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The basic sums of distinct factorial numbers are: $0! = 1!$ $0! + 1! = 2!$ There are no equal sums of distinct factorial numbers which don't boil down to one of those two. Proof: consider the numbers in the factorial number system. The factorials are $0_{10}!=10_!$, $1_{10}!=10_!$, $2_{10}!=100_!$, $3_{10}!=1000_!$, $\ldots$. Because $n > 2 \implies ...


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I'm assuming from your list of examples that what you are looking for is the number of partitions of $n$ into $k$ distinct parts with largest part equal to $m$ (this is different from the question in the title, but seems to fit your examples best). The number of partitions of $n$ with at most $k$ parts, each of length at most $m$, is the coefficient of ...


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The partition function is asymptotically $\frac{1}{4n\sqrt{3}}e^{\pi\sqrt{2n/3}}$ and the Fibonacci numbers are asymptotically $\frac{1}{\sqrt{5}}\varphi^n$, where $\varphi$ is the golden ratio. The quotient of these two is less than $1$ if $n$ is large enough: $$\frac{\sqrt{5}e^{\pi\sqrt{2n/3}}}{4n\sqrt{3}\varphi^n} = ...


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Not a "theoretical area" per se, but IIRC, ATMs use partitions of integers to detemerine the number and denominations of currency notes that must be used to achieve the sum (of money) requested by the customer.


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The number of partitions of $n$ into $k_2$ positive parts, no part exceeding $k_1$, is the coefficient of $x^n$ in $(x+x^2+x^3+\cdots+x^r)^s$, where I have written $r$ for $k_1$, and $s$ for $k_2$. This is the coefficient of $x^t$ in $(1+x+x^2+\cdots+x^{r-1})^s$, where I have written $t$ for $n-s$. Now ...


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Keywords here are "block design", "Steiner triple system" and "resolvable". We want to find $8$ resolution classes: each partitioning $\{1,2,\ldots,30\}$ into $10$ sets of size $3$. This is going to be an ugly case, since $30 \not\equiv 1,3 \pmod 6$ (a condition for the existence of a Steiner triple system). [The question would be more mathematically ...


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For a closed form formula for the question: given n cents, how many ways can I make change using pennies, nickels, dimes, quarters see either: Making change of n cents by William Gasarch http://arxiv.org/abs/1406.5213 This just uses recurrences. OR Graham-Knuth-Patashnik in their book Concrete Mathematics got a closed form using generating functions. ...


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I am assuming by intersection of the two partitions you mean for each element $x$ in the set we make it the element of a new block of the new partition: the partitions of elements that belong to Block $a$ in the first partition, and block $\alpha$ in the second. Is this correct? Consider the following partition of the set $\{1,2,3,4,5,6,7,8\}$ ...



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