Heronian triangles and Somos-5 sequences? (Part 1)

Question

While solving this post, I came across the Somos-5 sequence again (since it appeared in another post). Given the Heronian triangle with half-sides $(a,b,c)$, area $A$, and medians $(m_1,m_2,m_3)$,

$$A=\sqrt{(a + b + c) (b + c - a) (a + c - b) (a + b - c)}$$ $$m_1=\sqrt{2 (b^2 + c^2) - a^2}$$ $$m_2=\sqrt{2 (c^2 + a^2) - b^2}$$ $$m_3=\sqrt{2 (a^2 + b^2) - c^2}$$

Let $(a,b,c)$ be integers. By solving an elliptic curve, it is possible to make all $m_k$ as integers. However, if $A$ is to be a non-zero integer as well, that is still an open problem.

If we start with the area $A$, Schubert conjectured in 1905 that solving the first three equations in non-zero integers was impossible, but was proven wrong by Buchholz's 1989 thesis.

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I. Three quadratics

We do the cyclic substitution $(a,b,c) = (p+q,\,q-r,\,r-p)$ to simplify the equations,

\begin{align} |16\, p\, q\, r\, (-p - q + r)| &=\, A^2\\[5pt] (p - q)^2 - 4 (p + q) r + 4 r^2 &= m_1^2\\[5pt] (2 p + q)^2 - 2 (2 p - q) r + r^2 &= m_2^2\\[5pt] (p + 2 q)^2 + 2 (p - 2 q) r + r^2 &= m_3^2 \end{align}

So four quadratics in $r$ to be made squares. For this system, three at a time is infinitely solvable.

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II. Somos-5 Sequences

Define the two Somos-5 sequences (starting with $n=0$),

$S_n = 1,\, 1,\, 1,\, 2,\, 3,\, 5,\, 11,\, 37,\, 83,\, 274,\, 1217,\; 6161,\; 22833,\dots$ (A006721)

$T_n = 0, 1, -1, 1, 1, -7, 8, -1, -57, 391, -455, -2729, 22352,\dots$ (A360381)

Then in page 6 of Andrew Hone's 2022 paper, we find the formulas (modified),

$$A = |4\alpha\beta|$$

$$\alpha = S_{n}\,S_{n+1}\,S_{n+2}^2\, S_{n+3}\, S_{n+4}$$ $$\beta = T_{n}\,T_{n+1}\,T_{n+2}^2\, T_{n+3}\, T_{n+4}$$

and the $(p,q,r)$ as,

$$p_n = S_{n+1}\,S_{n+2}^3\, S_{n+3}\, T_{n+2}$$

$$q_n = \;S_{n}^2\, S_{n+1}\, T_{n+3}\, T_{n+4}^2$$

$$r_n = T_{n+1}\, T_{n+2}^3 \,T_{n+3}\, S_{n+2}$$

$$s_n = -\,T_{n}^2\, T_{n+1}\, S_{n+3}\, S_{n+4}^2$$

with the formula for $s_n$ added by yours truly. So these $(A,p,q,r,s)$ are highly factorable, being products of 12 and 6 terms, respectively (counting multiplicity). Specifically,

$$p_n = -2,\, 24,\, 270, -28875, 1969880, -46246189,\dots$$

$$q_n\; =\; 1,\; 49,\; -896,\; 96,\; -146205, -2396409675,\dots$$

$$r_n\; = -1,\, -2,\; -21, -13720,\; 39424,\; 16872,\dots\;\quad$$

$$\quad s_n \,=\; 0,\; 75, -605, -15059, 1784251, -2442672736,\dots$$

Example:

$(p_1,q_1,r_1) = (24,\,49,-2)$ solves the first three quadratics above while $s_n \to r_n$ $(p_1,q_1,s_1) = (24,\,49,\,75)\,$ solves the first two and the fourth. And so on for all $n$.

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III. Question

Q: What are the recurrence relations of the $(p,q,r,s)$? Are they also Somos-5 sequences but with different initial values?

P.S. Part 2 in in this post.

Answer 1

The context of this question is about generalized Somos sequences. They are usually defined by initial values and a single homogeneous bilinear recurrence somewhat similar to the definition of generalized Fibonacci sequences which also satisfy infinite families of linear recurrences. Similarly, generalized Somos sequences are best characterized by satisfying infinite families of homogeneous bilinear recurrences.

For example, the Somos-5 sequence $S_n$ and its companion sequence $T_n$ are defined by the same recurrence

$$ S_{n+3}S_{n-2} = S_{n+2}S_{n-1} + S_{n+1}S_n,\\ T_{n+3}T_{n-2} = T_{n+2}T_{n-1} + T_{n+1}T_n \tag1 $$

with initial values $\,S_0 = S_1 = S_2 = 1$ and $\,T_0 = 0,\, T_1 = T_3 = T_4 = 1,\, T_2 = -1.$ Note that $\,S_{-n} = S_n,\, T_{-n} = -T_n$ for all integer $n$.

These recurrences are special cases of the infinite family of recurrences

$$ S_{n+m+1}S_{n-m}=-T_{m+1}T_m S_{n+2}S_{n-1}+T_{m+2}T_{m-1}S_{n+1}S_n,\\ T_{n+m+1}T_{n-m}=-T_{m+1}T_mT_{n+2}T_{n-1}+T_{m+2}T_{m-1}T_{n+1}T_n\tag2$$

valid for all integers $n$ and $m$. Notice that the indices $\,(n+m+1,n-m)$ have opposite even/odd parity. Similarly, the Somos-4 sequence $U_n$ and its companion sequence $V_n$ are defined by the same recurrence

$$ U_{n+2}U_{n-2} = U_{n+1}U_{n-1} + U_{n}U_n,\\ V_{n+2}V_{n-2} = V_{n+1}V_{n-1} + V_{n}V_n \tag3 $$

with inital values $\,U_0 = U_1 = U_2 = U_3 = 1$ and $\,V_0 = 0,\,V_1 = V_2 = 1,\, V_3 = -1,\, V_4 = -5.$ Note that $\,U_{-n} = U_{n+3},\, V_{-n} = -V_n$ for all integer $n$.

These recurrences are special cases of the infinite family of recurrences

$$ U_{n+m}U_{n-m}=V_{m}V_m U_{n+1}U_{n-1}-V_{m+1}V_{m-1}U_{n}U_n,\\ V_{n+m}V_{n-m}=V_{m}V_m V_{n+1}V_{n-1}-V_{m+1}V_{m-1}V_{n}V_n\tag4$$

valid for all integers $n$ and $m$. Notice that the indices $\,(n+m,n-m)$ have the same even/odd parity. Notice also the two sequences satisfy the infinite family of recurrences

$$ U_{n+m+1}U_{n-m}=V_{m+1}V_m U_{n+2}U_{n-1}-V_{m+2}V_{m-1}U_{n+1}U_n,\\ V_{n+m+1}V_{n-m}=V_{m+1}V_m V_{n+2}V_{n-1}-V_{m+2}V_{m-1}V_{n+1}V_n\tag5$$

valid for all integers $n$ and $m$. Notice the similarity of equation $(4)$ to equations $(2)$ except for sign changes. Is there an variation of equation $(4)$ for the Somos-5 sequences? Yes, it is

$$ S_{n+m}S_{n-m}=e_{n,m}T_{m}T_m S_{n+1}S_{n-1}- f_{n,m}T_{m+1}T_{m-1}S_{n}S_n,\\ T_{n+m}T_{n-m}=g_{n,m}T_{m}T_m T_{n+1}T_{n-1}- h_{n,m}T_{m+1}T_{m-1}T_{n}T_n\tag6$$

where $\,e,f,g,h$ depend only on the parity of both $n$ and $m$. There are other recurrences whose coefficients do not depend on the parity of $n$ and $m$ but they have more terms. The question is

Q: What are the recurrence relations of the $(p,q,r,s)$? Are they also Somos-5 sequences but with different initial values?

The recurrence relations for the $p$ and $r$ sequences are the same and similarly for the pair of sequences $q$ and $s$. For example

$$ 133p_{n+7}p_{n-6} = \sum_{k=1}^6 d_k p_{n+k}p_{n-k+1} \tag7 $$

where $\,d_6 = -2131349,\,\dots,\,d_1 = 605797403016587130$ and

$$ q_{n+8}q_{n-7} = \sum_{k=1}^7 e_k q_{n+k}q_{n-k+1} \tag8 $$

where $\,e_7 = -175111,\,\dots,\,e_1 = 1620919496354218862230$.

The recurrence relations for the $\,a,b,c$ sequences are the same. For example:

$$ a_{n+5}a_{n-4} = \sum_{k=1}^4 f_k a_{n+k}a_{n-k+1} \tag9 $$

where $\,f_4 = -37,\, f_3 = 165585,\, f_2 = -175047,\, f_1 = 2567356.$

Initial values of the sequences (starting with $n=-2$),

$$ a_n = 1,\, 0, -1, 73, -626, -28779, 1823675, -2442655864,\dots \\ b_n = 1, -1, 2, 51, -875, 13816, -185629, -2396426547, \dots \\ c_n = 0, -1, 1, -26, -291, 15155, -1930456, 46263061, \dots $$

where $\,a_{-2+n} = b_{-2-n}$ and $\,c_{-2+n} = -c_{-2-n}$ for all integer $n$.