540 Mr. A. Cayley on a Question in the Theory of Numbers. 
g-N!, that is, the several numbers each of which has a given 
number N! of numbers less than and prime to itself. This may 
be effected as follows: let @ be any prime number, and multiply 
together all the different series of the form 
1+(a—1) [a] +a(a—1)[a?]... +a*"(a—1)[a*]+.., 
where the bracketed factors are to be multiplied together by 
enclosing the product in a bracket; the general term of the 
product is obviously 
at B8-1, . (a—1)(b—1) .. [a*d?..]. 
Or putting a*b@..=N, the general term is ¢N[N]; and if 
@N=N’, then the general term is N'[g-1N']. Hence in the 
product in question, each of the bracketed numbers which are 
multiplied by the coefficient N’ will be a value of $—"N’, that is, 
a number having N! numbers less than and prime to itself. The 
only values of a which need be considered are obviously those 
for which a—1 is a divisor of N’; and each series need only be 
continued during so long as the coefficient a*—\(a—1) is a divisor 
of N’.. Thus if N'/=12, the divisors of N! are 1, 2, 3, 4, 6 and 
12, and the values of @ are 2, 3 (4 is not a prime number), 5, 7 
and 12: the series to be multiplied together are 
141[2] +2[27] +4 [2°], 
1+42[3]+6[37], 
144[5], 
1+6(7], 
1+12[13], 
and the product contains the terms 
12 [2737] +12 [277] +12 [2.3.7] +12[2.13] +12[8.7] +12[13], 
i. e. the required numbers are 36, 28, 42, 26,21 and 13; orin 
order of magnitude, 13, 21, 26, 28, 36 and 42. The rule may 
be also stated as follows: write down all the numbers of the 
form a*-1(a—1) where a is prime, which are divisors of N‘, and 
combine these divisors in every possible way so as to give the 
product N'; the corresponding values of a*0* .. are the required 
series of values. The question is in fact one of the partition or 
- decomposition of N! into factors of a given form. 
2 Stone Buildings, W.C. 
June 25, 1857. 
