The identity of the difference circles 1 2 4 and 3 6 5 consists in this, that either of them can be made to take the form of the other, by setting down, instead of single elements, syllables of contiguous elements; thus 1 2 4 is

3 6 5 is

(1 + 2, 2 + 4, 4 + 1) = 3, 6, 5, and

(3 + 6, 65, 5 + 3) = 2, 4, 1.

And by virtue of this identity they will both give the same arrangements of triads. Thus using 3 6 5 for our difference circle instead of 1 2 4, we obtain the same two sets of triads as before;

If C stands for the circle 1 2 4, we may represent the above results by the equations (to modulus 7),

the truth of which depends on the fact that every number below seven is of the form 2a 36 to modulus 7, thus

2.3°, 213o, 2o31, 223o, 2o31, 2.31, are (to modulus 7) the numbers 1, 2, 3, 4, 5, 6. Or since

3=-22, (modulus 7)

all those six numbers are of the form 2m to that modulus.

It is readily found by trial that 1 2 6 4 is a perfect partition of 13. It becomes a matter of interest to examine whether the difference circles obtained by multiplying this by the successive numbers 2, 3, . . . 12, will be the same or different circles. Now

2 (1, 2, 6, 4) = 2, 4, 12, 8

3(1, 2, 6, 4,) = 3, 6, 5, 12.

The circle 3 6 5 12 is 1 2 6 4, because

(3, 6, 5, 12) = (1 + 2, 6, 4 + 1, 2 + 6 + 4), and

(1, 2, 6, 4) = (3 + 6 + 5, 12 + 3, 6, 5 + 12);

but 2 and 4 are contiguous elements in 2, 4, 12, 8, and non-contiguous in 1, 2, 6, 4; wherefore these circles are not the same. In fact 2, 4, 12, 8, transformed by a regular course about it, to shew syllables whose sum is 13, is

(2, 4 + 12, 8 + 2 + 4, 12 + 8) = 2, 3, 1, 7 ;

and cannot be thrown into four syllables by a regular circuit, whose values are 1, 2, 6, 4. Also,

2 (2, 3, 1, 7) = 4, 6, 2, 1.

If C stand for 1 2 6 4, and D for (2, 4, 12, 8 = ) 172 3, we have thus2C = D,

1, 3, 4, 9, 10, 12 are of the form 22m 3k and 2, 5, 6, 7, 8, 11 are of the form 22m+13k,

to modulus 13. It follows that if the circle 1 2 6 4 be multiplied by any of the former six numbers, it remains unaltered, and if by any of the latter six, it becomes 1 7 2 3.

And no other circle can be deduced by such multiplication from either of these. This does not prove that no other perfect partition of 13 exists; but it is easy to see that no more are possible, from the consideration that both 1 and 2 must be elements in every perfect partition, and that no sum of unlike figures greater than 2 can have a sum = 10, except 6, 4 and 7, 3.

We can therefore make four and only four arrangements of quadruplets

with 13 symbols, so as to exhaust the duads once and once only. These are obtained from the difference circles, thus:

In these arrangements no triad is repeated. And I think it will be found impossible to form another quadruplet which shall not contain a triad before employed.

These results might be thrown into the shape of an amusing puzzle thus:

A invites a family of 13 persons to dine with him, four at once, so long as no two shall dine together a second time. They manage to eat 13 dinners. A continues his invitation to the same family to dine with him four at once, so long as no three, reckoning from the first invitation, shall dine a second time together. They manage to eat 39 more dinners. How did they arrange themselves, and contrive it also in such a manner, that every member of the family ate the same number of dinners?

In a memoir in the fifth volume of the Cambridge and Dublin Journal, I have proved that, whenever r is prime, 2+r+1 symbols can be arranged in (r+1)-plets, so that every duad shall be once employed and once only; but the method given of forming the multiplets is not so elegant as this deduced from the difference circles here discussed. It was a matter of surprise to me to discover by accident a perfect partition of 21 = 4o + 4 + 1, thus proving that needs not be prime. This parti tion is 1, 3, 10, 2, 5.

to

All the numbers prime to 21 are of the form 5.2a, i. e., ± 2a, modulus 21. If then this difference circle is unchanged by multiplication by 2, or 5, we know that no multiplier can change it. Now

2 (1, 3, 10, 2, 5) = (2, 6, 20, 4, 10)

=(2,5+1, 3+10+2 +5, 13, 10);

5.(1, 3, 10, 2, 5) = (5, 15, 8, 10, 4)

= (5,2 + 10+ 3, 1 + 5 + 2, 10, 3 + 1);

But this

which are evidently the same circle 1, 3, 10, 2, 5. We conclude that no other perfect partition of 21 can be deduced from this one. does not prove that no other exists.

We know, however, that if 1,2 be not contiguous in a perfect partition, it must contain 3; and that if 1,2 be contiguous, it cannot contain 3. And it is easily proved, by examination of the 5-partitions of 21 that have five different elements, that no perfect one exists except 1, 3, 10, 2, 5. There are thus only two ways of arranging 21 things in 5-plets, so as to exhaust the duads once and once only.

In these no triad is repeated, and I think no 5-plet can be formed beside without repeating a triad.

I found a perfect partition of 31 by chance, simply by writing 2 2 2 2 2 under 1 3 10 2 5, which gave me 3 5 12 4 7,' and this by a small change became 2, 1, 5, 12, 4, 7, a perfect partition, exhibiting the numbers 1 2 3 30 in the proper form, thus:

1, 2, 2 + 1, 4, 5, 1 + 5, 7, 2 + 1 + 5, 2 + 7, 1 + 2 + 7, 4 + 7, 12, 4 + 7 + 2, 4 + 7 + 2 + 1, 7 + 2 + 1 + 5, 12 + 4, &c., the remaining numbers being simply the complements of those here written.

To examine whether we can obtain from this any other perfect partition, we write

2 (2, 1, 5, 12, 4, 7) = (4, 2, 10, 24, 8, 14)

which we see at once to be a new circle, because 2 and 4 on the right are contiguous syllables; it is, in fact, reduced to syllables whose sum is 31, (14, 4, 2, 10 + 4, 8 + 14 + 4 + 2 + 10, 24 + 8,) = 14, 4, 2, 3, 7, 1.Next,

3 (2, 1, 5, 12, 4, 7) = (6, 3, 15, 5, 12, 21);

this has 2+1, 5 contiguous on one side, and 3. . 5 non-contiguous on the other; we have therefore two partitions before us, and the new one is

(6, 3, 15 +5 +12, 21+ 6 + 3 + 15, 5, 12 +21)=(6, 3, 1, 14, 5, 2). Again,

4 (2, 1, 5, 12, 4, 7) = (8, 4, 20, 17, 16, 28), showing different partitions, by 2+1+5, 4 non-contiguous on the left, and 8, 4 contiguous on the right. The new one is(8+4+ 20, 17 + 16, 28 +8, 4, 20 + 17, 16 + 28) = (1, 2, 5, 4, 6, 13). The next step is

5 (2, 1, 5, 12, 4, 7) = (10, 5, 25, 29, 20, 4)

a repetition of our first difference circle; for it is simply

(7+2+1, 5, 12 + 4 +7 +2, 1+ 5 + 12 + 4 +7, 2+1 + 5 + 12, 4). We need not try 6 times; for if