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For any natural number and chosen base p, the number admits a unique expression of the form $a_np^n + ... + a_2p^2 + a_1p^1 + a_0$, where $a_k < p$ for all k. This property is effectively what makes our Hindu-Arabic positional numeral notation system possible.

When you multiply two polynomials (or power series) together, you're essentially taking the discrete convolution of their ordered sequences of coefficients. This is essentially what we're doing when we multiply numbers in a positional numeral system, except that there's an additional "carry" step involved to put the number back into its canonical representation.

Another useful type of infinite series is a Dirichlet series, which looks like $ ... + a_3 3^p + a_2 2^p + a_1$; p is now the exponent rather than the base. Could this sort of series also be the foundation for a numeral system?

For instance, is there a way to canonically represent any number as a finite sum $a_n n^p + ... + a_3 3^p + a_2 2^p + a_1$? When multiplying two numbers in this format, would multiplication resemble Dirichlet convolution rather than ordinary convolution, with some other type of "carry" to make the whole thing work out?

I'm interested to see how this would work - or if it would fail, and if so, why!

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$$\begin{align*}42&=1\times 1^2+4\times 2^2+1\times 3^2+1\times 4^2\\&= 4\times 1^2+1\times 2^2+2\times 3^2+1\times 4^2\\&= 5\times 1^2+3\times 2^2+1\times 3^2+1\times 4^2\\&= 9\times 1^2+2\times 2^2+1\times 3^2+1\times 4^2\\\end{align*}$$ –  Guess who it is. May 2 '13 at 9:36
$\begin{align*}1234&=1\times 10^3+2\times 10^2+3\times 10^1+4\times 10^0\\&=0\times 10^3+12\times 10^2+3\times 10^1+4\times 10^0\\&=0\times 10^3+0\times 10^2+123\times 10^1+4\times 10^0\\&=0\times 10^3+0\times 10^2+0\times 10^1+1234\times 10^0\\\end{align*}$ –  Mike Battaglia May 2 '13 at 18:34

1 Answer 1

up vote 2 down vote accepted

There's a bit of subjectivity in this problem, so I'm just presenting my thoughts/opinions.

I'll use numbers in brackets to represent finite Dirichlet series. If $p=1$, then to represent $n$, $[1000\ldots 0]$ works and is arguably the nicest representation. If $p=2$, then the squares are 1,4,9,16,25,36... Since $9+16=25$, then $[1100]=[10000]$, and we have to decide which we'd want.

Since the convention used in base $\varphi$ notation is to avoid adjacent 1's, we should probably prefer $[10000]$. Now, if our sole priority is to minimize the number of nonzero entries, then every non-squarefree number has a bunch of very boring $p=2$ representations. To make it more like a standard base notation, one idea I had was that we should try to minimize the sum of our digits.

The first several positive integers have a nice pattern for a while: $[1]$,$[2]$,$[3]$,$[10]$,$[11]$,$[12]$,$[13]$,$[20]$,$[100]$,$[101]$,$[102]$ But then we get to $[30]$=$[103]$. The problem is that aesthetically, $[30]$ might be bad because we lose the idea that once you're using digits in a certain place, you'll need that place (or higher) for all higher numbers. If you accept $[30]$ because it has a smaller digit sum, then there might be a unique way to write everything with minimal digit-sums; I haven't given it much thought.

But if you're like me, and you don't accept it, then you might want to put caps on the values in each place: e.g. the 4's place shouldn't have anything more than 2 in it since 2*4+1=9. This naturally leads to a sort of greedy algorithm, where the largest square less than the number gets the highest digit it can, and then the largest square less than what's left does, etc. You get a slightly weird continuation like $[103]$,$[110]$,$[111]$,$[112]$,$[1000]$ (better than $[113]$),$[1001]$,$[1002]$ (since $[200]$ is not allowed), $[1003]$,$[1010]$,$[1011]$,$[1012]$,$[1013]$,$[1020]$,$[10000]$ (since this is better than $[1100]$). You can allow this greedy algorithm to give representations for any $p$, but I don't know how satisfying they'd be to you.

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