408 
REV. T. P. KIRKMAN ON A CLASS OF POLYEDRA, 
differing- in the manner of cutting- the triangle through which the axis of reversion 
does not pass, giving two reversible 11-edra. Three irreversibles can be cut from it, 
one by three, another by four, a third by five sections, as is evident on a moment’s 
consideration, giving a lO-edron, 11-edron, and 12-edron, 
We can now collect into one group all the formulae above deduced, which con- 
tain the complete solution of our problem ; to find the number of these x-edra on an 
(x— l)-g-onal base. There is no ambiguity in the case of two (x— l)-gonal faces, for 
the figure is always identical with itself whichever be considered the base, and can 
have only two triangles. 
Let Il(x, k, 1) or IR(x, k, 1) be the number of irreversible (x+Zc+Zj-edra that can 
be cut to have {k-\-l) triangles from an irreversible or reversible x-edron having k 
triangles, the capital on the right denoting- the subject of operation : {k-\-l)<x. 
n(x, k, l) =li(x, k, /), 
lR(2x+l, 2k^\, 1) 
lTR^(4x+l, Ak, 1) =ii{x-\-\, k, \l), 
ITT(4x-l-l, Ak, 1) =i{n(2x+l, 2k, \l) — ii{x-\-\, k, 
RR^(4x+l, Ak, 1) =ii{2x-\-\, 2k, \l) — ii{x-\-\, k, \l), 
IR^(4x+ 1, 4/f, /) n{Ax-\-\, Ak, l)-\-2n{^-\-\, k,^-^—iiii{2x-\-\ , 2k, \l) ; 
R^R^(6x+l, Qk, 1) =^7(x-f 1, k, 1/), R^R«(7, 3, 3)=1, 
lTT(6x+ 1, 6/i-, 1) =^{/7(2x+l, 2k, i/) — k, ^l)], 
