% This paper has been transcribed in Plain TeX by
% David R. Wilkins
% School of Mathematics, Trinity College, Dublin 2, Ireland
% (dwilkins@maths.tcd.ie)
%
% Trinity College, 2000.
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\centerline{\Largebf ON A THEOREM RELATING TO THE}
\vskip12pt
\centerline{\Largebf BINOMIAL COEFFICIENTS}
\vskip24pt
\centerline{\Largebf By}
\vskip24pt
\centerline{\Largebf Charles Graves}
\vskip12pt
\centerline{\Largebf and}
\vskip 12pt
\centerline{\Largebf William Rowan Hamilton}
\vskip24pt
\centerline{\largerm (Proceedings of the Royal Irish Academy,
9 (1867), pp.\ 297--302.)}
\vskip36pt
\vfill
\centerline{\largerm Edited by David R. Wilkins}
\vskip 12pt
\centerline{\largerm 2000}
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\null\vskip36pt
\centerline{ON A THEOREM RELATING TO THE BINOMIAL COEFFICIENTS.}
\vskip 12pt
\centerline{Charles Graves}
\vskip12pt
\centerline{and}
\vskip 12pt
\centerline{Sir William Rowan Hamilton.}
\vskip12pt
\centerline{Communicated 26 June, 1865.}
\vskip12pt
\centerline{[{\it Proceedings of the Royal Irish Academy},
vol.~ix (1867), pp.\ 297--302.]}
\bigskip
The President read the following paper, with a {\sc Note} by the
late Sir W.~R. Hamilton, LL.D.:---
\bigbreak
\centerline{\sc On a Theorem relating to the Binomial Coefficients.}
\nobreak\bigskip
Towards the end of March, I communicated the following theorem to
Sir William Rowan Hamilton:---
\bigbreak
Putting
$$\eqalign{
s_0 &= n_0 + n_3 + n_6 + \hbox{\&c.},\cr
s_1 &= n_1 + n_4 + n_7 + \hbox{\&c.},\cr
s_2 &= n_2 + n_5 + n_8 + \hbox{\&c.},\cr}$$
where $n_0$, $n_1$, \&c., are the coefficients of the development
$$(1 + x)^n = n_0 x^0 + n_1 x^1 + n_2 x^2 + \hbox{\&c.},$$
and $n$ is a positive whole number; we shall find that, of the
three quantities $s_0$, $s_1$, $s_2$, two are always equal, and
the third differs from them by unity.
I mentioned at the same time that I had arrived at theorems,
analogous, but less elegantly expressed, by summing the series
formed by taking every {\it fourth\/} or {\it fifth\/}
coefficient, and so on, in the binomial development; and I asked
Sir William R. Hamilton whether he remembered to have seen these
theorems stated anywhere. I thought it likely that the well-known
elementary theorem respecting the equality of the sums of the
alternate coefficients in the binomial development would have
suggested research in this direction. In a note, written on the
day on which he received mine, Sir William stated that my theorem
was new to him, and that he had proved it by the help of
imaginaries and determinants. The following day he wrote again
to me, furnishing me with the following more precise statement of
my theorem:---
\bigbreak
``Let $\nu$ and $N$ be the following (whole) functions of $n$,
$$\nu = (-1)^n,\quad
N = {\textstyle {1 \over 3}} (2^n - \nu);$$
then $N$, $N$ and $N + \nu$ are {\it always\/} the value of the
{\it three sums}, if suitably {\it arranged\/}; and the {\it
singular sum\/} is $s_0$, or $s_1$, or $s_2$, according as $n$, or
$n + 1$ or $n + 2$ is a multiple of $3$.''
\bigbreak
I communicated the following demonstration of my theorem to Sir
William, in a letter of the 29th March:---
\bigbreak
Using the notation employed above, we know that
$$(n + 1)_r = n_r + n_{r-1},\quad
(n + 1)_{r-1} = n_{r-1} + n_{r-2},$$
and
$$(n + 1)_r - (n + 1)_{r-1} = n_r - n_{r-2}.$$
Now, putting
$$\eqalign{
s_r &= \ldots + n_{r-m} + n_r + n_{r+m} + \ldots,\cr
s_r' &= \ldots + (n + 1)_{r-m} + (n + 1)_r + (n + 1)_{r+m} + \ldots,\cr}$$
($m$ being any positive integer), we have, from equation~(1.),
$$s_r' - s_{r-1}' = s_r - s_{r-2};$$
and, in the particular case under consideration, viz. $m = 3$,
$$s_2' - s_1' = s_2 - s_0,\quad
s_1' - s_0' = s_1 - s_2,\quad
s_0' - s_2' = s_0 - s_1.$$
Thus it appears that the {\it differences\/} of the quantities
$s_0'$,~$s_1'$,~$s_2'$, are equal in magnitude, but of opposite
signs to those of $s_1$,~$s_2$,~$s_0$; and if we form these
differences for successive values of $n$, they will arrange
themselves in a cycle of six. Thus, if
$$s_2 - s_1 = \Delta_0,\quad
s_1 - s_0 = \Delta_2,\quad
s_0 - s_2 = \Delta_1,$$
we might form the following Table.
$$\vbox{\offinterlineskip
\hrule
\halign{&\vrule#&\strut\quad\hfil $#$\quad\cr
height 2pt&\omit&&\omit&&\omit&&\omit&\cr
&n && \Delta_0 && \Delta_2 && \Delta_1 &\cr
height 2pt&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height 2pt&\omit&&\omit&&\omit&&\omit&\cr
& 1 && -1 && 0 && 1 &\cr
& 2 && -1 && 1 && 0 &\cr
& 3 && 0 && 1 && -1 &\cr
& 4 && 1 && 0 && -1 &\cr
& 5 && 1 && -1 && 0 &\cr
& 6 && 0 && -1 && 1 &\cr
& 7 && -1 && 0 && 1 &\cr
& 8 && -1 && 1 && 0 &\cr
& \ldots && \ldots && \ldots && \ldots &\cr
height 2pt&\omit&&\omit&&\omit&&\omit&\cr}
\hrule}$$
Combining this result with the well-known theorem---
$$s_0 + s_1 + s_2 = n_0 + n_1 + n_2 + \ldots = 2^n,$$
we arrive at formulae for $s_0$,~$s_1$,~$s_2$.
\bigbreak
A couple of days later, I communicated to Sir William Hamilton
my statement and proof of the corresponding theorem respecting
the four sums obtained by adding every fourth binomial
coefficient.
\bigbreak
The theorem is as follows:---
\bigbreak
``Writing $\nu = (-1)^i$ where $i$ is any positive integer,
\noindent
If $n$ is of the form $4i$,
$$\eqalign{
s_0 &= 2^{n-2} + \nu 2^{n-2 \over 2},\cr
s_1 &= 2^{n-2},\cr
s_2 &= 2^{n-2} - \nu 2^{n-2 \over 2},\cr
s_3 &= 2^{n-2};\cr}$$
If $n$ is of the form $4i+1$,
$$\eqalign{
s_0 &= 2^{n-2} + \nu 2^{n-3 \over 2},\cr
s_1 &= 2^{n-2} + \nu 2^{n-3 \over 2},\cr
s_2 &= 2^{n-2} - \nu 2^{n-3 \over 2},\cr
s_3 &= 2^{n-2} - \nu 2^{n-3 \over 2};\cr}$$
If $n$ is of the form $4i+2$,
$$\eqalign{
s_0 &= 2^{n-2},\cr
s_1 &= 2^{n-2} + \nu 2^{n-2 \over 2},\cr
s_2 &= 2^{n-2},\cr
s_3 &= 2^{n-2} - \nu 2^{n-2 \over 2};\cr}$$
If $n$ is of the form $4i+3$,
$$\eqalign{
s_0 &= 2^{n-2} - \nu 2^{n-3 \over 2},\cr
s_1 &= 2^{n-2} + \nu 2^{n-3 \over 2},\cr
s_2 &= 2^{n-2} + \nu 2^{n-3 \over 2},\cr
s_3 &= 2^{n-2} - \nu 2^{n-3 \over 2}.\hbox{''}\cr}$$
The proof of this rests upon the equations
$$\eqalign{
s_3' - s_2' &= s_3 - s_1,\cr
s_2' - s_1' &= s_2 - s_0,\cr
s_1' - s_0' &= s_1 - s_3,\cr
s_0' - s_3' &= s_0 - s_2,\cr}$$
combined with $s_0 + s_1 + s_2 + s_3 = 2^n$.
\bigbreak
Though the theorems which I have now stated or indicated are not
devoid of interest, I should hardly have brought them under the
notice of the Academy if they had not led Sir William R. Hamilton
to discuss the more general question treated of in the Note
appended to this paper. It is at his suggestion that I have
communicated the substance of the letter which I addressed to him
on this subject.
I may be allowed to add, that the first theorem stated in this
paper was suggested by the investigation of a very simple
geometrical problem, and that I have found that it admits of
being very curiously illustrated by means of my theory of
algebraic triplets.
\nobreak\bigskip
\centerline{\vbox{\hrule width 144pt}}
\bigbreak
\centerline{\largesc Extract \largeit from a recent Manuscript
Investigation, suggested by a Theorem of}
\nobreak\vskip 3pt
\centerline{\largesc Dean Graves, \largeit which was contained in
a letter received by me a week ago.}
\nobreak\bigskip
1.
Let $n_r$, for any whole value not less than zero of $n$, and for
{\it any\/} whole value of $r$, be defined to be the (always
whole) coefficient of the power~$x^r$, in the expansion of
$(1 + x)^n$ for an arbitrary~$x$; so that we have always
$n_0 = 1$, but $n_r = 0$ in each of the cases $r < 0$, $r > n$.
\bigbreak
2.
Let $p$ be any whole number $> 0$; and let the sum of all the
coefficients $n_m$, for which $m \equiv r$ (mod.~$p$), the value
of $n$ being given, by denoted by the symbol,
$$s_{n,r}^{(p)};$$
which thus represents, when $n$ and $p$ are given, a periodical
function of $r$, in the sense that
$$s_{n,r}^{(p)} = s_{n,r+tp}^{(p)},$$
if $t$ be {\it any\/} whole number (positive or negative).
\bigbreak
3.
A fundamental property of the binomial coefficient~$n_r$ is
expressed by the equation,
$$(n + 1)_r = n_r + n_{r-1};$$
from which follows at once this analogous {\it equation in
differences},
$$s_{n+1,r}^{(p)} = s_{n,r}^{(p)} + s_{n,r-1}^{(p)};$$
with the $p$ {\it initial values},
$$s_{0,0}^{(p)} = 1,\quad
s_{0,1}^{(p)} = 0,\quad
s_{0,2}^{(p)} = 0,\quad\ldots,\quad
s_{0,p-1}^{(p)} = 0.$$
\bigbreak
4.
Hence may be deduced the {\it general expression},
$$s_{n,r}^{(p)} = p^{-1} \Sigma x^{-r} (1 + x)^n;$$
in which the summation is to be effected with respect to the $p$
roots~$x$ of the binomial equation,
$$x^p - 1 = 0.$$
\bigbreak
5.
The summand {\it term},
$$x^{-r} (1 + x)^n,$$
usually involves {\it imaginaries}, which must however disappear
in the {\it result\/}; and thus the general expression for the
{\it partial sum},~$s$, may be reduced to the {\it real\/} and
{\it trigonometrical form},
$$s_{n,r}^{(p)}
= p^{-1} \Sigma
\left( 2 \cos {m \pi \over p} \right)^n
\cos {m (n - 2r) \pi \over p},$$
with the verification that
$$0 = \Sigma
\left( 2 \cos {m \pi \over p} \right)^n
\sin {m (n - 2r) \pi \over p};$$
each summation being performed with respect to an auxiliary
integer~$m$, from $m = 0$ to $m = 1$.
\bigbreak
6.
Accordingly, {\it without\/} using {\it imaginaries}, it is easy
to prove that this expression (5) satisfies all the recent
conditions~(3), and is therefore a correct expression for the
partial sum
$$s_{n,r}^{(p)};$$
while a similar proof of the recent equation $0 = $ \&c.
\bigbreak
7.
But to form {\it practically}, with the easiest possible
{\it arithmetic}, a {\it Table of Values\/} of $s$, for any given
{\it period},~$p$, we are led by No.~3 to construct a
{\it Scheme}, such as the following:
$$\vbox{\offinterlineskip
\hrule
\halign{&\vrule#&\strut\quad\hfil $#$\quad\cr
height 2pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
& && r = 5 && 4 && 3 && 2 && 1 && 0 && \hbox{Verification} &\cr
height 2pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
\noalign{\hrule}
height 2pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr
& n = 0 && s = 1 && 0 && 0 && 0 && 0 && 1 && \Sigma s = 1 &\cr
& 1 && 1 && 0 && 0 && 0 && 1 && 1 && 2 &\cr
& 2 && 1 && 0 && 0 && 1 && 2 && 1 && 4 &\cr
& 3 && 1 && 0 && 1 && 3 && 3 && 1 && 8 &\cr
& 4 && 1 && 1 && 4 && 6 && 4 && 1 && 16 &\cr
& 5 && 2 && 5 && 10 && 10 && 5 && 2 && 32 &\cr
& 6 && 7 && 15 && 20 && 15 && 7 && 7 && 64 &\cr
height 2pt&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&&\omit&\cr}
\hrule}$$
\bye