Somos 4 conjecture for a quadratic AA-matrix

About 7 years old · traced to

Let f(x)f(x) be determined by the quadratic equation

f(x)x=1+af(x)+bf(x)2+x(1+cf(x)+df(x)2),\frac{f(x)}{x}=1+af(x)+bf(x)^2+x\bigl(1+cf(x)+df(x)^2\bigr),

and let the Hankel transform be formed from the coefficient sequence of f(x)/xf(x)/x. A sequence is an (α,β)(\alpha,\beta) Somos 44 sequence when

snsn−4=αsn−1sn−3+βsn−22.s_n s_{n-4}=\alpha s_{n-1}s_{n-3}+\beta s_{n-2}^2.

Somos 4 conjecture. The Hankel transform of the expansion of f(x)/xf(x)/x is a

((b+ab+d)2, b4−b3(2+3a+a2−2c)+b(a+a2−ac−2d)d+(1+a−c)d2−b2(c+ac−c2+2d+3ad))\left((b+ab+d)^2,\ b^4-b^3(2+3a+a^2-2c)+b(a+a^2-ac-2d)d+(1+a-c)d^2-b^2(c+ac-c^2+2d+3ad)\right)

Somos 44 sequence.

This conjecture proposes a uniform Somos 44 recurrence for the family of quadratic AA-matrices considered in the paper; the supplied text does not state whether it has been proved or disproved.

References

Primary source

Paul Barry, “Riordan arrays, the A-matrix, and Somos 4 sequences”, arXiv:1912.01126 (2019).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.