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3

I don't currently have Sage installed but browsing the documentation seems to indicate that taking a closed interval (with one end of the interval being your partition in question and the other end being the finest, all-ones partition) of the IntegerPartitions poset should do the trick. EDIT: I just tested it on SageMathCell, an online Sage interface, and ...


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Good question. But the list of refinements of a composition is implemented. Thus you can do the following: def finer(p): # Return the list of all partitions refining the given partition ``p``. # # EXAMPLES:: # # sage: finer(Partition([3,2,1])) # [[1, 1, 1, 1, 1, 1], [2, 1, 1, 1, 1], [2, 2, 1, 1], [3, 1, 1, 1], [3, 2, 1]] ...


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As Doug M said what you are doing is obtaining the modulo $9$ of the number, it is also called the Digital root of the number. An excerpt from Wikipedia: The digital root (also repeated digital sum) of a non-negative integer is the (single digit) value obtained by an iterative process of summing digits, on each iteration using the result from the ...


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Let $a_n$ be the number of ways to partition $\{1,\dots, 2n\}$. Then $a_{n+1}=a_{n}+(2n)(2n-1)a_{n-1}$, because either $2n+2$ and $2n+1$ are in the same set, and we have $a_n$ ways to finish, or they are in different sets and then there are $(2n+2)(2n+1)a_{n-1}$ ways to finish. Now by induction it is easy to see that the answer is $(2n-1)!!=(2n-1)(2n-3)\dots ...


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Induction on $n$. For $n=1,2$ there is nothing to prove. Assume the result is proved for $<n$ and consider the case $n$. We establish the case $n$ by induction on $m$. If $n$ is odd the smallest possible value of $m$ is $m=n$. We have $n=(n-1)+1=(n-2)+2=\dots=\frac{1}{2}(n+1)+\frac{1}{2}(n-1)$, so the result is true for $m=n,k=\frac{1}{2}(n+1)$. If $n$...


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Very useful paper, though this only gives a good approximation to the number of partitions of n into exactly k parts each no larger than N due to Ratsaby (App. Analysis and Discrete M. 2008): http://www.doiserbia.nb.rs/img/doi/1452-8630/2008/1452-86300802222R.pdf



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