Another pentagonal number theorem

An earlier post presented Euler’s pentagonal number theorem. This post presents a similar theorem by N. J. Fine developed two centuries later.

Define the jth pentagonal number by

Pj = j(3j – 1) / 2

where j can be any integer, e.g. j can be negative.


Let De(n) is the number of distinct partitions of n of even length and Do(n) is the number distinct partitions of odd length. (See this post if any of these terms are unfamiliar.)

Euler’s pentagonal number theorem says

D_e(n) - D_o(n) = \left\{ \begin{array}{ll} (-1)^j & \mbox{if } n = P_j \\ 0 & \mbox{otherwise} \end{array} \right.


Euler discovered his theorem in 1741 and proved it a few years later. N. J. Fine published [1] the following variation on Euler’s theorem in 1948.

Define πe(n) to be the number of partitions of n into distinct numbers, the largest of which is even, and define πo(n) to be the number of partitions of n into distinct numbers, the largest of which is odd. Then Fine’s theorem says

\pi_e(n) - \pi_o(n) = \left\{ \begin{array}{ll} -1 & \mbox{if } n = P_j \mbox{ for } j > 0 \\ \phantom{-}1 & \mbox{if } n = P_j \mbox{ for } j < 0 \\ \phantom{-}0 & \mbox{otherwise} \end{array} \right.

Note that Euler and Fine have different ways of defining parity, the length of a partition versus the parity of the largest element. So, for example,

{8, 3, 2}

would count as an odd partition of 13 for Euler’s theorem, because it has length 3, but an even partition for Fine’s theorem, because the largest element is 8.

Both produce ±1 for pentagonal numbers and 0 otherwise. But the sign in Euler’s theorem depends on the parity of the index j. The parity in Fine’s theorem depends on the sign of j.


As in the previous two posts, we will illustrate the theorem here with Mathematica.

    distinctPartitionQ[list_] := Length[list] == Length[Union[list]]
    check[n_] := 
        Total[Map[(-1)^Max[#] &, 
              Select[IntegerPartitions[n], distinctPartitionQ]]] 

    Table[{n, check[n]}, {n, 1, 8}]

This returns

    {{1, -1}, {2, 1}, {3, 0}, {4, 0}, 
     {5, -1}, {6, 0}, {7, 1}, {8, 0}}

We get -1 for n = 1 = P1 and for n = 5 = P2.

We get 1 for n = 2 = P-1 and for n = 7 =  P-2.

We get 0 for all other inputs because they are not pentagonal numbers.


[1] I found this in “Euler’s Pentagonal Number Theorem” by George E. Andrews, Mathematics Magazine, Vol. 56, No. 5 (Nov., 1983), pp. 279-284. Andrews cites “Some new results on partitions” by N. J. Fine, Proc. Nat. Acad. Sci. U.S.A., 34 (1948) 616-618.


Leave a Reply

Your email address will not be published. Required fields are marked *