of Edinburgh, Session 1879 - 80 . 
401 
Write down the numbers 
1 2 1 . 
Put the double of the middle number to the right of it, and the 
next lower number to the left. Thus 
1 3 2 4 1. 
Operate in the same way on the numbers last introduced ) and we 
have 
15362748 1. 
Continue in this way, and arrange these groups in successive order, 
leaving out the final 1 from each. We thus have the series 
1, 2, 1, 3, 2, 4, 1, 5, 3, 6, 2, 7, 4, 8, 1, 9, 5, 10, 3, 11, 6, 12, 2, 13, 7, &c. 
Strike off the first n - 1 of these numbers ( n being the number of 
boys), and the next n represent the arrangement of the class after 
all have been displaced : the numbers designating the several boys 
by their original places. Hence we have the key for translating 
the series into the successive derangements. 
Another curious mode of getting this series is to begin with 1, 
then prefix 1, and insert 2, as below : — 
1 2 1. 
Again prefix 1, and insert 2, 3, 4, then 
1 2 1 3 2 4 1, 
and so on indefinitely. 
It is worthy of remark that this series gives the integral of the 
equation 
u ix + 1 == U X ’) 
with the conditions 
u 2x = x + 1, 
u x = 1 \ 
i.e., the solution of the following question : — 
“ Arrange an infinite row of numbers, those in the even places 
being 2, 3, 4, &c., so that if the first ( n — 1) be struck off (n being 
any integer) the next n may consist of all the natural numbers 
from 1 to n inclusive.” 
Another result which these numbers present is the following : — 
