Counting Summations ¶

We want $p(100) - 1$, where $p(n)$ is the partition function. We subtract 1 because $p(n)$ counts $n$ by itself as a partition of $n$, but we only want partitions composed of two or more numbers.

Guess what? SageMath has this built-in.

In [1]:

number_of_partitions(100) - 1

Out[1]:

190569291

Alternatively, if we think of this problem like problem 31 - just with coins of every possible denomination instead of only a few - we can adapt any of our approaches to that problem, where we construct a generating function.

In [2]:

R.<x> = PowerSeriesRing(ZZ, default_prec=101)
G = 1 / prod(1 - x^n for n in range(1, 100))
G[100]

Out[2]:

190569291

Relevant sequences¶

Partition numbers: A000041

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