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Let $\pi$ be the partition of $n=a_1+a_2+...+a_r$, where $a_1\geq a_2 \geq ....\geq a_r\gt0$. Prove that number of partition $\pi$ of $n$ with $a_r=1$ and $a_j-a_{j+1}$ = 0 or 1 for $1\leq j\leq r-1$ equals $p^d (n)$, i.e the number of partitions of n into distinct parts.

What i have so Far:

I can see that since $a_r=1$ and $a_j-a_{j+1}$=0 or 1 so putting j = r-1 we have $a_{r-1}=1$. Similarly all $a_j$ must be 1.This implies that no two $a_j$ are equal. Please help

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  • $\begingroup$ This sounds very similar to your previous question? Perhaps we/you should first solve the other one before posting new ones immediately. $\endgroup$ Feb 21, 2019 at 15:40
  • $\begingroup$ @Dietrich Burde Any help on either of them would be appreciated $\endgroup$
    – Ayan Shah
    Feb 21, 2019 at 16:06

1 Answer 1

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Set $$\epsilon_j=a_j-a_{j+1} \quad \text{ for } \quad 1\leq j\leq r-1.$$ Then we have
$$ a_{r-j}=1+\epsilon_{r-1}+\cdots +\epsilon_{r-j} \quad \text{ for } \quad 1\leq j\leq r-1,$$ so $$ n=r+(r-1)\epsilon_{r-1}+ (r-2)\epsilon_{r-2}+\cdots +2\epsilon_2+\epsilon_1 $$ defines a partition of $n$ with all distinct parts (ignoring the summands with $\epsilon_j=0$) and highest part being $r$. The correspondence thus defined is bijective: follow the same steps in reverse order.

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  • $\begingroup$ can you please point out if there is any mistake in my reasoning? $\endgroup$
    – Ayan Shah
    Feb 21, 2019 at 18:39
  • $\begingroup$ Also, can you elaborate on the equivalence of data you mentioned? $\endgroup$
    – Ayan Shah
    Feb 21, 2019 at 18:41
  • $\begingroup$ I do not understand it: $a_r=1$ and $a_{r-1}-a_r=a_{r-1}-1=0$ or $1$ only implies $a_{r-1}=1$ or $2$. $\endgroup$
    – user135826
    Feb 21, 2019 at 18:43
  • $\begingroup$ OOH...that was a mistake, thanks. Can you elaborate on the last line of your proof? $\endgroup$
    – Ayan Shah
    Feb 21, 2019 at 18:53
  • $\begingroup$ I edited the answer to clarify. $\endgroup$
    – user135826
    Feb 21, 2019 at 18:56

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