# General term of $1,2,3,10,11,12,13, \dots$

what is the 444th term of the series $1,2,3,10,11,12,13,20,21,22,23,30,31, \dots$

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One resource that comes often useful in these cases is The On-Line Encyclopedia of Integer Sequences oeis.org –  Andreas Caranti Feb 1 '13 at 10:44
In this particular case, cleavebooks.co.uk/scol/calnumba.htm may also be useful. –  Andreas Caranti Feb 1 '13 at 10:50

First of all, this is a sequence not a series. Ok, so how about this...

01,02,03,10,

11,12,13,20,

21,22,23,30,

31,…

Can you see the pattern? :)

If you do observe the pattern, the 444 term would be 1110

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Besides if this is not in base 10, then thats a different approach :) –  Roupam Ghosh Jan 20 '11 at 10:33

Based on comments on other answers, you may have guessed that the given information does not determine a unique answer. As to a hint at what the commenters might be thinking (which would also be my first guess), a number like 12571 is normally taken to mean $1\cdot 10^4+2\cdot 10^3+5\cdot 10^2+7\cdot 10^1+1\cdot 10^0$—that is, we normally interpret strings of digits to be in "base 10" where the digits read from the right represent coefficients of successively higher powers of 10. In base 10, we use digits 0-9. Your sequence only uses digits 0-3. Supposing that your sequence were not in base 10, what would this suggest?

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What do you think the next 10 terms are? Then how can you express the nth term in terms of n/4 and n mod 4?

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Of course, there is a potential trap if it hasn't been specified that the terms given are numbers in base 10... –  Peter Taylor Jan 20 '11 at 9:28

By observation, the $n$th term in the sequence may be given by $10\lfloor\frac{n}{4}\rfloor +k$ where $k$ is the remainder when $n$ is divided by $4$ and $\lfloor x\rfloor$ denotes the largest integer not larger than $x$. Since $444$ is a multiple of $4$, the $444$th term is simply $1110$.

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But what comes after 33? 100, surely? –  TonyK Jan 20 '11 at 10:19

12330 since 444 is 1$\times4^4+$ 2 $\times4^3+$ 3 $\times4^2+$ 3 $\times4+$ 0 $\times1$.

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Acknowledgements to @TonyK for this. –  Did Feb 1 '13 at 10:59