I want to express the following in sequence:

$S$ is a sequene of natural numbers such that $n$ is repeated $x$ times, where $x$ is a positive integer.

In the following example, $x = 2n-1$

$$S = \{\underbrace{1}_\textrm{x = 1},\underbrace{2,2,2}_\textrm{x = 3},\underbrace{3,3,3,3,3}_\textrm{x = 5},\underbrace{4,4,4,4,4,4,4}_\textrm{x = 7},\dots\}$$

  • 1
    $\begingroup$ Sets don't have repeated elements (compare with multiset). Therefore $$S=\{1,2,3,4,\ldots\}=\mathbb{N}.$$ $\endgroup$ – Zev Chonoles Aug 8 '13 at 0:13
  • $\begingroup$ @ZevChonoles: Point. A sequence is what I want. $\endgroup$ – JohnWO Aug 8 '13 at 0:17
  • $\begingroup$ What you've done already is probably best: define $S$ and subsequently refer to $S$. You could get clever, e.g. define $\bigoplus$ (or something else) as sequence concatenation and define $S=\bigoplus_{n \geq 1} (n)_{i=1}^{2n-1}$, but it's more likely to cause more harm than good. (NB. This is Sloane's A003059.) $\endgroup$ – Douglas S. Stones Aug 8 '13 at 0:27
  • 1
    $\begingroup$ Curiously, this sequence is $(\lceil \sqrt{n} \rceil)_{n \geq 1}$. $\endgroup$ – Douglas S. Stones Aug 8 '13 at 0:35
  • $\begingroup$ @DouglasS.Stones yes, and with $x = 2n+1$ the sequence is $(\left \lfloor \sqrt{n} \right \rfloor)_{n\geq1}$ $\endgroup$ – JohnWO Aug 8 '13 at 0:53

There are a few ways you could do this, one of the easiest would be to let your sequence contain ordered pairs of natural numbers $(n_i,m_i)$ which you just choose to interpret as "the number $n_i$ repeated $m_i$ times."

Alternatively if you really wanted to have each member of the sequence actually be some number repeated that many times, just give an explicit definition in words like you have. E.g.

Take the sequence with has $n$th term the natural number $n$ repeated $x_n$ times (where $x_n$ is something you define).

There is nothing wrong with sequences having members which aren't numbers or sets of numbers or whatever, as long as you clearly define your sequence you should be fine.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.