Seeds for Generalized Taxicab Numbers

TL;DR

This paper introduces the concept of seed numbers for generalized taxicab numbers, proves their existence, and provides explicit formulas for certain cases, advancing understanding of these special numbers.

Contribution

It establishes the existence of seed numbers for generalized taxicab numbers and derives explicit formulas for cases with two or three ways, proposing a conjecture for higher cases.

Findings

Existence of seed numbers for all positive integers m and t

Explicit formulas for seed numbers when t=2 or 3

Conjecture for seed numbers when t ≥ 4

Abstract

The generalized taxicab number $T(n,m,t)$ is equal to the smallest number that is the sum of $n$ positive $m$th powers in $t$ ways. This definition is inspired by Ramanujan's observation that $1729 = 1^3+ 12^3 =9^3 + 10^3 $ is the smallest number that is the sum of two cubes in two ways and thus $1729= T(2,3,2)$. In this paper we prove that for any given positive integers $m$ and $t$, there exists a number $s$ such $T(s+k,m,t) =T(s,m,t) +k$ for every $k \geq 0$. The smallest such $s$ is termed the seed for the generalized taxicab number. Furthermore, we find explicit expressions for this seed number when the number of ways $t$ is 2 or 3 and present a conjecture for $t \geq 4$ ways.

Equations

$$T(6,2,2)\le 21=4^2+1+1+1+1+1=2^2+2^2+2^2+2^2+2^2+1.$$

$$T(7,2,2)\le 22=4^2+1+1+1+1+1+1=2^2+2^2+2^2+2^2+2^2+1+1.$$

$$S_0=\underset{n}{\underbrace{t^m+\dots +t^m}}$$

$$S_i=\underset{{\alpha }_i{\delta }_i}{\underbrace{(t-i{)}^m+\dots +(t-i{)}^m}}+\underset{{\beta }_i{\delta }_i}{\underbrace{(t+i{)}^m+\dots +(t+i{)}^m}}.$$

$${\alpha }_i{\delta }_i+{\beta }_i{\delta }_i=({\alpha }_i+{\beta }_i){\delta }_i={\gamma }_i{\delta }_i=n$$

$$\begin{array}[]{lll}S_{i}&=&\alpha_{i}\delta_{i}(t-i)^{m}+\beta_{i}\delta_{i}(t+i)^{m}\\ &=&(\alpha_{i}(t^{m}-a_{i})+\beta_{i}(t^{m}+b_{i}))\delta_{i}\\ &=&((\alpha_{i}+\beta_{i})t^{m}+(\beta_{i}b_{i}-\alpha_{i}a_{i}))\delta_{i}\\ &=&((\alpha_{i}+\beta_{i})t^{m}+(l_{i}-l_{i}))\delta_{i}\\ &=&\gamma_{i}t^{m}\ {n\over\gamma_{i}}\\ &=&nt^{m}\end{array}$$

$$T^{\prime }(n_k+1)=T^{\prime }(n_0+x_k+1)\le T^{\prime }(n_0)+x_k+1-k.$$

$$\underset{\alpha }{\underbrace{4^m+4^m+\dots +4^m}}+\underset{n-\alpha }{\underbrace{1+1+\dots +1}}=\underset{n}{\underbrace{2^m+2^m+\dots +2^m}},$$

$$\alpha (\frac{3^m-1}{d})=n(\frac{2^m-1}{d}).$$

$$\underset{\alpha }{\underbrace{3^m+3^m+\dots +3^m}}+\underset{n-\alpha }{\underbrace{1+1+\dots +1}}=\underset{n}{\underbrace{2^m+2^m+\dots +2^m}}.$$

$$4^m+\underset{2^m}{\underbrace{1+\dots +1}}=\underset{2^m+1}{\underbrace{2^m+\dots +2^m}}=(2^m+1)2^m(1)$$

$$\underset{\frac{2^m-1}{d}}{\underbrace{3^m+\dots +3^m}}+\underset{\frac{3^m-2^m}{d}}{\underbrace{1+\dots +1}}=\underset{\frac{3^m-1}{d}}{\underbrace{2^m+\dots +2^m}}=(\frac{3^m-1}{d})2^m.(2)$$

$$a_1^m+a_2^m+\dots +a_n^m+\underset{2^m+1-n}{\underbrace{1+1+\dots +1}}>4^m+\underset{2^m}{\underbrace{1+\dots +1}}$$

$$\underset{2^m+1}{\underbrace{4^m+1+\dots +1}}+\underset{2^m+1}{\underbrace{4^m+1+\dots +1}}=\underset{2^m+1}{\underbrace{4^m+1+\dots +1}}+\underset{2^m+1}{\underbrace{2^m+\dots +2^m}}=\underset{2^m+1}{\underbrace{2^m+\dots +2^m}}+\underset{2^m+1}{\underbrace{2^m+\dots +2^m}}.\text{mbox}(3)$$

$$5^m+2(2^m+1)-1\le \sum _{i=1}^s{a}_i^m+(2(2^m+1)-s)\le T(2(2^m+1),m,3)$$

$$5^m+2(2^m+1)-1\le T(2(2^m+1),m,3)\le 2(2^m+1)2^m.$$

$$4^m+{\alpha }_3{3}^m+{\alpha }_2{2}^m+{\alpha }_1{1}^m=4^m+{\beta }_3{3}^m+{\beta }_1{1}^m$$

$${\alpha }_3{3}^m+{\alpha }_2{2}^m+{\alpha }_1{1}^m={\beta }_3{3}^m+{\beta }_1{1}^m.$$

$${\alpha }_2{2}^m=({\beta }_3-{\alpha }_3)3^m+({\beta }_1-{\alpha }_1)$$

$$\underset{l_4}{\underbrace{4^m+1+\dots +1}}=\underset{(2^m-1)/d)}{\underbrace{3^m+\dots +3^m}}+\underset{(3^m-2^m)/d)}{\underbrace{1^m+\dots +1^m}}+\underset{l_4-l_3}{\underbrace{2^m+\dots +2^m}}=\underset{{\text{l}}_4}{\underbrace{2^m+\dots +2^m}}.\text{mbox}(4)$$

$${\alpha }_3{3}^m+{\alpha }_2{2}^m+{\alpha }_1{1}^m={\beta }_3{3}^m+{\beta }_1{1}^m=s{2}^m(5)$$

$${\alpha }_2{2}^m=({\beta }_3-{\alpha }_3)3^m+({\beta }_1-{\alpha }_1)1^m$$

$${\alpha }_3{3}^m+{\alpha }_1{1}^m=(s-{\alpha }_2)2^m.$$

$$\underset{l_4}{\underbrace{4^m+1+\dots +1}}+\underset{l_3-l_4}{\underbrace{2^m+\dots +2^m}}=\underset{(2^m-1)/d)}{\underbrace{3^m+\dots +3^m}}+\underset{(3^m-2^m)/d)}{\underbrace{1^m+\dots +1^m}}=\underset{{\text{l}}_3}{\underbrace{2^m+\dots +2^m}}.\text{mbox}(6)$$

$${\alpha }_4{4}^m+{\alpha }_3{3}^m+{\alpha }_2{2}^m+{\alpha }_1{1}^m={\beta }_4{4}^m+{\beta }_3{3}^m+{\beta }_1{1}^m=s{2}^m.$$

$${\alpha }_4{4}^m+{\alpha }_3{3}^m+{\alpha }_2{2}^m+{\alpha }_1{1}^m=4^m+{\beta }_3{3}^m+{\beta }_1{1}^m=s{2}^m.$$

$${\alpha }_3{3}^m+{\alpha }_2{2}^m+{\alpha }_1{1}^m=4^m+{\beta }_3{3}^m+{\beta }_1{1}^m=s{2}^m.$$

$$\underbrace{3^{m}+\ldots+3^{m}}_{(2^{m}-1)/d)}+\underbrace{1^{m}+\ldots+1^{m}}_{(3^{m}-2^{m})/d)}+\underbrace{3^{m}+\ldots+3^{m}}_{(2^{m}-1)/d)}+\underbrace{1^{m}+\ldots+1^{m}}_{(3^{m}-2^{m})/d)}=\ \ \ \ \ \ \ \$$

$$\underset{(2^m-1)/d)}{\underbrace{3^m+\dots +3^m}}+\underset{(3^m-2^m)/d)}{\underbrace{1^m+\dots +1^m}}+\underset{{\text{l}}_3}{\underbrace{2^m+\dots +2^m}}=\underset{2{\text{l}}_3}{\underbrace{2^m+\dots +2^m}}.\text{mbox}(7)$$

$${\alpha }_3{3}^m+{\alpha }_2{2}^m+{\alpha }_1{1}^m={\beta }_3{3}^m+{\beta }_1{1}^m=s{2}^m.(8)$$

Full text

1 Introduction

Hardy relays the following story about visiting Ramanujan during his illness (see [4], p. xxxv):

I remember once going to see him when he was lying ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavorable omen. No, he replied, it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways.

Indeed, $1729=1^{3}+12^{3}$ and $1729=9^{3}+10^{3}$ and it is the smallest such number that is the sum of two cubes in two different ways. In honor of the Ramanujan–Hardy conversation, the smallest number expressible as the sum of two cubes in $t$ different ways is known as the $t^{th}$ taxicab number and is denoted $\operatorname{Taxicab}(t)$ . Therefore, with this notation, $\operatorname{Taxicab}(2)=1729$.

There has been quite a bit of effort expended in finding these taxicab numbers. The interested reader is referred to [1] for information about these numbers. That paper contains interesting information about the history of the problem as well as a discussion about the techniques used to find certain values of $\operatorname{Taxicab}(t)$. Further information about taxicab numbers and their variants can also be found at [2, 3, 5, 6, 7]. Basically, $\operatorname{Taxicab}(t)$ is known for $2\leq t\leq 6$ and upper bounds for $\operatorname{Taxicab}(t)$ have been given for $7\leq t\leq 22$.

In this paper we will generalize the definition of taxicab numbers, as there is really nothing special about using exactly two cubes (except for historical reasons). We will be concerned with finding the smallest number that is the sum of $n$ positive $m^{th}$ powers in at least $t$ ways.

Let $T(n,m,t)$ denote the least number that is the sum of $n$ positive $m^{th}$ powers in at least $t$ ways provided such a number exists.111It should be noted that in Wikipedia, the generalized taxicab number $\operatorname{Taxicab}(k,j,n)$ is the smallest number which can be expressed as the sum of $j$ $k^{th}$ positive powers in $n$ different ways, however since there have been no published papers with this notation we will use the notation in the definition given above. So as noted above, $T(2,3,2)=1729$ and in general $T(2,3,t)$ is the taxicab number $\operatorname{Taxicab}(t)$. It is also easy to verify that

$T(2,2,2)=50=5^{2}+5^{2}=7^{2}+1,$

$T(3,2,2)=27=5^{2}+1+1=3^{2}+3^{2}+3^{2},$

$T(4,2,2)=28=5^{2}+1+1+1=3^{2}+3^{2}+3^{2}+1,$  and

$T(5,2,2)=20=4^{2}+1+1+1+1=2^{2}+2^{2}+2^{2}+2^{2}+2^{2}.$

Note that by adding 1 to the two sums in the last example we obtain

[TABLE]

Doing this again we obtain

[TABLE]

In fact it is indeed true that $T(6,2,2)=21$ and $T(7,2,2)=22$.

Considering cubes now, it is straightforward to verify that

$T(2,3,2)=1729=12^{3}+1=10^{3}+9^{3},$

$T(3,3,2)=251=6^{3}+3^{3}+2^{3}=5^{3}+5^{3}+1,$

$T(4,3,2)=219=6^{3}+1+1+1=4^{3}+4^{3}+4^{3}+3^{3},$

$T(5,3,2)=157=5^{3}+2^{3}+2^{3}+2^{3}+2^{3}=4^{3}+4^{3}+3^{3}+1+1$

$T(6,3,2)=158=5^{3}+2^{3}+2^{3}+2^{3}+2^{3}+1=4^{3}+4^{3}+3^{3}+1+1+1$

$T(7,3,2)=131=5^{3}+1+1+1+1+1+1=4^{3}+3^{3}+2^{3}+2^{3}+2^{3}+2^{3}+2^{3}$

$T(8,3,2)=132=5^{3}+1+1+1+1+1+1+1=4^{3}+3^{3}+2^{3}+2^{3}+2^{3}+2^{3}+2^{3}+1$

$T(9,3,2)=72=4^{3}+1+1+1+1+1+1+1+1=2^{3}+2^{3}+2^{3}+2^{3}+2^{3}+2^{3}+2^{3}+2^{3}+2^{3}$

$T(10,3,2)=73=4^{3}+1+1+1+1+1+1+1+1+1=2^{3}+2^{3}+2^{3}+2^{3}+2^{3}+2^{3}+2^{3}+2^{3}+2^{3}+1$

One also can check that $T(11,3,2)=74$ and that this solution comes about by adding 1 to both of the (equal) sums in the case of $T(10,3,2)$.

From the examples above it seems plausible that there exists a number, say $s_{0}$, such that $T(n+1,m,t)=T(n,m,t)+1$ for all $n\geq s_{0}$ or equivalently, that $T(s_{0}+k,m,t)=T(s_{0},m,t)+k$ for all $k\geq 0$ . This motivates the following definition.

Definition. If $s_{0}$ is the smallest positive integer such that $T(n+1,m,t)=T(n,m,t)+1$ for all $n\geq s_{0}$, then we call $s_{0}$ the seed number for $m^{th}$ powers in $t$ ways and denote this number by $S(m,t)=s_{0}$. We also call $T(S(m,t),m,t)$ the seed value of $m^{th}$ powers in $t$ ways and denote it by $V(m,t)$.

This paper proceeds as follows. In Section 2 we will show that for every $m$ and $t$ there exists a seed number. In Section 3 we will give an explicit value for the seed of the sum of $m^{th}$ powers in 2 ways. We will prove there that the seed number for squares in 2 ways is indeed 5 and the seed value is 20. In our notation, this says $S(2,2)=5$ and $V(2,2)=20$ and hence $T(5+k,2,t)=20+k$ for all $k\geq 0$. In Section 4 we will give an explicit value of $V(m,3)$. In Section 5, we end with a general theorem and a conjecture about the seed for $m^{th}$ powers in $t$ ways for all $t\geq 2$.

2 Seeds exist

In this section we will prove that for every $m$ and $t$ there exists a seed number. We will first show that for every $m$ and $t$ there exist some positive integer $n$ and some value $v$ such that $v$ is the sum of $n$ $m^{th}$ powers in $t$ ways. From that we will then show that there is a least such $n$ and hence there will exist a seed number (and a seed value).

Lemma 2.1

For all positive integers $m,t\geq 1$, there exist positive integers $n$ and $v$ such that $v$ is the sum of $n$ $m^{th}$ powers in $t$ ways.

Proof: If $m=1$ or $t=1$, the result is obvious, so assume that $m,t>1$.

We will give a direct construction of $t$ different sums of $m^{th}$ powers, each sum having the same number of terms. The first sum will be $t^{m}+t^{m}+\ldots+t^{m}$ for a suitable number of terms. For each $1\leq i\leq t-1$ we will construct the sum $S_{i}$ from the terms $(t-i)^{m}$ and $(t+i)^{m}$. So each $S_{i}=\underbrace{(t-i)^{m}+\ldots+(t-i)^{m}}_{x_{i}}+\underbrace{(t+i)^{m}+\ldots+(t+i)^{m}}_{y_{i}}$ for suitable $x_{i}$ and $y_{i}$.

In order to define the sums, we first need to define some values. For each $1\leq i\leq t-1$ define

[TABLE]

Using these values, define

[TABLE]

and for each $1\leq i\leq t-1$ define

[TABLE]

Obviously all of these sums are different (i.e. no two contain the same terms). We must prove two things: first, that each of the sums $S_{0},S_{1},\ldots,S_{t-1}$ contain the same number of terms and second, that $S_{0}=S_{1}=\ldots=S_{t-1}$.

To show the first, we note that $S_{0}$ contains $n$ terms and for each $1\leq i\leq t-1$, $S_{i}$ contains $\alpha_{i}\delta_{i}+\beta_{i}\delta_{i}$ terms. But now

[TABLE]

as desired.

Next we compute the sums. Clearly $S_{0}=nt^{m}$. For each $1\leq i\leq t-1$,

[TABLE]

Thus we conclude that $S_{0}=S_{1}=\ldots=S_{t-1}=nt^{m}$. This completes the proof.

By adding a 1 to each multiset of terms in the above sums, we see that $T(n+1,m,t)\leq T(n,m,t)$+1. As an immediate consequence of this observation and Lemma 2.1 we can conclude that $T(n,m,t)$ exists for every $m,t\geq 1$ when $n$ is large enough.

In the next theorem we will prove that there is a seed number for every $m$ and $t$.

Theorem 2.2

For every $m,t\geq 1$, there is a smallest number $s_{0}$ such that $T(s_{0}+k,m,t)=T(s_{0},m,t)+k$ for every $k\geq 0$. Hence for every $m,t\geq 1$ the seed number $s_{0}=S(m,t)$ exists.

Proof: Fix $m$ and $t$. Note that for some $n$ large enough, $T(n,m,t)$ exists by Lemma 2.1. Now let $n_{0}$ be the smallest integer for which $T(n_{0},m,t)$ exists. To shorten notation let $T^{\prime}(n)=T(n,m,t)$. Notice that for any $n$, $T^{\prime}(n)\geq\underbrace{1^{m}+1^{m}+\ldots+1^{m}}_{n}=n$. So a (very naive) lower bound for $T^{\prime}(n)$ is $n$. The gap between the value of $T^{\prime}(n_{0})$ and the naive lower bound on $T^{\prime}(n_{0})$ is $g=T^{\prime}(n_{0})-n_{0}\geq 0$. We also have that $T^{\prime}(n_{0}+k)\leq T^{\prime}(n_{0})+k$ for every $k\geq 0$. Thus $n_{0}+k\leq T^{\prime}(n_{0}+k)\leq T^{\prime}(n_{0})+k$.

We say that the function $T^{\prime}$ drops at $n$ if $T^{\prime}(n+1)<T^{\prime}(n)+1$ (note that $n$ is the location of the “drop” and not the amount $T^{\prime}$ drops). Let $D=\{n\geq n_{0}\ |\ T^{\prime}(n+1)<T^{\prime}(n)+1\}$ be the set of all drops of $T^{\prime}$. We claim that $D$ is a finite set and in fact we will show that $|D|\leq g$. Let $D=\{n_{1},n_{2},\ldots\}$. (We should note that possibly $D=\emptyset$, in which case $S(m,t)$ is just $n_{0}$.) If $D$ is nonempty, say that $n_{k}=n_{0}+x_{k}$, then we see that since $T^{\prime}$ drops by at least one at each $n_{i}$ we have that

[TABLE]

So since $n_{0}+x_{k}+1\leq T^{\prime}(n_{0}+x_{k}+1)$, we have that $n_{0}\leq T^{\prime}(n_{0})-k$ and hence that $k\leq T^{\prime}(n_{0})-n_{0}=g$. This implies that $|D|\leq g$ and hence there are at most $g$ drops in the function $T^{\prime}$. So if $D=\{n_{1},n_{2},\ldots n_{i}\}$ is the set of all drops, then since $n_{i}$ is the last drop in the function $T^{\prime}$, we have that $T^{\prime}(n_{i}+1+k)=T^{\prime}(n_{i}+1)+k$ for all $k\geq 0$ and hence $n_{i}+1$ is the seed number $S(m,t)$.

As an example of the above theorem we consider the values of $T(2,3,2),T(3,3,2),\ldots,$ $T(10,3,2)$ given in Section 1. Notice that (from Ramanujan) $n_{0}=2$. We also see that $n_{1}=2,n_{2}=3,n_{3}=4,n_{4}=6,$ and $n_{5}=8$. We will prove in Section 4 that indeed $D=\{2,3,4,6,8\}$. Thus we will conclude that $S(3,2)=9$ and hence that $T(9+k,3,2)=T(9,3,2)+k=72+k$ for all $k\geq 0$.

3 Seeds for two ways

In this section we will give the explicit value for the seed numbers for two ways. We assume that all variables are positive integers except where noted. We begin with three easy lemmas that hold for any number of sums. The first lemma says that if all the sums share a common term, then that term must be equal to 1.

Lemma 3.1

If $x=T(n,m,t)$ and $x=\sum_{i=1}^{n}a_{i}^{m}=\sum_{i=1}^{n}b_{i}^{m}=\ldots=\sum_{i=1}^{n}t_{i}^{m}$, and if $a_{i_{1}}=b_{i_{2}}=\ldots=t_{i_{t}}$ for some choice of $i^{\prime}s$, then $a_{i_{1}}=b_{i_{2}}=\ldots=t_{i_{t}}=1$.

Proof: If not, then replace each of $a_{i_{1}},b_{i_{2}},\ldots,t_{i_{t}}$ with a 1 and get a contradiction to $x=T(n,m,t)$ since the new sums of the $m^{th}$ powers will still all be the same and will be less than before, a contradiction.

The next lemma says that the $t$ sums adding to the seed value can’t all have a 1 as a term.

Lemma 3.2

If $x=V(m,t)$ and $x=\sum_{i=1}^{n}a_{i}^{m}=\sum_{i=1}^{n}b_{i}^{m}=\ldots=\sum_{i=1}^{n}t_{i}^{m}$, and if $a_{i_{1}}=b_{i_{2}}=\ldots=t_{i_{t}}$ for some choice of $i^{\prime}s$, then $a_{i_{1}}=b_{i_{2}}=\ldots=t_{i_{t}}\neq 1$.

Proof: If each sum has a 1 as a term, then by simply deleting the 1 in each of these sums we would obtain a smaller seed value, a contradiction.

In the next lemma we show that any seed value for $m^{th}$ powers in $t$ ways must always be greater or equal to $n2^{m}$ where $n$ is the seed number. This will essentially say that we can always assume that one of the sums is $\underbrace{2^{m}+2^{m}+\ldots+2^{m}}_{n}$. This fact will be of fundamental importance in finding seeds for 2 and 3 ways.

Lemma 3.3

If $V(m,t)=\sum_{i=1}^{s_{0}}a_{i}^{m}=\sum_{i=1}^{s_{0}}b_{i}^{m}=\ldots=\sum_{i=1}^{s_{0}}t_{i}^{m}$ is the seed value for $m^{th}$ powers in $t$ ways, then $V(m,t)\geq s_{0}2^{m}$. Further, if $n$ is any number which provides a solution to $\sum_{i=1}^{n}a_{i}^{m}=\sum_{i=1}^{n}b_{i}^{m}=\ldots=\sum_{i=1}^{n}t_{i}^{m}=n2^{m}$, then $V(m,t)=s_{0}2^{m}$, where $s_{0}$ is the smallest such $n$ (and hence $s_{0}=S(m,t)$).

Proof: Let $V(m,t)=\sum_{i=1}^{s_{0}}a_{i}^{m}=\sum_{i=1}^{s_{0}}b_{i}^{m}=\ldots=\sum_{i=1}^{s_{0}}t_{i}^{m}$ be the seed value for $m^{th}$ powers in $t$ ways and assume that each sum is written in nonincreasing order. Now if $a_{n}=1$, then from Lemma 3.2 we have without loss of generality that $b_{n}\neq 1$. Thus for all $1\leq i\leq n$ it must be that $b_{i}\geq 2$. Hence in this case we have that $V(m,t)\geq{s_{0}}2^{m}$. If $a_{n}\geq 2$, then since $a_{i}\geq a_{i+1}$ for all $1\leq i\leq n-1$, then clearly $V(m,t)\geq{s_{0}}2^{m}$. So we have that $V(m,2)\geq{s_{0}}2^{m}$.

The second part of this lemma now follows immediately.

Lemma 3.3 says that the seed value is the sum of $s_{0}$ $2^{m}$’s (where $s_{0}$ will be equal to the seed number $S(m,t)$). So we are interested in this value for the sum. In the next two lemmas we consider two different sums that are equal to the sum of $2^{m}$’s. The verification of the first is straightforward.

Lemma 3.4

If $\alpha 4^{m}+(n-\alpha)=n2^{m}$, then $n=(2^{m}+1)\alpha$ and hence $n\geq(2^{m}+1)$.

A comment is in order concerning Lemma 3.4. This lemma deals with the case when two sums are equal and one of the sums is all 2’s and the other is 4’s and 1’s. It says that if

[TABLE]

then $n\geq(2^{m}+1)$. The next lemma deals with the case when two sums are equal and one of the sums is all 2’s and the other is 3’s and 1’s.

Lemma 3.5

If $\alpha 3^{m}+(n-\alpha)=n2^{m}$ and if $d=\gcd(3^{m}-2^{m},2^{m}-1)$, then ${3^{m}-1\over d}\ |\ n$ and hence $n\geq{3^{m}-1\over d}$.

Proof: Assume that $\alpha 3^{m}+(n-\alpha)=n2^{m}$ and that $d=\gcd(3^{m}-2^{m},2^{m}-1)$, then

[TABLE]

Now since $d=\gcd(3^{m}-2^{m},2^{m}-1)=\gcd(3^{m}-1,2^{m}-1)$, then $1=\gcd({3^{m}-1\over d},{2^{m}-1\over d})$ and thus we have that ${2^{m}-1\over d}\ |\ \alpha.$ So $\alpha({d\over 2^{m}-1})$ is an integer. Now since $\alpha({d\over 2^{m}-1})\ ({3^{m}-1\over d})=n$ it follows that $({3^{m}-1\over d})\ |\ n$ and hence that $n\geq{3^{m}-1\over d}$.

The application of this lemma will be similar to that of Lemma 3.4. In this case we have the situation where

[TABLE]

So here we will have that $n=\alpha({d\over 2^{m}-1})\ ({3^{m}-1\over d})$ and our main application will be that in this case $n\geq{3^{m}-1\over d}$. We now obtain our characterization of the seed number and the seed value for sums in two ways.

Theorem 3.6

Let $d=\gcd(3^{m}-2^{m},2^{m}-1)$. The seed number $S(m,2)=\min({3^{m}-1\over d},2^{m}+1)=s_{0}$ and the seed value $V(m,2)=s_{0}2^{m}$. Hence $T(s_{0}+j,m,2)=s_{0}2^{m}+j$ for every $j\geq 0$.

Proof: Consider the two equations

[TABLE]

and

[TABLE]

In view of Equations (1) and (2) and Lemma 3.3 if $V(m,t)=a_{1}^{m}+a_{2}^{m}+\ldots+a_{n}^{m}=b_{1}^{m}+b_{2}^{m}+\ldots+b_{n}^{m}$ is the seed value for $m^{th}$ powers in 2 ways, then we can assume that $b_{i}=2$ for all $i$. We next show that $a_{1}\leq 4$.

Assume that $V(m,t)=a_{1}^{m}+a_{2}^{m}+\ldots+a_{n}^{m}=2^{m}+2^{m}+\ldots+2^{m}$ is the seed value for $m^{th}$ powers in 2 ways with $a_{i}\geq a_{i+1}$ for all $1\leq i\leq n-1$. Assume that $a_{1}\geq 5$. Clearly, if $n\geq 2^{m}+1$, then in view of Equation (1) above this is a contradiction (since $a_{1}^{m}+a_{2}^{m}+\ldots+a_{n}^{m}>4+1+\ldots+1$). Assume $n<2^{m}+1$, then

[TABLE]

which is again a contradiction to the assumption that $V(m,t)=a_{1}^{m}+a_{2}^{m}+\ldots+a_{n}^{m}$. Assuming that $a_{1}=4$ and $a_{2}>1$ yields a similar contradiction.

So either $\{a_{1},a_{2},\ldots a_{n}\}=\{1,4\}$ or $\{a_{1},a_{2},\ldots a_{n}\}=\{1,3\}$ since by Lemma 3.1 no $a_{i}$ can equal any $b_{i}=2$. In the first case we obtain Equation (1) since no smaller sum can have only $4^{m}$’s and 1’s as its terms. In the second case we can assume that $s3^{m}+(n-s)=n2^{m}$ for some $s$. From Lemma 3.5, the minimum value of $n$ is ${3^{m}-1\over d}$, which leads to Equation (2). The seed number will therefore be the minimum length of the sums in either Equation (1) or Equation (2). Thus the minimum of $2^{m}+1$ and $3^{m}-1\over d$ will be the seed number $S(m,2)$.

In the following table we compute seeds for $m^{th}$ powers in 2 ways for $m\leq 20$.

Example 3.7

In the following table we give explicit values from Theorem 3.6. Remember that $S(m,2)$ is the number of terms in the seed, while $V(m,2)$ is the exact value of the seed.

[TABLE]

The interested reader may note that $S(m,2)=2^{m}+1$ in every case above except when $m=1,4,12$. This says that $2^{m}+1\leq(3^{m}-1)/d$ for every $m\leq 20$ with $m\neq 1,4,12$. We computed values of $2^{m}+1$ and $(3^{m}-1)/d$ for all $m\leq 200,000$ and found that $2^{m}+1\leq(3^{m}-1)/d$ for all $m$ in that range except for $m=1,4,12$ and 36. We do not conjecture that this is true for all $m>36$, it sure appears to be true, however.

4 Seeds for three ways

In this section we will give an explicit value for $V(m,3)$, the seed value for the smallest number that can be written as the sum of $m^{th}$ powers in 3 ways. We first need a preliminary lemma that says that no term in a sum that is a seed value (for powers $m\geq 4$) can exceed the number 4.

Lemma 4.1

The seed value $V(m,3)=\sum_{i=1}^{s}a_{i}^{m}=\sum_{i=1}^{s}b_{i}^{m}=s2^{m}$ where $s=S(m,3)$ is the seed number. If $m\geq 4$, then $a_{i},b_{i}\leq 4$ for all $1\leq i\leq s$.

Proof: We first note following equation:

[TABLE]

From this equation and Lemma 3.3 we have $V(m,3)=s2^{m}$ for $s=S(m,3)$. We also see from this equation that the taxicab number $T(2(2^{m}+1)+j,m,3)\leq 2(2^{m}+1)2^{m}+j$ for all $j\geq 0$ and that the seed number $S(m,3)\leq 2(2^{m}+1)$. So in particular, when $j=0$ we have that $T(2(2^{m}+1),m,3)\leq 2(2^{m}+1)2^{m}$.

Now, assume $V(m,3)=\sum_{i=1}^{s}a_{i}^{m}=\sum_{i=1}^{s}b_{i}^{m}=s2^{m}$ where $s=S(m,3)$ is the seed number. Then $s\leq 2(2^{m}+1)$. Assume that $a_{1}\geq 5$. Since $a_{i}\geq 1$ for all $i>1$, when extending the sums to have $2(2^{m}+2)$ terms by adding sufficinetly many 1’s, we get that

[TABLE]

and hence

[TABLE]

Thus

\begin{array}[]{rcll}5^{m}+2(2^{m}+1)-1&\leq&2(2^{m}+1)2^{m}&\\ 5^{m}+2(2^{m}+1)-1&\leq&(2^{m}+1)2^{m+1}&\\ 5^{m}&\leq&(2^{m}+1)(2^{m+1}-2)+1&\\ 5^{m}&\leq&2^{2m+1}-1&\\ 5^{m}&\leq&2\times 4^{m}-1.&\\ \end{array}

This last inequality implies that $m=1,2,$ or 3, but by hypothesis $m\geq 4$, so we obtain a contradiction. Hence $a_{i}\leq 4$ (similarly $b_{i}\leq 4$) for all $1\leq i\leq s$.

We are now in position to obtain our characterization of the seed number and the seed value for sums in three ways. We begin with the small values of $m$.

Theorem 4.2

(a) $S(1,3)=3$ and the seed value $V(1,3)=3\times 2^{1}=6$, (b) $S(2,3)=8$ and the seed value $V(2,3)=8\times 2^{2}=32$, (c) $S(3,3)=18$ and the seed value $V(3,3)=18\times 2^{3}=144$.

Proof: The sums are given below. It is straightforward to check that they are minimal.

(a) $6=4+1+1=3+2+1=2+2+2$

(b) $32=4^{2}+2^{2}+2^{2}+2^{2}+1+1+1+1=3^{2}+3^{2}+3^{2}+1+1+1+1+1=\underbrace{2^{2}+\ldots+2^{2}}_{8}$

(c) $144=4^{3}+4^{3}+\underbrace{1+\ldots+1}_{16}=4^{3}+\underbrace{1+\ldots+1}_{8}+\underbrace{2^{3}+\ldots+2^{3}}_{9}=\underbrace{2^{3}+\ldots+2^{3}}_{18}$

Theorem 4.3

Assume that $m\geq 4$ and let $d=\gcd(3^{m}-2^{m},2^{m}-1)$. Also let $l_{3}={3^{m}-1\over d}$ and $l_{4}=2^{m}+1$. Given the four values $l_{3},l_{4},2l_{3},2l_{4}$, the second smallest of these values is the seed number $S(m,3)$ and the seed value $V(m,3)=S(m,3)\times 2^{m}.$

Proof: Considering Equation (3) in the proof of Lemma 4.1, in all cases the seed number $S(m,3)\leq 2l_{4}$. Also, we can assume that the seed value $V(m,3)=\sum_{i=1}^{s}a_{i}^{m}=\sum_{i=1}^{s}b_{i}^{m}=s2^{m}$ where $s=S(m,3)$ is the seed number and (from Lemma 4.1) that $a_{i},b_{i}\leq 4$ for all $i$. Let $A=\{a_{1},a_{2},\ldots a_{s}\}=\{1^{\alpha_{1}},2^{\alpha_{2}},3^{\alpha_{3}},4^{\alpha_{4}}\}$ be the multiset containing all the terms in the sum $\sum_{i=1}^{s}a_{i}^{m}$ (so $A$ contains the term $i^{m}$ exactly $\alpha_{i}$ times for $1\leq i\leq 4$), and let $B=\{b_{1},b_{2},\ldots b_{s}\}=\{1^{\beta_{1}},2^{\beta_{2}},3^{\beta_{3}},4^{\beta_{4}}\}$ be the multiset containing the terms in the sum $\sum_{i=1}^{s}b_{i}^{m}$. From Lemma 3.1 we can assume without loss of generality that $\beta_{2}=0$.

Case 1.)   If $2l_{4}$ is the second smallest value, then $2l_{4}\leq l_{3}$. Assume that $s=S(m,3)<2l_{4}$. We see first that $0\leq\alpha_{4},\beta_{4}\leq 1$, since if (say) $\alpha_{1}\geq 2$, then $\sum_{i=1}^{s}a_{i}^{m}\geq 4^{m}+4^{m}+(s-2)1^{m}$ and so $\sum_{i=1}^{s}a_{i}^{m}+(2l_{4}-s)\geq 4^{m}+4^{m}+(2l_{4}-2)$ which (because of Equation (3)) says that $\sum_{i=1}^{s}a_{i}^{m}$ can not be a seed value unless $\alpha_{3}=\alpha_{2}=0$ in which case we are led to one of the sums in Equation (3). However since we assumed that $s<2l_{4}$ this is a contradiction.

Now, if $\beta_{4}=0$ we obtain the equation $\beta_{3}3^{m}+\beta_{1}1^{m}=s2^{m}$. By Lemma 3.5 we thus have that $s\geq l_{3}$. So $s\geq l_{3}>2l_{4}>s$ a contradiction. If $\alpha_{4}=0$, then by subtracting $\alpha_{2}$ $2^{m}$’s from each side of the equation $\sum_{i=1}^{s}a_{i}^{m}=s2^{m}$ we obtain a similar contradiction. Hence we can assume that $\alpha_{4}=\beta_{4}=1$.

So we have that

[TABLE]

subtracting $4^{m}$ from both sides yields

[TABLE]

This implies that

[TABLE]

Now since $(\beta_{3}-\alpha_{3})+(\beta_{1}-\alpha_{1})=\alpha_{2}$ and since $\alpha_{2}>0$ (else the equation is degenerate), then by Lemma 3.5 we have that $\alpha_{2}\geq l_{3}$, a clear contradiction.

So in this case we have that $s\geq 2l_{4}$. Equation (3) then proves that indeed in this case that $s=S(m,3)=2l_{4}$ and hence the seed value $V(m,3)=2l_{4}2^{m}$.

Case 2.) If $l_{4}$ is the second smallest value, then $l_{3}\leq l_{4}\leq 2l_{3}$. First consider the following equation:

[TABLE]

From this equation we see that in this case $s=S(m,3)\leq l_{4}$. Assume $s=S(m,3)<l_{4}$.

If $\alpha_{4}\geq 1$ (or $\beta_{4}\geq 1$), then $\sum_{i=1}^{s}a_{i}^{m}\geq 4^{m}+(s-1)1^{m}$ and so $\sum_{i=1}^{s}a_{i}^{m}+(l_{4}-s)\geq 4^{m}+(l_{4}-1)$ which (because of Equation (4)) says that $\sum_{i=1}^{s}a_{i}^{m}$ can not be a seed value, unless $\alpha_{3}=\alpha_{2}=0$ in which case we are led to the first sum in Equation (4). However, since we assumed that $s<l_{4}$ we see that this is a contradiction. So $\alpha_{4}=\beta_{4}=0.$

Hence

[TABLE]

subtracting $\alpha_{3}3^{m}$ and $\alpha_{1}1^{m}$ from the first two sums yields

[TABLE]

and so by Lemma 3.5 we have that $\alpha_{2}\geq l_{3}.$ Also, subtracting $\alpha_{2}2^{m}$ from the first and third sums in Equation (5) we have

[TABLE]

So, again by Lemma 3.5 we have $s-\alpha_{2}=\alpha_{1}+\alpha_{3}\geq l_{3}$. Thus $s=\alpha_{1}+\alpha_{2}+\alpha_{3}\geq 2l_{3}$ a clear contradiction to our assumption that $s<l_{4}\leq 2l_{3}$.

So in this case we have $s\geq l_{4}$. Equation (4) then proves that in this case that $s=S(m,3)=l_{4}$ and hence the seed value $V(m,3)=l_{4}2^{m}$.

Case 3.) If $l_{3}$ is the second smallest value, then we have $l_{4}\leq l_{3}\leq 2l_{4}$. First note the following equation:

[TABLE]

From this we see that $s=S(m,3)\leq l_{3}$. Assume that $s=S(m,3)<l_{3}$. So similar to Case 1, the fact that $s<2l_{4}$ implies that $\alpha_{4},\beta_{4}\leq 1$. We show this by considering $T(2l_{4},m,3)$. This value is equal to $\sum_{i=1}^{s}a_{i}^{m}+(2l_{4}-s)$ since $\sum_{i=1}^{s}a_{i}^{m}$ is the seed value. Note $\sum_{i=1}^{s}a_{i}^{m}+(2l_{4}-s)<2l_{4}2^{m}$. However, if $\alpha_{4}\geq 2$, then $4^{m}+4^{m}+(l_{4}-2)1^{m}=2l_{4}2^{m}$, a contradiction.

So we can assume that $0\leq\alpha_{4},\beta_{4}\leq 1$. We have that

[TABLE]

If $\beta_{4}=0$, then we have $\beta_{3}3^{m}+\beta_{1}1^{m}=s2^{m}$, but from Lemma 3.5 this implies that $s\geq l_{3}$, a contradiction our assumption that $s<l_{3}$. So now we have that

[TABLE]

If $\alpha_{4}=1$, then by subtracting $4^{m}$ from the first two sums in the equation above, we obtain $\alpha_{2}2^{m}=(\beta_{3}-\alpha_{3})3^{m}+(\beta_{1}-\alpha_{1})1^{m}$. But from Lemma 3.5 we have $\alpha_{2}\geq l_{3}$, a contradiction. So $\alpha_{4}=0$. So now we have

[TABLE]

Finally, from $\alpha_{3}3^{m}+\alpha_{2}2^{m}+\alpha_{1}1^{m}=s2^{m}$ subtract $\alpha_{2}2^{m}$ from each side to obtain $\alpha_{3}3^{m}+\alpha_{1}1^{m}=(s-\alpha_{2})2^{m}$. From Lemma 3.5 we obtain $s-\alpha_{2}\geq l_{3}$ a clear contradiction to our assumption that $s<l_{3}$.

So we have shown that $s\geq l_{3}$. Equation (6) then proves that in this case that $s=S(m,3)=l_{3}$ and hence the seed value $V(m,3)=l_{3}2^{m}$.

Case 4.) Finally, if $2l_{3}$ is the second smallest value, then $2l_{3}\leq l_{4}$. Consider the following equation:

[TABLE]

[TABLE]

From this we see that $s=S(m,3)\leq 2l_{3}$. Assume that $s=S(m,3)<2l_{3}$.

Since $s<l_{4}$ (as in the proof of Case 2) we have that $\alpha_{4}=\beta_{4}=0$. So we have that

[TABLE]

Subtracting $\alpha_{3}3^{m}$ and $\alpha_{1}1^{m}$ from the first two sums yields

[TABLE]

and so by Lemma 3.5 we have that $\alpha_{2}\geq l_{3}.$ Also, subtracting $\alpha_{2}2^{m}$ from the first and third sums in Equation (8) we have

[TABLE]

So, again by Lemma 3.5 we have $s-\alpha_{2}=\alpha_{1}+\alpha_{3}\geq l_{3}$. Thus $s=\alpha_{1}+\alpha_{2}+\alpha_{3}\geq 2l_{3}$ a contradiction to our assumption that $s\leq 2l_{3}$.

So in this case $s\geq 2l_{3}$. Equation (7) then proves that in this case $s=S(m,3)=2l_{3}$ and hence the seed value $V(m,3)=l_{4}2^{m}$.

As was done after the proof of Theorem 3.6 we wish to compute the exact value of $S(m,3)$ using the results of Theorem 4.3. We found that for every $1\leq m\leq 200,000$ with $m\neq 1,2,4,6,12,36$, that $S(m,3)=2l_{4}$ and hence that $V(m,3)=2l_{4}2^{m}.$ This is Case 1 above and says that $2l_{4}\leq l_{3}$ for all powers $36<m\leq 200,000$. Again we do not conjecture that $2l_{4}\leq l_{3}$ for all $m>36$, but certainly the evidence is very strong.

When $m=4$ we are in Case 2, so $l_{3}\leq l_{4}\leq 2l_{3}$ and hence $S(4,3)=l_{4}=2^{4}+1$. We should note that $m=1$ also has the property that $l_{3}\leq l_{4}\leq 2l_{3}$, and although it doesn’t follow from the general proof, it is indeed true that $S(1,3)=l_{4}=3$ and so $V(1,3)=3\times 2^{1}=6$, since $6=1+2+3=4+1+1=2+2+2.$

When $m=6$ we are in Case 3 so $l_{4}\leq l_{3}\leq 2l_{4}$, and hence $S(6,3)=l_{3}=104$. Also note that $m=2$ also has the property that $l_{4}\leq l_{3}\leq 2l_{4}$, and indeed $S(2,3)=l_{3}=8$ and so $V(1,3)=8\times 2^{2}=32$, since $32=4^{2}+2^{2}+2^{2}+2^{2}+1+1+1+1=3^{2}+3^{2}+3^{2}+1+1+1+1+1=2^{2}+2^{2}+2^{2}+2^{2}+2^{2}+2^{2}+2^{2}+2^{2}$. As a bonus here we see that $32=5^{2}+7\times 1^{2}$ and hence four different sums of 8 squares are equal to 32. So we get $S(2,4)=8$ and hence $T(8,2,4)=T(8,2,3)=32$.

Finally when $m=12$ or $m=36$, we have $2l_{3}<l_{4}$ and thus both these values fall into case 4.

5 More than 3 ways

In this section we present a general theorem and a conjecture about seeds. Again let $d=\gcd(3^{m}-2^{m},2^{m}-1)$, $l_{3}={3^{m}-1\over d}$ and $l_{4}=2^{m}+1$. Also let the sums

[TABLE]

Theorem 5.1

Given $t$, there exists a number $m_{0}$, such that if $m\geq m_{0}$, then the seed number $S(m,t)$ is bounded above by the $t-1$st smallest of the values $al_{3}+bl_{4}$ over all $a,b\geq 0$. The seed value $V(m,t)=S(m,t)\times 2^{m}.$

Proof: Let $n=\min(l_{3},l_{4})$ and let $m_{0}=\max\{m\ |\ 5^{m}<(t-1)n2^{m}\}$. Let $n_{0}$ be the $t-1$st smallest of the values $al_{3}+bl_{4}$ over all $a,b\geq 0$. Finally define the sum $aS_{1}+bS_{2}+\overline{2^{m}}$ to be $a$ copies of $S_{1}$ added to $b$ copies of $S_{2}$ added to $n_{0}-(al_{3}+bl_{4})$ copies of $2^{m}$. Now it is clear that for all $al_{3}+bl_{4}\leq n_{0}$, that the sum $aS_{1}+bS_{2}+\overline{2^{m}}=n_{0}2^{m}$. Thus the $t-1$ sums $aS_{1}+bS_{2}+\overline{2^{m}}$ with the sum of $m_{0}$ $2^{m}$’s are all equal, proving our upper bound. Furthermore, we note that if $m\geq m_{0}$, then $k^{m}\geq(t-1)n2^{m}\geq n_{0}2^{m}$ for all $k\geq 5$ and hence no sum with $n_{0}$ terms and equal to $n_{0}2^{m}$ can contain any $k^{m}$ for $k\geq 5$.

Note that in the proof of the previous theorem we did not need to prove that no sum could contain a $k^{m}$ for any $k\geq 5$ in order to obtain an upper bound. We included that fact in order to add credence to our conjecture below. Indeed we believe that the number presented in Theorem 5.1 is the actual seed number. We state this in the following conjecture. One can see that both Theorem 3.6 and Theorem 4.3 follow from this conjecture.

Conjecture 5.2

Given $t$, there exists a number $m_{0}$, such that if $m\geq m_{0}$, then the $t-1$st smallest of the values $al_{3}+bl_{4}$ over all $a,b\geq 0$ is the seed number $S(m,t)$ and the seed value $V(m,t)=S(m,t)\times 2^{m}.$

6 Conclusion

The generalized taxicab number $T(n,m,t)$ is equal to the smallest number that is the sum of $n$ $m^{th}$ powers in $t$ ways. This definition is inspired by Ramanujan’s observation that $1729=1^{3}+12^{3}=9^{3}+10^{3}$ is the smallest number that is the sum of two cubes in two ways and thus $1729=T(2,3,2)$. In this paper we first proved that for any given positive integers $m$ and $t$, there exist a seed for the generalized taxicab number, i.e. there exists a number $s=S(m,t)$ such $T(s+k,m,t)=T(s,m,t)+k$ for every $k\geq 0$. We then found explicit expressions for this seed number when the number of ways $t$ is 2 or 3. We ended with a general theorem and conjecture about the seed number $S(m,t)$ for all $t$.

Addendum: Research for this paper was mostly undertaken while the authors were together in their first year of graduate school in mathematics at The Ohio State University in 1974. This paper should have appeared shortly after that, but at least it is finally finished now. (It only took another 45 years).