Non-trivial zero(s) of Akiyama-Tanigawa triangle

Introduced in 1997, the Akiyama-Tanigawa triangle is a doubly-indexed recursion that encodes the Bernoulli numbers, among other sequences. It is defined as follows: let $a:\mathbb{N^0}\times\mathbb{N^+}\to \mathbb{R}$ (this indexing is to agree with that of the Bernoulli numbers) be given by: $$ a_{0,j}=\frac{1}{j}\qquad a_{i,j} = j(a_{i-1,j}-a_{i-1,j+1}) $$Here is a table for $1\le i+1,j\le10$: $$ \begin{array}{c|cccccccccc} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline 0 & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} & \frac{1}{10} \\ 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} & \frac{1}{10} & \frac{1}{11} \\ 2 & \frac{1}{6} & \frac{1}{6} & \frac{3}{20} & \frac{2}{15} & \frac{5}{42} & \frac{3}{28} & \frac{7}{72} & \frac{4}{45} & \frac{9}{110} & \frac{5}{66} \\ 3 & 0 & \frac{1}{30} & \frac{1}{20} & \frac{2}{35} & \frac{5}{84} & \frac{5}{84} & \frac{7}{120} & \frac{28}{495} & \frac{3}{55} & \frac{15}{286} \\ 4 & -\frac{1}{30} & -\frac{1}{30} & -\frac{3}{140} & -\frac{1}{105} & \color{red}{0} & \frac{1}{140} & \frac{49}{3960} & \frac{8}{495} & \frac{27}{1430} & \frac{125}{6006} \\ 5 & 0 & -\frac{1}{42} & -\frac{1}{28} & -\frac{4}{105} & -\frac{1}{28} & -\frac{29}{924} & -\frac{7}{264} & -\frac{28}{1287} & -\frac{87}{5005} & -\frac{27}{2002} \\ 6 & \frac{1}{42} & \frac{1}{42} & \frac{1}{140} & -\frac{1}{105} & -\frac{5}{231} & -\frac{9}{308} & -\frac{343}{10296} & -\frac{1576}{45045} & -\frac{27}{770} & -\frac{205}{6006} \\ 7 & 0 & \frac{1}{30} & \frac{1}{20} & \frac{8}{165} & \frac{5}{132} & \frac{295}{12012} & \frac{67}{5720} & \frac{4}{6435} & -\frac{6}{715} & -\frac{75}{4862} \\ 8 & -\frac{1}{30} & -\frac{1}{30} & \frac{1}{220} & \frac{7}{165} & \frac{200}{3003} & \frac{1543}{20020} & \frac{3997}{51480} & \frac{464}{6435} & \frac{1539}{24310} & \frac{775}{14586} \\ 9 & 0 & -\frac{5}{66} & -\frac{5}{44} & -\frac{44}{455} & -\frac{629}{12012} & -\frac{41}{12012} & \frac{133}{3432} & \frac{140}{1989} & \frac{1113}{12155} & \frac{9597}{92378} \\ \end{array} $$As claimed (I could provide a proof, if there's interest, but it's not relevant to my question), $a_{i,1}=B_i$, the $i^{th}$ Bernoulli number, with $B_1=1/2$. As such, $a_{2i+1,1}=0$ for $i>1$; call these trivial zeros. Note $a_{4,5}=0$ as well: my conjecture is that this is the only such non-trivial zero.

I have verified this claim for $a_{i,j}$ for $1\le i+1,j\le 100$. I also believe that for fixed $j$, $a_{n,j}>0$ for some $n=n(j)$. Additionally, for several fixed $j$ there is a closed-form by using the recursion. However, I'm not sure which of any of these lines is a feasible plan-of-attack, or if miraculously a closed-form for $a_{i,j}$ exists.

Here I obtain a closed form of the recurrence relation with a view to later proving your conjecture, if possible. Using Mathematica's FindSequenceFunction I inferred the possible ansatz $$a_{i,j}=\frac{p_i(j)}{(j)_{i+1}},$$ where $(x)_n$ is the Pochhammer symbol and $p_i(j)$ is some polynomial in $j$.

Now using

a[0, j_] := 1/j
a[i_, j_] := j (a[i - 1, j] - a[i - 1, j + 1])
i := 6
FindSequenceFunction[Table[a[i, j] Pochhammer[j, i + 1], {j, 1, 10}],
  j]

I was able to find the following numerator polynomials $p_i(j)$:

$$\begin{array}{l} p_0(j)=1\\ p_1(j)=j\\ p_2(j)=j^2\\ p_3(j)=(-1 + j) j^2\\ p_4(j)=-5 j^3 + j^4\\ p_5(j)=(-1 + j) (-4 j^2 - 15 j^3 + j^4)\\ p_6(j)=42 j^3 + 119 j^4 - 42 j^5 + j^6\\ p_7(j)=(-1 + j) (120 j^2 + 518 j^3 + 659 j^4 - 98 j^5 + j^6)\\ p_8(j)=-2160 j^3 - 7250 j^4 - 6189 j^5 + 3721 j^6 - 219 j^7 + j^8\\ p_9(j)=(-1 + j) (-12096 j^2 - 57720 j^3 - 98906 j^4 - 57735 j^5 + 15241 j^6 - 465 j^7 + j^8)\\ \end{array} $$

My first port of call was the OEIS but the coefficients of each polynomial do not appear to be listed.

Let's try working out generating functions for each row $(a_{i,j})_{j=1}^\infty$ instead:

$$P_i(x):=\sum_{j=1}^\infty\frac{p_i(j)}{(j)_{i+1}}x^j.$$

Mathematica tells me the following:

$$\begin{array}{l} P_0(x)=-\log (1-x)\\ P_1(x)=-\frac{x+\log (1-x)}{x}\\ P_2(x)=\frac{(x-2) \log (1-x)-2 x}{x^2}\\ P_3(x)=\frac{3 (x-2) x-\left(x^2-6 x+6\right) \log (1-x)}{x^3}\\ P_4(x)=\frac{\left(x^3-14 x^2+36 x-24\right) \log (1-x)-4 x \left(x^2-6 x+6\right)}{x^4}\\ P_5(x)=\frac{5 x \left(x^3-14 x^2+36 x-24\right)-\left(x^4-30 x^3+150 x^2-240 x+120\right) \log (1-x)}{x^5}\\ P_6(x)=\frac{\left(x^5-62 x^4+540 x^3-1560 x^2+1800 x-720\right) \log (1-x)-6 x \left(x^4-30 x^3+150 x^2-240 x+120\right)}{x^6}\\ P_7(x)=\frac{7 x \left(x^5-62 x^4+540 x^3-1560 x^2+1800 x-720\right)-\left(x^6-126 x^5+1806 x^4-8400 x^3+16800 x^2-15120 x+5040\right) \log (1-x)}{x^7}\\ P_8(x)=\frac{\left(x^7-254 x^6+5796 x^5-40824 x^4+126000 x^3-191520 x^2+141120 x-40320\right) \log (1-x)-8 x \left(x^6-126 x^5+1806 x^4-8400 x^3+16800 x^2-15120 x+5040\right)}{x^8}\\ P_9(x)=\frac{9 x \left(x^7-254 x^6+5796 x^5-40824 x^4+126000 x^3-191520 x^2+141120 x-40320\right)-\left(x^8-510 x^7+18150 x^6-186480 x^5+834120 x^4-1905120 x^3+2328480 x^2-1451520 x+362880\right) \log (1-x)}{x^9}\\ \end{array}$$ The coefficients of the polynomials appear to be related to the sequence A090582 and on further investigation we find that $$P_i(x)=(-1)^{i+1}\frac{ixu_{i-1}(x)-u_i(x)\log(1-x)}{x^i},$$ where $$u_i(x)=\sum _{k=1}^i (-1)^{k+1} k! \mathcal{S}_i^{(k)} x^{i-k},$$ where $\mathcal{S}_i^{(k)}$ is the Stirling number of the second kind.

Now we'd like to find out when the coefficient of $x^j$ in $P_i(x)$ is zero, where $j>1$. The case $j=1$ relates to the Bernoulli numbers in the first column and their trivial zeros, defined by the OP.

We can now use Cauchy's integral formula,

$$P_i^{(j)}(0)=\frac{j!}{2\pi\sqrt{-1}}\oint\frac{P_i(z)}{z^{j+1}}dz,$$

to compute the entries in your original matrix (note I used $\sqrt{-1}$ to avoid confusion with the index $i$), e.g. $a_{4,5}=0$ corresponds to $$\oint\frac{P_4(z)}{z^6}dz=\oint\frac{\left(z^3-14 z^2+36 z-24\right) \log (1-z)-4 z \left(z^2-6 z+6\right)}{z^{10}}=0.$$ Expanding the general integrand we have $$\frac{(-1)^{i+1}iu_{i-1}(z)}{z^{j+i}}+(-1)^{i+1}u_i(z)\sum_{k=1}^\infty\frac{z^{k-(j+i+1)}}{k}.$$ Substituting $u_i(x)$ we have $$(-1)^{i+1}i\sum _{k=1}^{i-1} (-1)^{k+1} k! \mathcal{S}_{i-1}^{(k)} \frac{1}{z^{j+k+1}}+(-1)^{i+1}\sum _{\ell=1}^i (-1)^{\ell+1} \ell! \mathcal{S}_i^{(\ell)}\sum_{k=1}^\infty\frac{1}{k}\frac{1}{z^{j+1-k+\ell}}.$$ Since the first sum has no $1/z$ term that just leaves the double sum.

Now $i$ and $j$ are effectively fixed (they correspond to the row and column in your original matrix) and since $1\leq\ell\leq i$ then the coefficients of $1/z$ occur when $j+1-k+\ell=1\implies k=j+\ell$, hence the sum of coefficients gives you the closed form you were wondering about,

$$a_{i,j}=(-1)^i\sum _{\ell=1}^i (-1)^{\ell} \ell! \mathcal{S}_i^{(\ell)}\frac{1}{j+\ell},$$ which again passes the $a_{4,5}=0$ test. Now we just need to prove $a_{i,j}=0$ if and only if $i=4$ and $1<j=5$, but I've not been able to do this yet...

It's worth taking a look at the list of polynomials $p_i(j)$. Notice how every odd polynomial from $p_3(j)$ onward has a factor of $(j-1)$ which corresponds to $B_n=0$ (odd Bernoulli numbers) when you plug in $j=1$ !

You can also see why $a_{4,5}=0$ since $-5j^3+j^4$ has a root at $j=5$.

If your conjecture is true then these polynomials must be irreducible over $\mathbb{Z}$, excluding the $(j-1)$ factor of the odd polynomials. Factorising the polynomials, $$\begin{array}{l} p_0(j)=1\\ p_1(j)=j\\ p_2(j)=j^2\\ p_3(j)=(-1 + j) j^2\\ p_4(j)=j^3(-5 + j)\\ p_5(j)=(-1 + j) j^2(-4 - 15 j + j^2)\\ p_6(j)=j^3(42 + 119 j - 42 j^2 + j^3)\\ p_7(j)=(-1 + j) j^2(120 + 518 j + 659 j^2 - 98 j^3 + j^4)\\ p_8(j)=j^3(-2160 - 7250 j - 6189 j^2 + 3721 j^3 - 219 j^4 + j^5)\\ p_9(j)=(-1 + j) j^2(-12096 - 57720 j - 98906 j^2 - 57735 j^3 + 15241 j^4 - 465 j^5 + j^6)\\ \end{array} $$

Proving $p_0,p_1,p_2,p_3\neq 0$ is easy for $j>1$. It's also easy to see that $p_4=0$ if and only if $1<j=5$.

Let's try to use the Rational Root Theorem (RRT) to see if each polynomial is irreducible. First of all notice that the coefficient of the highest $j$ power in each un-factored term is $1$ so any rational root $r=n/d$ with $\gcd(n,d)=1$ must have be such that $d\mid 1$, so the denominator can only every be $1$ which implies $r\in\mathbb{Z}$. RRT also tells us that $n\mid c_0$ where $c_0$ is the constant term of each un-factored polynomial.

So for $p_5$ the candidate roots are $r\in\pm\{1,2,4\}$ and since $p_5(r)\neq 0$ then $p_5$ is irreducible, and you will never find a zero.

For $p_6$ the candidate roots are $r\in\pm\{1,2,3,7,42\}$ and since $p_6(r)\neq 0$ then $p_6$ is irreducible, and you will never find a zero.

For $p_7$ the candidate roots are $r\in\pm\{1,2,4,8,3,5,120\}$. Again no zeros.

For $p_8$ the candidate roots are $r\in\pm\{1,2,4,8,16,3,9,27,5,2160\}$. Again no zeros.

For $p_9$ the candidate roots are $r\in\pm\{1,2,4,8,16,32,64,3,9,27,7,12096\}$. Again no zeros.

So in general after fully factoring the polynomial as far as possible, you may be able to find a proof based on an analysis of the constant term and whether or not it's $\pm$ divisors are roots of the polynomial factor.

Another perspective is this: the un-factored polynomial terms of each $p_i(j)$ are $O(j^{i-3})$ so there always exists an $N(j)$ such that $a_{i,j}\neq 0$ for all $j>N(j)$, so that covers an infinite number of cases for each row $i$.