all
multipartite numbers (p1p2p2 ...) into exactly two parts. We find

h2^2 = h4 - h3h1 + (h2)^2

h3^2 = h6 - h5h1 + h4h2

h4^2 = h8 - h7h1 + h6h2 + h5h3 + (h4)^2,

and paying attention to the fact that in the expression of h_r2 the term
(h_r)^2 is absent when r is uneven, the law is clear. The generating
function is

h2x^2 + h2h1x^3 + (h4 + h2^2)x^4 + (h4h1 + h3h2)x^5 + (h6 + 2h4h2)x^6
+ (h6h1 + h6h2 + h4h3)x^7 + (h8 + 2h6h2 + h4^2)x8 + ...

Taking h4 + h2^2 = h4 + {(2) + (1^2)}^2

= 2(4) + 3(31) + 4(2^2) + 5(21^2) + 7(1^4),

the term 5(21^2) indicates that objects such as a, a, b, c can be
partitioned in five ways into two parts. These are a|a, b, c; b|a; a, c;
c|a, a, b; a, a|b, c; a, b|a, c. The function h_{r^s} has been studied.
(See MacMahon, _Proc. Lond. Math. Soc._ vol. xix.) Putting x equal to
unity, the function may be written (h2 + h4 + h6 + ...)(1 + h1 + h2 + h3
+ h4 + ...), a convenient formula.

Method of differential operators.

The method of differential operators, of wide application to problems of
combinatorial analysis, has for its leading idea the designing of a
function and of a differential operator, so that when the operator is
performed upon the function a number is reached which enumerates the
solutions of the given problem. Generally speaking, the problems
considered are such as are connected with lattices, or as it is possible
to connect with lattices.

To take the simplest possible example, consider the problem of finding
the number of permutations of n different letters. The function is
here x^n, and the operator (d/dx)^n = [delta]_x^n, yielding
[delta]_x^n.x^n = n! the number which enumerates the permutations. In
fact--

[delta]_x.x^n = [delta]_x. x. x. x. x. x. ...,

and differentiating we obtain a sum of n terms by striking out an x
from the product in all possible ways. Fixing upon any one of these
terms, say x. [x]. x. x. ..., we again operate with [delta]_x by
striking out an x in all possible ways, and one of the terms so
reached is x. [x]. x. [x]. x. .... Fixing upon this term, and again
operating and continuing the process, we finally arrive at one
solution of the problem, which (taking say n = 4) may be said to be in
correspondence with the operator diagram--
([x] = striken-out x)

or say
+-------+-------+----