Looking for examples that illustrate the use of the $\operatorname{seq}$ notation

Question

This question is about the $\operatorname{seq}$ "notation" (for lack of a better word) defined in

https://math.stackexchange.com/a/312915/13675

Can someone give some concrete examples illustrating the $\operatorname{seq}$ notation?

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ASIDE

For what it's worth, in a response to an earlier request for clarification of $\operatorname{seq}$, the author of the cited answer writes (in a comment):

$\operatorname{seq}(a, b, k, x)$ means "$x$ is the $k$th term in the somewhat strangely defined sequence $a \operatorname{mod} b + 1, a \operatorname{mod} 2b + 1, a \operatorname{mod} 3b + 1, \dots$". Remarkably, any finite sequence can be represented this way.

I don't understand what the author means by the last remark. How, for example, does one use "this way" (whatever that means) to represent the finite sequence $(3, 1, 4, 1, 5, 9, 2, 6, 5, 3)$? Also, what values of $a$, $b$, $k$, and $x$ would one use in $\operatorname{seq}(a, b, k, x)$ to assert that "3 is the 10-th element of $(3, 1, 4, 1, 5, 9, 2, 6, 5, 3)$"?

NOTE, however, the point of this question is not to decipher the author's comment cited above. That is secondary. What I'm interested here is in examples that illustrate the meaning of $\operatorname{seq}$.

Answer 1

I personally think what the author had in his twisted mind was this:

To encode the finite sequence $a_1,a_2, \ldots, a_n$ "this way", we first need to find a suitable $b$. In order to apply the Chinese Remainder Theorem, the numbers $b+1,2b+1,\ldots, nb+1$ should be coprime. But any common divisor of $ib+1$ and $jb+1$, $1\le i<j\le n$, must also divide their difference $(j-i)b$; in fact, as $b$ is coprime to each of these, said common factor must divide $j-i$, a number $\le n-1$. Is there an easy way to prevent $kb+1$ from having any divisors $<n$ at all? Sure, just make $kb$ a multiple of these. In other words, it suffices to pick $b=(n-1)!$ or $b=\operatorname{lcm}(1,2,3,\ldots, n-1)$ in order to guarantee that our intended application of the Chinese Remainder Theorem will succeed.

Of course, there is another constraint: We must have $a_k<kb+1$ for all $k$. But of course we can just let $b$ a sufficiently high multiple of $\operatorname{lcm}(1,2,3,\ldots, n-1)$.

So in your concrete example of a ten-term sequence, we can take $$b=\operatorname{lcm}(1,2,3,\ldots,9)=2520.$$ Solving the simultaneous congruence $a\equiv 3\pmod {2521}$, $a\equiv 1\pmod {5041}$, $\ldots$, $a\equiv 3\pmod {25201}$ tells us that $$a=23362866138146210613972657326044344349283$$ works. To summarize: The numbers $a,b$ found here, have the property that - at least for $1\le k\le 10$ - the statement $\operatorname{seq}(a,b,k,x)$ is true if and only if $x$ is the $k$th decimal of $\pi$. In particular, $$\operatorname{seq}(23362866138146210613972657326044344349283,2520,10,3) $$ expresses that the tenth decimal of $\pi$ is $3$. Of course, no actual knowledge of $\pi$ flew into the computation of $a,b$ above, only the given and otherwise arbitrary sequence of terms. Also, the fact that $$\operatorname{seq}(23362866138146210613972657326044344349283,2520,11,2416) $$turns out to be true should lead no-one to believe that the next digit of $\pi$ is $2416$.

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Incidentally, my original answer already contained a useful example in the context of "is a power of ten": With $a=41366298973$ and $b=250$, we verify that $$ \begin{align}a\bmod 251&= 1\\ a\bmod 501&= 10\\ a\bmod 751&= 100\\ a\bmod 1001&= 1000 \end{align}$$ So that for these values of $a,b$, the claim $$ a\bmod ((k+1)b+1)=10\cdot(a\bmod (kb+1))$$ holds at least for $k\in\{1,2,3\}$, i.e., the first few terms are in a geometric progression with the factor $10$. The fact "$1000$ is a power of $10$" is thus expressed as "there exists a (finite) geometric sequence with factor $10$ that starts with $1$ and ends in $1000$".

Answer 2

There's a bit of an elaboration in Andrés E. Caicedo's answer to the same question you linked to. I'll base off his ideas to work through an example. You'll see that the numbers involved can get quite large, so let's stick to a relatively basic example: the sequence $(3,1,0)$.

We choose some number $t$ which is at least as big as all the terms in the finite sequence, and also at least as big as the number of terms in the sequence. So, for this example, where there are $3$ terms in the sequence and the highest term is $3$, we can choose $t=3$. We set $b = t!$; in our case, $b=6$. Then we choose $a$ so that the following relations hold:

$$ \begin{align} a &\equiv 3 \mod 7 \\ a &\equiv 1 \mod 13 \\ a &\equiv 0 \mod 19 \end{align}$$ The numbers $7$,$13$,$19$ will be in general $b+1,2b+1,3b+1,\dots$ The reason for the particular choice of $b$ is that these numbers are pairwise relatively prime: if we choose any pair $mb+1,m'b+1$, the difference will be $nb$ (where $n=|m-m'|$). Any common factor of $mb+1$ and $m'b+1$ must also be a factor of $nb$. Clearly none of the factors of $b$ divide $mb+1$, so it must be a factor of $n$. But all the factors of $n$ divide $b$, since $n$ is $\leq$ than the number of terms and thus $\leq t$, and $b=t!$. So there can be no common factor.

The Chinese Remainder Theorem guarantees that this system of congruence equations, in the case that the moduli are pairwise relatively prime, always has a solution. In this case, $a=703$ gives a solution.

Conversely, given the pair of numbers $a=703,b=6$, we can recover the sequence by doing the divisions and taking the remainders. Thus, in Hagen von Eitzen's notation, the following sentences hold: $$ \operatorname{mod}(703,7,3)\\ \operatorname{mod}(703,13,1)\\ \operatorname{mod}(703,19,0) $$ And thus, by definition, the following sentences hold: $$ \operatorname{seq}(703,6,0,3)\\ \operatorname{seq}(703,6,1,1)\\ \operatorname{seq}(703,6,2,0) $$ In practice, when you're using the sequence in Peano arithmetic, you don't care what $a$ and $b$ are or how to find them, so you'll usually be using something like the following sentence (which also holds): $$ \exists a \exists b : \operatorname{seq}(a,b,0,3) \land \operatorname{seq}(a,b,1,1) \land \operatorname{seq}(a,b,2,0) $$

I hope this example was illustrative. A few closing remarks:

• The choices of $a$ and $b$ are not unique. All we need to encode the sequence is that they exist. Thus you'll see $\operatorname{seq}$ used with $\exists$s.
• I've used ideas external to Peano arithmetic, such as the Chinese Remainder Theorem, in this exposition. That's okay; I'm just trying to prove that the sequence is encodable within Peano arithmetic, i.e., that the last sentence above holds. That sentence doesn't choose $a$ and $b$ explicitly, it just says that there exists some choice which encodes the sequence, which is true. If you're trying to prove within Peano arithmetic that the encoding always exists, you have to work to develop these ideas within Peano arithmetic; Andrés E. Caicedo discusses this in his answer, but I believe it's beyond the scope of this question.