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.

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    $\begingroup$ I have displayed the first few entries as a square array, though the recursion is clearer if written as a left-justified triangle. Those familiar with Pascal's matrix will not object to this convention, I think. $\endgroup$
    – Integrand
    Sep 13, 2020 at 13:51
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    $\begingroup$ For those interested the table read by antidiagonals is given by A051714(n)/A051715(n), two OEIS sequences. It is the Akiyama-Tanigawa algorithm for Bernoulli numbers. $\endgroup$
    – Somos
    Sep 13, 2020 at 14:33
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    $\begingroup$ I have a non-recursive form for each column in terms of Stirlingnumbers 1st kind and consecutive Bernoulli-numbers; this does so far nothing for a proof of existence/nonexistence of "nontrivial zeros". If this is interesting anyway I can compose a short answer of it. $\endgroup$ Feb 27, 2021 at 10:38
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    $\begingroup$ Ah, I see - this is even known to wikipedia... (en.wikipedia.org/wiki/…) $\endgroup$ Feb 27, 2021 at 12:00
  • $\begingroup$ Wow, encoding the Bernoulli numbers. $\endgroup$
    – user822157
    Feb 28, 2021 at 0:16

1 Answer 1


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}],

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:


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 $$u_i(x)=\sum _{k=1}^i (-1)^{k+1} k! S_i^{(k)} x^{i-k},$$ where $S_i^{(k)}$ 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,


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 $$(-1)^{i+1}i\sum _{k=1}^{i-1} (-1)^{k+1} k! S_{i-1}^{(k)} \frac{1}{z^{j+k+1}}+(-1)^{i+1}\sum _{\ell=1}^i (-1)^{\ell+1} \ell! S_i^{(\ell)}\sum_{k=1}^\infty\frac{1}{k}\frac{1}{z^{j+1-k+\ell}}.$$ 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,

$$a_{i,j}=(-1)^i\sum _{\ell=1}^i (-1)^{\ell} \ell! S_i^{(\ell)}\frac{1}{j+\ell},$$ 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 $i$.

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    $\begingroup$ Impressive @Pixel. $\endgroup$ Feb 26, 2021 at 22:29
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    $\begingroup$ I have never seen this use of the Cauchy's Formula; that's very tricky! I quote rural reader: impressive. $\endgroup$ Feb 26, 2021 at 22:40
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    $\begingroup$ Thanks for the detailed, well-reasoned answer. I recognize many of the polynomials and the closed-form, having played around with it myself for a while. I think this answer is worthy of the bounty but of course I will wait for a few more days. $\endgroup$
    – Integrand
    Feb 28, 2021 at 16:09

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