# Computing Partitions

Euler gave us the following recursion formula for $$p(n)$$ via his Pentagonal Number Theorem :

$$p(n) = p(n-1) + p(n-2)-p(n-5)-p(n-7) + p(n-12) + p(n-15)-…$$

Note the generalised pentagonal numbers in the above.

Define  $$p(n) = 0$$ if $$n \lt 0$$ and let $$p(0) = 1$$.

\begin{align} p(1) & = p(0) & = 1 & = 1\\ p(2) & = p(1) + p(0) & = 1 + 1 & = 2\\ p(3) & = p(2) + p(1) & = 2 + 1 & = 3\\ p(4) & = p(3) + p(2) & = 3 + 2 & = 5\\ p(5) & = p(4) + p(3)-p(0) & = 5 + 3-1 & = 7\\ p(6) & = p(5) + p(4)-p(1) & = 7 + 5-1 & = 11\\ p(7) & = p(6) + p(5)-p(2)-p(0) & = 11 + 7-2-1 & = 15\\ p(8) & = p(7) + p(6)-p(3)-p(1) & = 15 + 11-3-1 & = 22\\ p(9) & = p(8) + p(7)-p(4)-p(2) & = 22 + 15-5-2 & = 30\\ \end{align}

This is the formula that MacMahon used to compute $$p(n)$$ up to $$n = 200$$ by hand.

$$p(200) = 3,972,999,029,388$$

Ramanujan conjectured that $$p(14,031)\equiv 0\pmod {11^4}$$.

This has been verified by Lehmer.  The number has $$127$$ digits.

He used the asymptotic formula of Ramanujan-Rademacher :

$$\displaystyle p(n) \sim \frac {e^{\pi \sqrt {2n/3}}}{4n \sqrt 3}$$

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