let define $\{x_n\},\{y_n\}$ such $$x_{1}=1,x_{n}+y_{n}=an-1,a\in \mathbb{N}^{+}.a\ge 5$$ and such $x_{n+1}$ is smallest positive integer except $x_1,x_2,\ldots,x_n,y_1,y_2,\ldots,y_n$

Find the $x_{n}$ and $y_{n}$

this problem see it is interesting.But I can't find it.c

My try: since $$y_{1}=a-1-1=a-2\ge 3$$ and $$x_{2}+y_{2}=2a-1$$ and $$x_{2}>x_{1},y_{1}\Longrightarrow x_{2}>a-2\Longrightarrow x_{2}\ge a-1$$ since $x_{2}$ is smallest positive integer except $x_{1},y_{1}$,so $$x_{2}=a-1\Longrightarrow y_{2}=a$$ and $$x_{3}>y_{2}=a\Longrightarrow x_{3}=a+1\Longrightarrow y_{3}=3a-1-x_{3}=2a-2$$ and so on

It is said this two sequence called Complementary sequences.

But I can't solve this problem.

Thank you

  • $\begingroup$ Can you find the first few terms? Does any pattern emerge? $\endgroup$ – Gerry Myerson Mar 21 '14 at 11:41
  • $\begingroup$ Hello,@GerryMyerson,I Post the first few terms,and I can't following $\endgroup$ – user94270 Mar 21 '14 at 12:02
  • $\begingroup$ Maybe you should take some particular value of $a$, and calculate more terms than that. $\endgroup$ – Gerry Myerson Mar 21 '14 at 12:04

By writing down the first few terms of the series, we see that the series can be written as

$x_{2n-1} = a(n-1) +1, \quad y_{2n-1} = an-2$

$x_{2n}= an-1, \quad y_{2n} = an$

Lets prove it by induction. The base case is indeed true. Assume the statement is true for $n=k,k-1, \dots,2$ i.e

$x_{2k-1} = a(k-1) +1, \quad y_{2k-1} = ak-2$

$x_{2k}= ak-1, \quad y_{2k} = ak$

Then clearly $y_k$ is also increasing and hence $x_{k+1}=y_k+1$

Then $x_{2k+1} = y_{2k}+1 = ak+1 = a(k+1-1) + 1$ and hence $y_{2k+1} = a(2k+1) - 1 -(ak+1) = a(k+1)-2$

$x_{2k+2} = y_{2k+1}+1 = a(k+1)-2+1 = a(k+1)-1$ and $y_{2k+2} = a(2k+2) -1- (ak+a-1) = a(k+1)$

So we have proved the required statement.


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