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There are exactly 18 partitions of the integer 13 into 4 parts, as on the left of the table, and also 18 partitions into 5 parts, as on the right of the table:

$$\begin{array}{c|c} 10+1+1+1 & 9+1+1+1+1 \\ 9+2+1+1 & 8+2+1+1+1 \\ 8+3+1+1 & 7+3+1+1+1 \\ 8+2+2+1 & 7+2+2+1+1 \\ 7+4+1+1 & 6+4+1+1+1\\ 7+3+2+1 & 6+3+2+1+1\\ 7+2+2+2 & 6+2+2+1+1\\ 6+5+1+1 & 5+5+1+1+1\\ 6+4+2+1 & 5+4+2+1+1\\ 6+3+3+1 & 5+3+3+1+1\\ 6+3+2+2 & 5+3+2+2+1\\ 5+5+2+1 & 5+2+2+2+2\\ 5+4+3+1 & 4+4+3+1+1\\ 5+4+2+2 & 4+4+2+2+1\\ 5+3+3+2 & 4+3+3+2+1\\ 4+4+4+1 & 4+3+2+2+2\\ 4+4+3+2 & 3+3+3+3+1\\ 4+3+3+3 & 3+3+3+2+2 \end{array}$$

Is there some reason, such as a natural bijection, that one might have expected there to be the same number of partitions of 13 into either 4 or 5 parts? Or is it a mere coincidence, the Law of Small Numbers at work?

Finding a bijection is of course trivial, but I have not been able to find one that seems to respect the structure of the partitions. The most natural sort of bijection would be a member of an infinite family of such bijections.

Or perhaps there is an explanation of this sort: “So-and-so's theorem states that under certain conditions $p(n, k) = c\cdot p(n, f(k))$, and as it happens this is the single case of so-and-so's theorem where $c=1$.”

Similarly, there are 23 partitions of 14 into 4 parts, and also 23 partitions of 14 into 5 parts. Coincidence, or is there something interesting going on?

MJD
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  • 13 is a prime and 14 is in the form pq for p,q prime. Have you noticed this to some other random number not prime and not of some specific form as 14? – Kal S. Dec 29 '13 at 03:31
  • Isn't every number of some specific form such as $p$, $pq$, $pq+1$, $p^2q$, etc.? Is there any reason to believe this occurrence is related to the primeness of 13 or the $pq$-ness of 14? – MJD Dec 29 '13 at 03:38
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    It's possible that it would be fruitful to look at this not as dividing 13 into 4 parts each $\geq$ 1 but as dividing 9 into 4 parts each $\geq$ 0, or counting the distinct ways of putting 9 items in 4 buckets. Nothing leapt out at me when I tried it (why are there the same number of ways to put 9 items in 4 buckets as 8 items in 5 buckets?) but it may enable the use of other tools in your toolbox. – dfan Dec 29 '13 at 03:42
  • No I have no clue if this is related to the specific form of 13 or 14. I was just curious whether you noticed this to occur in numbers whose prime factorization is not $p$ or $pq$ or $p^2q$ or $p_1^1p_2^1...p_k^1$. Maybe this occurrence is related to the form of the prime factorization. – Kal S. Dec 29 '13 at 03:44
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    Further such coincidences might include: $p(34, 4) = p(34,17) = 297$; $p(43,9) = p(43,11)=5708$; $p(60,4) = p(60,36) = 1575$, after which I find nothing of interest for $n\le 200$. – MJD Dec 29 '13 at 03:54
  • I vote for coincidence. Why 4/5 parts and not 2/3 or 3/4 parts, for example? – Greg Martin Dec 29 '13 at 04:53
  • There are also examples with 2/3 and 3/4: $p(5,2) = p(5,3)$ and $p(6,2) = p(6,3)$, and $p(8,3) = p(8,4)$. I didn't consider them surprising enough or interesting enough to mention, but I suppose they might be related. – MJD Dec 29 '13 at 04:56

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