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It is easy to see that every natural number $n$ can be written in a unique way $n = a+b$ where $gcd(a,b)=1$, $b>a$ and $b-a$ is minimal with this property. For instance if $n$ is odd the representation is given by $n = \frac{n-1}{2}+\frac{n+1}{2}$. If $n=0(2)$, $n\neq0(4)$ then this representation is given by $n= (n/2+2)+(n/2-2)$. If $n=0(4)$ then the representation is given by $n = (\frac{n}{2}+1)+(\frac{n}{2}-1)$. Define for every natural number $n$ the following polynomial:

If $n=0(3)$, then $p_n(x) = \frac{n}{3}x+\frac{n}{3}$

If $n=1(3)$, then $p_n(x) = \frac{n-1}{3}x+\frac{n+2}{3}$

If $n=2(3)$, then $p_n(x) = \frac{n+1}{3}x+\frac{n-2}{3}$

(1) Is it true, that if $c\ge3$, $c=a+b$ is the decomposition in coprime integers then $p_c(x) = p_a(x) + p_b(x)$?

Edit: Yes, The proof is a bit lengthy, but working $mod(12)$ gives the desired result as Gerry Myerson suggested.

(2) This might seem a bit far-fetched but I did some computer experiments and it seems that in the situation of (1) we have: $c \le rad(abc)-1$ Does somebody know, if this inequality has a chance to be true or not?

I did some search at OEIS and it seems that:

The sequence $p_n(0)$ appears in https://oeis.org/A008611

The sequence $p_n(1)$ appears in https://oeis.org/A004396

The sequence $p_n(2) = n$ are the natural numbers.

The sequence $p_n(3)$ appears in https://oeis.org/A042965

For $m\ge 3$ it seems that the sequence $p_n(m)$ consists of exactly those natural numbers which are congruent to $\{0,1,m\}$ mod $m+1$

For instance $p_n(4)$ appears in https://oeis.org/A008854

I think I found a way how to define $p_n(x)$ more "compact": Let $p(t,x) := \frac{t^2}{(t-1)^2(1+t+t^2)}x+\frac{t(1-t+t^2)}{(t-1)^2(1+t+t^2)} = \sum_{n=0}^\infty {(a_n x+b_n) t^n} = \sum_{n=0}^\infty { p_n(x) t^n} $

Thanks for your help!

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  • $\begingroup$ If you know what $n$ is modulo 12, then you know what $a$ and $b$ are modulo 3, so you can work out $p_a(x)$ and $p_b(x)$ and see whether (1) holds, right? $\endgroup$ – Gerry Myerson Feb 13 '17 at 21:56
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    $\begingroup$ Dear @GerryMyerson: Thank you for your comment. Working modulo 12 is really helpfull. I proved the equality when n is odd! I think that when n is even a case by case inspection will work.Thanks. $\endgroup$ – orgesleka Feb 14 '17 at 7:45

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