260 REPORT — 1875. 



tudes equal to each other and superior to those of the other main branches 

 (if any); or else (as in the right-hand figure) a hieentre AB, viz. two contigu- 

 ous knots, such that issuing from A (but not counting AB), and issuing from 

 B (but not counting BA), wo have two or more main branches, one at least 

 from A and one at least from B, of altitudes equal to each other and superior 

 to those of the other main branches in question (if any). The theorem once 

 understood, is proved without difficulty : we consider two terminal knots, 

 the distance of which, measured by the number of intermediate branches, is 

 greater than or equal to that of any other two terminal knots ; if, as in the 

 left-hand figure, the distance is even, then the central knot A is the centre 

 of the tree ; if, as in the right-hand figure, the distance is odd, then the two 

 central knots AB form the bicentrc of the tree. 



In the former case, observe that if G, H are the two terminal knots, the 

 distance of which is =2\, then the distance of each from A is = X, and there 

 cannot be any other terminal knot I, the distance of which from A is greater 

 than X (for if there were, then the distance of I from G or else from H would 

 be greater than 2\) ; there cannot be any two terminal knots I, J, the dis- 

 tance of which is greater than 2X ; and if there are any two knots I, J, the 

 distance of which is =2\, then these belong to different main branches, the 

 distance of each of them from A being = X ; whence, starting with I, J (in- 

 stead of G, H), we obtain the same point A as centre. Similarly in the 

 latter case there is a single bicentre AB. 



Hence, since in any tree there is a unique centre or bicenti-e, the question 

 of finding the number of distinct trees with n knots is in fact that of finding 

 the number of centre- and bicentre- trees -^ith n knots ; or say it is the problem 

 of the " general centre- and biceutre-trees with n knots:" general, inasmuch 

 as the number of branches from a knot is as yet taken to be without limit ; 

 or since (as will appear) the number of the biccntrc-trees can be obtained 

 without difliculty when the problem of the root-trees is solved, the problem 

 is that of the " general centre-trees -with n knots." It will appear that the 

 solution depends upon and is very readily derived from that of the foregoing 

 problem of general root-trees, so that this last has to bo considered, not 

 only for its own sake, but with a view to that of the centre-trees. And in 

 each of the two problems we doubly divide the whole system of trees 

 according to the number of the main branches (issuing from the root or centre 

 as the case may be), and according to the altitude of the longest main branch 

 or branches, or say the altitude of tho ti'ce ; so that the problem really is, for 

 a given number of knots, a given number of main branches, and a given 

 altitude, to find the number of root-trees, or (as the case may be) centre- 

 trees. 



We next introduce the restriction that the number of brandies from any 

 knot is equal to a given number at most ; viz. according as this number is 

 = 2, 3 or 4, we have, say oxygon-trees, boron-trees*, and carbon-trees re- 

 spectively ; and these are as before root-trees or centre- or bicentre-trees, as 

 the case may be. Tho case where the number is 2 presents no difficulty : in 

 fact if the number of knots be =n, then the number of root-trees is either 

 ^ (n-f-1) or |«; viz. «=3 and » = 4, the root-trees are 



* I should have said nitrogen-trees ; but it appears to me that nitrogen is of necessity 

 ."j-valent, as shown by the compound, Ammonium-Chloride, =NH4 CI : of course the word 

 boron is used simply to stand for a .S-Talent' element. 



