=Σ_{r≥0}τ_{n,r}x^r/r!. Coefficient extraction yields, for every r≥0, τ_{n,r}=n^rT_n−T_{n−1}Σ_{j=1}^r C(r,j)b_j n^{r−j}.

Both U_n=T_n and V_n=T_{n−1} satisfy F(E)U=F(E)V=0. More generally, if F(E)U=0 and p(n) is a polynomial of degree at most d, write (f₀,f₁,f₂)=(Q,−P,1). For every shift h≥0, F(E)(p(n)E^hU) =Σ_{s=0}² f_s[p(n+s)−p(n)]E^{h+s}U. Indeed the omitted term p(n)E^hF(E)U vanishes. Every polynomial difference on the right has degree at most d−1. Therefore F(E) lowers by at least one the polynomial degree in the span of polynomial multiples of shifts of U. Consequently F(E)^{d+1}(p(n)U_n)=0. The same argument applies to V.

Since the displayed formula for τ_{n,r} is a sum of polynomial multiples of U_n and V_n with degrees at most r, it follows that F(E)^{r+1}τ_{n,r}=0 for every r≥0. Writing F(X)^{r+1}=Σ_{j=0}^{2r+2}f_{r,j}(P,Q)X^j gives exactly Σ_{j=0}^{2r+2} f_{r,j}(P,Q)τ_{n+j,r}=0. Thus the conjectured recurrence holds in every order, including repeated characteristic roots. In fact the argument works for any formal series B(x), without requiring the source's additional Riccati equation.

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Shivam Patel · 2 days ago