If we divide the integers into groups of 3 ("ternary groups"), beginning with 3, we have group one $= 3,4,5$; group two $= 6,7,8$; and so on. If we ask which multiples of 5 belong to which groups, the answer is: 5 in group one, none in group two, 10 in group three, none in group four, and 15 in group five. So we have this pattern: $2\ \_\ 1\ \_\ 0$, where each value is the multiple of 5 modulo 3, and each index of the pattern corresponds to a ternary group. There are 5 places in the pattern, so the length of one cycle period is $5$. The blank value ("_") means no multiple of 5 is in the corresponding group. This pattern repeats for the next three multiples of 5, then the next three, on to infinity. The same applies to all values of $n$, e.g. the pattern for 7 is $\_\ 1\ \_\ 2\ \_\ \_\ 0$. The cycle is always of length $n$, and always ends with $3n \pmod 3$ (meaning it always end with 0, which corresponds to the third multiple of $n$). My question is, what is the formal rubric for this cycle? Where should I look to learn more about this phenomenon?
Pattern in arithmetic progression modulo 3
arithmetic-progressionsmodular arithmeticnumber theory
Related Solutions
A cyclic group is defined as a group which can be generated by a single element.
For example, let $G = \Bbb Z/5\Bbb Z = \lbrace 0, 1, 2, 3, 4\rbrace$ with addition. Then
$$\begin{align} 1 &\equiv 1 \bmod 5\\ 1 + 1 &\equiv 2 \bmod 5\\ 1+1+1 &\equiv 3 \bmod 5\\ 1+1+1+1 &\equiv 4 \bmod 5\\ 1+1+1+1+1 &\equiv 0 \bmod 5 \end{align}$$
and since these are all of the elements of $G$, we conclude that this group is cyclic, and is generated by $1$, which can be written $G = \langle 1\rangle$ (although this requires some context, simply writing $\langle 1\rangle$ doesn't give enough information).
The way the elements cycle comes from the way the group $\Bbb Z/n\Bbb Z$ is defined. The "number" in the set $\lbrace 0, 1, 2, \dots, n-1\rbrace$ actually refers to the remainder when an integer is divided by $n$. For example, in the case $n = 5$, we see that $23 = 4 \times 5 + 3$, so $23 \equiv 3 \bmod 5$. Since the only possible candidates for remainders after dividing by $5$ are the numbers $0, 1, 2, 3, 4$ (convince yourself of this; for example, if I wrote $23 = 3 \times 5 + 8$, I could take another $5$ from the $8$ and have $23 = 3 \times 5 + 5+ 3 = 4\times 5 + 3$), these are the elements of the group.
You are looking at numbers of the form $2n$.
Looking at your enumeration, the most obvious thing is that every second number is divisible by $2$ only once. That this happens is clear, as these are the numbers where $n$ from above is odd.
The second pattern you note is that every $3+4k$th term is divisible by $2$ only twice, and these are the only numbers divisible by $2$ twice. In order to make this more clear, remove all the numbers of the form $2(2k+1)$ from the list to get
$$0,2,3,2,4,2,...$$
These $2$'s correspond to numbers of the form $2(2 (2k+1))=8k+4$, which gives you every number divisible by $2$ exactly twice.
If we strip these two obvious patterns, what you write is
$$3,4,3,5,3,4,3,6,3,4,3,4$$
and here there is an error, the last term should not be a $4$ but at $5$ (it corresponds to $96=2^5\cdot 3^1$).
So we actually find the same pattern as in the other two instances, just that everything got a $+2$.
What happens is this: You are removing from the sequence all odd terms and then you recover the same sequence but with a $+1$. Thus the pattern you see is an artefact of the odd numbers having no $2$ divisors and the rule "remove every second term and you receive the same list but with $+1$ to everything".
Best Answer
This is an example of the Chinese Remainder Theorem. If you have coprime moduli $a$ and $b$ you can solve the equations $$x \equiv c \pmod a\\ x \equiv d \pmod b$$ for any given $c,d$. The solutions will recur at intervals of $ab$. In your example when you note that $5$ is the middle of a group of $3$ you are solving $$x\equiv 2 \pmod 3 \\ x\equiv 0 \pmod 5$$ As you noticed the first solution is $x=5$ and the next is $15$ higher or $20$ and so on.