On certain ratios regarding integer numbers which are both triangulars and squares

TL;DR

This paper studies the ratios between successive numbers that are both triangular and perfect squares, revealing new insights about the limits of their ratios and differences, and proposing related conjectures.

Contribution

It introduces a novel investigation into the limit of the ratio of differences of successive triangular-square numbers, showing it matches the known ratio limit.

Findings

The limit of the ratio of successive numbers is known.

The paper proves the limit of the ratio of differences coincides with the known ratio.

A conjecture about related limits is formulated.

Abstract

We investigate integer numbers which possess at the same time the properties to be triangulars and squares, that are, numbers $a$ for which do exist integers $m$ and $n$ such that $ a = n^2 = \frac{m \cdot (m+1)}{2} $. In particular, we are interested about ratios between successive numbers of that kind. While the limit of the ratio for increasing $a$ is already known in literature, to the best of our knowledge the limit of the ratio of differences of successive ratios, again for increasing $a$, is a new investigation. We give a result for the latter limit, showing that it coincides with the former one, and we formulate a conjecture about related limits.

Equations

$$\bullet$$

$$T_n:=\frac{n\cdot (n+1)}{2}=\left(\genfrac{}{}{0ex}{}{n+1}{2}\right)$$

$$\bullet$$

$$P_{m,n}=\frac{n[(m-2)n-(m-4)]}{2}$$

$$n^2=\frac{m\cdot (m+1)}{2}$$

$$n^2$$

$$2n^2-(m^2+m)$$

$$2n^2-\left(m^2+m+\frac{1}{4}\right)$$

$$2n^2-{\left(m+\frac{1}{2}\right)}^2$$

$$8n^2-(2m+1{)}^2$$

$$t^2-8n^2$$

$$a_n$$

$$48\text{mbox}024\text{mbox}900\cdot 33.97056279144=1\text{mbox}631\text{mbox}432\text{mbox}881.00263$$

$$a_n$$

$$t^2-2s^2=1$$

$$\sqrt{2}=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots }}}$$

$$i$$

$$(t_{i-1}+s_{i-1}\sqrt{2})(3+2\sqrt{2})$$

$$3t_{i-1}+2\sqrt{2}{t}_{i-1}+3\sqrt{2}{s}_{i-1}+4s_{i-1}$$

$$3t_{i-1}+4s_{i-1}+(2t_{i-1}+3s_{i-1}\sqrt{2})$$

$$\{\begin{array}{rlr}{t}_i& =& 3t_{i-1}+4s_{i-1}\\ s_i& =& 2t_{i-1}+3s_{i-1}\end{array}$$

$$\{\begin{array}{rl}{s}_i& =2t_{i-1}+3s_{i-1}\\ t_i& =\sqrt{2}{s}_i+l_i{s}_i\end{array}$$

$$s_i$$

$$s_i^2$$

$$4n_i^2$$

$$n_i^2$$

$$a_{2,j}=\frac{a_{1,j+1}}{a_{1,j}}-\frac{a_{1,j}}{a_{1,j-1}}$$

Full text

1 Preliminaries

We recall some basic definitions from elementary number theory.

Definition 1.1**.**

A non-negative integer is said to be triangular if it can be the number of objects in a set able to form a triangle, right or equilateral.

[TABLE]

A triangular number has the form

[TABLE]

where $n$ is a natural number.

Definition 1.2**.**

Similarly, a non-negative integer is said to be square if it can be the number of objects in a set able to form a square.

[TABLE]

It is straightforward to say that square numbers have the form $n\cdot n=n^{2}$, where $n$ is a natural number.

We can also define a generic polygonal number as an integer that can be the number of objects in a set able to form a regular polygon having a certain number of sides.

Definition 1.3**.**

A number is said to be $m$-gonal if it can be the number of objects in a set able to form a regular $m$-gon. The $n$-th $m$-gonal number has the form:

[TABLE]

2 Basic computations

With reference to square and triangular numbers’ definitions, by imposing equality, we obtain:

[TABLE]

that can be algebrically transformed to:

[TABLE]

by setting at the end $t:=2m+1$.

This allows us to say that $(t,n)$ should solve a Pell equation, assuming that $t$ is odd; we can also set $s:=2n$, in order to write $t^{2}-2s^{2}=1$, that is the more classical Pell equation, in which we need $s$ even.

3 A numerical approach

We treat here the problem empirically, by using a spreadsheet.

The idea is to create a table, where in the first column is listed a certain number of positive integers, in the second one the respective triangular number, in the third one its square root; in the fourth column we chop the square root at the lower integer, while in the fifth and last column, we do the difference between the third one and the fourth one, obtaining its decimal part. The first ten rows of the table give:

[TABLE]

and so on.

It is straightforward to say that the considered triangular number is also a squadre if and only if, for its row, the fifth column is 0. Exceptions can arise due to finite arithmetic errors, but it’s not the case at least for the moment, because all numbers appearing are not too large to generate machine-caused loss of precision.

By extending with the table until $65534$ (if we use a spreadsheet with $2^{16}-1=65535$ rows, using the first for naming columns), we can directly found some of these numbers:

[TABLE]

A property that can be observed is that the ratio between two successive numbers, both triangular and square, seems to be the same, case after case. By explicit computation:

[TABLE]

we can see how this ratio seems to rapidly converge to a fixed value. But we can say more of that, and this is why we kept 11 digits instead of 5 in the last step: also the ratio between differences of subsequent ratios converge to the same value. In fact:

[TABLE]

So, we can create new rows in the table, by following these steps:

• •

we divide the latest value $b_{n}$ obtained, i.e. $0.0000480584221$, for the analogous from the sequence of ratios which in the table lies on its right, so $33.97185$;

• •

we obtain $0.0000014146543$, and we subtract it from the latest value available in the second column, specifically $33.97056420609$;

• •

we obtain $33.97056279144$, and we multiply it for the latest number written, $48\mbox{ }024\mbox{ }900$; if what we conjectured is correct, we should obtain a number (real, not necessarily integer) well-approximating a new both triangular and square number.

In fact, we compute:

[TABLE]

and $1\mbox{ }631\mbox{ }432\mbox{ }881$ is both triangular and square. That allows us to add a line into the table:

[TABLE]

We used a backward completion: by observing that the value in the last column tends to stabilize, we estimate, by accepting an error margin, that it is constant starting from the considered row, and we complete the row by calculating all values in the previous columns.

By taking account of the fact that we know the exact value of the new number both triangular and square, we can rectify the table, moving from backward completion to forward completion: if we know the numbers having this property, we can derive ratios and differences.

[TABLE]

It seems we can approximate the limit of the left ratio, using 5 digits, to $33.97056$; if we try to take the same number as the right ratio, hoping to find new numbers, we can proceed:

• •

$0.0000014147063/33.97056=0.0000000416451$;

• •

$33.97056279139-0.0000000416451=33.97056274974$;

• •

$1\mbox{ }631\mbox{ }432\mbox{ }881\cdot 33.97056274974=55\mbox{ }420\mbox{ }693\mbox{ }055.99960$

Any CAS allows us to consider $55\mbox{ }420\mbox{ }693\mbox{ }056$ as:

• •

the $332928$-th triangular number: if $c$ is the number, the algebraic equation $n^{2}+n-2c=0$ has that value of $n$ as positive root;

• •

the $235416$-th square number, just by calculating its square root.

These data allows us, again, to update and rectify the table:

[TABLE]

[TABLE]

We have in some sense fastened the procedure: in fact, by following what we done before, we would have taken as value in the last column $33.97060$, i.e. the last value available, while we took instead $33.97056$, assuming that ratios in the fourth column converge at the same quantity ratios in the second column do.

We note that we obtain an almost exact value: a rectify in the first column doesn’t change anything in the others, with respect to the number of digits considered; we can also observe, as seen in the next table, that by using the same number of digits, we would obtain the same result by taking $33.97060$ as right ratio, while an increment in the number of digits would likely result in a difference, in which the lower precision lies in the choice of that value.

[TABLE]

By applying again the method, we obtained another couple of numbers: $1\mbox{ }882\mbox{ }672\mbox{ }131\mbox{ }025$ and $63\mbox{ }955\mbox{ }431\mbox{ }761\mbox{ }796$.

We can also note that:

• •

if we define $c_{n}:=\frac{b_{n-2}}{b_{n-1}}-\frac{b_{n-1}}{b_{n}}$, even $\frac{c_{n-1}}{c_{n}}$ tends to the same value; we can conjecture that it happens every time we iterate in this way, that is, if we denote $a_{1,n}:=a_{n},\mbox{ }a_{2,n}:=b_{n},\mbox{ }a_{3,n}:=c_{n}$, and we define for every $i\geq 3$ a corresponding $a_{i,n}:=\frac{a_{i-1,n-2}}{a_{i-1,n-1}}-\frac{a_{i-1,n-1}}{a_{i-1,n}}$, we can say that, again for every $i\geq 3$, while $n$ tends to infinity, $\frac{a_{i,n-1}}{a_{i,n}}$ tends to the value.

• •

if we use more digits for the ratios, and we assume correct the conjecture, we can consider one of the ratios, call $d$ the difference between a value and the previous one, $q$ the recurring value of about $33.97056$, and say that the subsequent difference will be approximable by $\frac{d}{q}$, the next one by $\frac{d}{q^{2}}$, and so on. The sum of the difference from there to infinity will be approximable by $\frac{d}{q}+\frac{d}{q^{2}}+\frac{d}{q^{3}}+\ldots=\frac{d}{q-1}=\frac{d}{32.97056}$, that allows us to obtain a gain in the relative precision of at least $32$ times every single step, and at least $1000$ times every two steps, that corresponds to three digits.

On the other hand, we need a certain machine precision: with $15$ digits, that corresponds to a relative precision of about $2^{-52}$, the standard of the double type, we report a loss of precision in the computation of the biggest number found before, a $14$-digit integer. If we multiply that number for $q$, we obtain a $16$-digit integer, and in general we can’t exactly write a $16$-digit integer as a 64-bit real value.

4 Exact approach with Pell equations

It is widely known from Pell equations’ theory that, for solving:

[TABLE]

we start by write $\sqrt{2}$ as a continuous fraction, that is:

[TABLE]

The first convergent is $\frac{3}{2}$, and $(t,s)=(3,2)$ does in fact solve the equation, i.e. $3^{2}-2\cdot 2^{2}=9-8=1$.

By the relation $s=2n$, we have $n=1$, and $n^{2}=1$, that is the first number both triangular and square.

Successive integers can be found in a traditional way, involving well-estabilished theory:

[TABLE]

and again:

[TABLE]

and so on; we can generalize:

[TABLE]

and, by recurrence:

[TABLE]

4.1 Ratio limit: first ratio

By observing that $n=\frac{s}{2}$ implies $n_{i}^{2}=\frac{s_{i}^{2}}{4}$, we can express $s_{i}$ as a function of $s_{i-1}$ and not of $t_{i-1}$.

If we define, for every $i$, $t_{i}=k_{i}s_{i}$, $\lim_{i\rightarrow+\infty}k_{i}=\sqrt{2}$ holds (it is straightforward to prove), and we can set $l_{i}:=k_{i}-\sqrt{2}$, so $t_{i}=\sqrt{2}s_{i}+l_{i}s_{i}$, and $\lim_{i\rightarrow+\infty}l_{i}=0$. Now, from equations:

[TABLE]

we obtain:

[TABLE]

from which:

[TABLE]

and, for $i\rightarrow\infty$:

[TABLE]

4.2 Ratio limit: second ratio, first method

We prove now in two ways that, if we define:

[TABLE]

then also the ratio $a_{2,j-1}/a_{2,j}$ tends at the same value for diverging $j$.

Here is the first one.

We will write alternatively $a_{1,j}$ or $a_{j}$ for the $j$-th term of the OEIS sequence A001110 (see also [1, 2, 3, 4, 5, 6, 7, 8, 9] and some references therein).

We have:

[TABLE]

Since $a_{j}=\frac{s_{j}^{2}}{4}$, where $s_{j}$ is the $j$-th value of $s$ which is solution, for a certain value of $t$ (namely $t_{j}$), of $t^{2}-2s^{2}=1$, we can operate a substitution, implicitely simplifying a $4$ in every fraction:

[TABLE]

Now is:

[TABLE]

[TABLE]

[TABLE]

where $l_{j}=t_{j}/s_{j}-\sqrt{2}$, and $l_{j}\rightarrow 0$ for $j\rightarrow+\infty$.

This lead to the ratios:

[TABLE]

We can now rewrite $L_{2}$ by using the ratios:

[TABLE]

Now the square differences can be rewritten as a product of a sum and a difference:

[TABLE]

Considering the fact that $l_{j}$ tends to zero for diverging $j$, we can both approximate $6+4\sqrt{2}+2(l_{j-1}+l_{j})$ and $6+4\sqrt{2}+2(l_{j}+l_{j+1})$ with $6+4\sqrt{2}$. Then:

[TABLE]

and so:

[TABLE]

where $k_{j}=t_{j}/s_{j}$.

[TABLE]

By proceeding with calculations we can state:

[TABLE]

where $s_{j+1}/s_{j-1}=(s_{j+1}/s_{j})\cdot(s_{j}/s_{j-1})$, and the limit of both factors is equal to $(3+2\sqrt{2})$.

For the remaining limit, we consider just the denominator:

[TABLE]

where the factor in brackets is equal to $1$ for every $j$, because $(t_{j},s_{j})$ is a solution of the Pell equation $t_{j}^{2}-2s_{j}^{2}=1$. In particular the same result is obtaining by considering the numerator, because it is just the denominator with indices shifted by one. Then the ratio is constant and equal to $1$; so is the limit for $j\rightarrow 0$, and:

[TABLE]

as we wanted to prove.

4.3 Ratio limit: second ratio, second method

We will see now an alternate way to get that result.

We know the solutions of the Pell equation to be $t_{j}+s_{j}\sqrt{2}=(3+2\sqrt{2})^{j}$, and also that $t_{j}-s_{j}\sqrt{2}=(3-2\sqrt{2})^{j}$. Observed $(3-2\sqrt{2})=(3+2\sqrt{2})^{-1}$, and defined $\beta:=(1+\sqrt{2})$, hence $(3+2\sqrt{2})=\beta^{2}$, $(3-2\sqrt{2})=\beta^{-2}$, by respectively summing and subtracting:

[TABLE]

This allows us to write a closed formula, from which we can generate numbers which are both triangulars and squares:

[TABLE]

by setting $\alpha=\beta^{4}=(1+\sqrt{2})^{4}$.

We can obtain via these calculations the well-known result:

[TABLE]

considering that $|\alpha|>1$ and so other terms are trascurable for $j\rightarrow+\infty$.

In an analogue way we can compute:

[TABLE]

by implicitely simplifying a $32$ in every fraction.

The use of standard algebra techniques gives the subsequent results.

[TABLE]

By rearranging:

[TABLE]

by semplifying in the right factor:

[TABLE]

and by explicitely calculating the left factor:

[TABLE]

Considered the fact that there is the limit operator, we can consider just the elements depending on $j$, in which the coefficents of them at the exponential are positive, because the others are trascurabile with respect to them, considering the operations we are doing. This finally gives:

[TABLE]

again, as we wanted to prove.

5 Open points

We conjecture that the result holds for every $h$-th ratio, $h\geq 3$, defined by:

[TABLE]

This means that it holds:

[TABLE]

but we are not able to either prove or disprove it, at the moment.

On the other hand, it can be investigated whether similar results can be written for other sequences of integers figurate in more than one way, like both triangular and pentagonal, both square and pentagonal, and so on.