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we transform it into

+

+

the middle term of which becomes by transposition of the first two rows, and the subsequent transposition of the first two columns,

*

Consequently we have

(5)=x1(4)+x2(4)+Xs(4),

and it is easily seen that a similar transformation is possible in every case, giving

¥(n) = x1(n − 1) + x,(n − 1) +x,(n − 1) + ... . + Xn − 2(n − 1).

Expressing X2, X3, in terms of x, by means of what precedes, we

have

....

¥(n) = x(n−1) + x (n − 2) + x.(n − 3)

1

+xo (n − 1)+x(n − 2)+x(n − 3)+x(n − 4)+x(n − 5)

+x(n − 1)+x(n − 2) + x(n − 3) + Xo(n − 4) + x(n − 5)
+x(n-1)+x(n-2)+x(n − 3),

and now using (a), to express x, in terms of y, we find

¥(n) = ¥(n−1) + 2¥(n − 2) + 3 {V(n − 3) +
+ ¥(n − 1) + 2¥(n − 2) + 3¥(n − 3) + 4¥(n − 4)

+Y(3)}

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+(n-1)+2(n-2)+3(n-3)+4(n4)

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where, on the first line the coefficient of the third and all the following terms is 3, on the second line the coefficient of the fifth and all the following terms is 5, on the third line the coefficient of the seventh and all the following terms is 7, and so on, the middle term (when such occurs) having a 1 superadded.

Hence, for the determination of (n) when (n - 1), (n-2), ... are known, we have

ч(n) = (n − 2)¥(n − 1)+(2n − 4)(n − 2)+(3n - 6)¥(n-3)

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where the coefficients proceed for two terms with the common difference n 2, for the next two terms with the common difference n-4, for the next two terms with the common difference n - 6, and

so on.

And as it is self-evident that (2) = 0, we obtain

¥(3) = 14(2)+1

Y(4) = 2V(3)

¥(5) = 3¥(4)+6¥(3) +1

¥(6) = 4V(5)+8¥(4)+12¥(3)

Y(7)=5¥(6)+10¥(5)+15Y(4) + 18¥(3) + 1

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¥(8) = 6¥(7) +12¥(6)+18¥(5)+22¥(4)+26¥(3) = 4738

and so forth.

To the foregoing Professor Cayley has kindly made the following additions:

The investigation may be carried further: writing for shortness &c., in place of Y(3), Y(4), &c., the equations are

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x3

du

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we have

1

u – u ′ (1 — x)2 = 1 − x2

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=

1

+(Ug+Uμx+U5x2 .. )(2x + 6x2 + 12x3 + 18x1..)

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+u(2x+6x2 + 12x3 +18x + ...);

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But for the purpose of calculation it is best to integrate by a series the differential equation for Q: assume

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u = (1 − x2)(1 + 2x+14x2 + 82x3 +593x1 +4820x5 +

.),

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In the more simple problem, where the arrangements of the n things are such that no one of them occupies its original place, if u be the number of arrangements, we have

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