MATHEMATICS OF COMPUTATION, VOLUME 32, NUMBER 144
OCTOBER 1978, PAGES 1271-1273
A Sum of Binomial Coefficients
By Lajos Takács
Abstract.
An explicit expression
coefficients
is derived for the sum of the (k + l)st binomial
in the nth, (n - m)th, (n - 2m)th, . . . row of the arithmetic
triangle.
In combinatorial analysis and in probability theory we occasionally encounter
the problem of calculating the sum
(1)
t-^
<Xn,k,m)=
Z
In -im
0<j<n/m\
k
for n = 0, 1, 2, . . . where k and m are given positive integers. If n is large, the sum-
mation in (1) is time-consuming and it is desirable to derive some simple formulas for
ß(n, k, m) which make it possible to determine Q(n, k, m) for any n in an easy way.
For m = 1 and »2 = 2 such formulas are
(2)
Q(n,k,l)=(
/w + 1
\k + 1
and
(3)
Q(n,k,2)=i(
» + 2)tlï-\l77J7JÏL]tï
i
Our aim is to derive analogous expressions for any m.
We shall prove that if n = r (mod m) where 0 < r < m, then Q(n, k, m) is a
polynomial of degree k + 1 in the variable n. In this polynomial every term is independent of r except the constant term which does depend on r.
More specifically, we have the following result.
Theorem.
(4)
If n=r
(mod m) where 0 < r < m, then
QXn,k, m)=P(n
+ m, k, m) - Pir, k, m)
for n > 0, k > 1, m > 1 where
(5)
Pix, k,m) = - Z (X)Aim, k + 1 -/)
and Aim, j) (j = 0, 1, . . . , k + 1) are determined by the generating function
(6)
-^-
=±Aim,j)xi,
(1 + x)m - 1
Received September
;=o
1, 1977.
AMS (MOS) subject classifications (1970).
Primary 39A05.
Copyright
© 1978, American
1271
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Mathematical
Society
1272
LAJOS TAKÂCS
which is convergent if \x\ < 2 sin(n/m). In another form we have
k I x/m\
m
¡
. .(j\(mi
*■*■'-£(/+JS'^CX*
Note. We define (*) = 1 for any x and (*) = x(x - 1) . . . (x -/ + 1)//!
for any x and / = 1, 2, . . . .
Proof. We observe that Q{n, k, m) is the coefficient of**4"1 in the polynomial
[(1 + x)r + (1 + x)r+m
(8)
+■••+(!+
|"(i +x)n+m
"L
x)n)x
-(i
+jcy"|
*
mx
(1 + x)m - 1
Ju
Consequently, (4) is true if Zt)c, A:,m) is defined by (5). It remains to determine
A(m, j) for / = 0, 1, 2, . . . . By expanding (6) into partial fractions, we get
mx
(9)
m-l
e„x
V
+ l-e, r
= 1+ V
(1 +x)m - 1
r— 1
where e, = e2rni'm for 1 <r < m - 1. Therefore, ,4(m, 0) = 1 and
m—1
-iv'-i
(10) ¿(m. ;•) = (-!)
g
_"^,1
r=i(l-ery
cos((2r/ + m; - 4r)ir/2m)
(2 %in(rirlm))i
r=\
for / = 1, 2, . . . . Formula (10) is an explicit expression for A(m, /); however, it
is more convenient to determine A(m, j) for /= 1, 2, ... by the recurrence formula
ID
Á+.1/m\
(10
A(m, j + 1 - 0 = 0,
starting from the initial condition A(m, 0) = 1. To prove (11) we multiply both
sides of (6) by ( 1 + x)m - 1 and form the coefficient of x'+ x.
For m < 12 and/ < 10 the following table contains the numbers
A(m, /)n_|mpl/^p_1^'
where p = 2, 3, 5,1, . . . are prime numbers. A Texas
SR 52 calculator was programmed to obtain the entries of this table.
10
MX.I)
0
A(2,f)2>
I
-1
1
/)(3,/)3l"21
2
-I
1
0
0
0
-1
0
1
0
-1
0
-1
I
0
0
0
I
- i
-I
I
-1
1
A(4,f)2'
5
-5
7
-15
15
1
-33
.4(5,/)5l"4l
2
-1
-1
4
-3
11
-II
35
-119
567
-1765
4
-35
-2
0
-3355
21
-21
/l(6,/)2'3l"2l
A(l.i)^'"']
A(S,¡)2'
yl(9,/)3l"2l
20
yl(10,/)2'5l//41
33
-10
-33
ii(ll,/)ll""0l
10
-5
143
-143
/l(12,/)2'3l'721
1
-2
-63
-62
-891
-17
-3575
41041
-41041
65
3
-249613
4
-8
-29
39
0
-52
231
-15
-1521
3073
4319
-29631
3160
3333
-755
-37125
1699
3003
188441
1568743
-20332
-5091303
28
25
-110
20
317
29982095
180388429
I OX
80
11583 87659
-673387
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41041
-4467
A SUM OF BINOMIAL COEFFICIENTS
1273
Now we are going to prove (7). Let us denote the right-hand side of (7) by
Rix, k, m). By Newton expansion we obtain
for any x. If we add (12) for x = 0, 1, . . . , s, we get
(.a,
a—, .'-)=|o(;i;)¿e.r-(;)('"/;')
for 0 < r < m. If, in particular, n = 0 (mod m), that is, r = 0, then by (13) we have
QXn,k, m) = A(n + m, k, m). We have demonstrated that QXn,k, m) is a polynomial
in the variable n and only the constant term depends on r. Accordingly, for any n =
r (mod m), QXn, k, m) differs from A(n + m, k, m) by a constant. If we put first
r = 0 and then x = r/m in (12), then we obtain that Q(r, k, m) = R(r + m, k, m) R(r, k, m). This implies that if n = r (mod m) and 0 < r < m, then
(14)
QXn,k, m) = R(n + m, k, m) - R(r, k, m),
where R(x, k, m) is the right-hand side of (7). Since both P(0, k, m) and R(0, k, m)
are 0, therefore R(x, k, m) = P(x, k, m) for all x. This completes the proof of the
theorem.
We note that formulas (4) and (7) are more advantageous than (13) because in
(7) the second sum does not depend on r. If we use (13), the second sum should be
calculated for every r = 0, 1, . . . , m - 1. Actually, in (7) the second sum is A'(mkx)
taken at x = 0. This can be determined from the sequence {(*?*), x = 0, 1, 2, . . . }
by forming repeated differences.
By using (7) we can derive another formula for A(m, j). By (5) we have
05)
A(m,j) = mP(I,i,m),
where the right-hand side is given by (7).
We remark also that from (4) and (7) it follows that
(16)
lim QXn,m, k)/nk+x = 1/ik + l)!m.
Finally, I would like to thank the referee for calling my attention to the paper
of L. Carlitz [1]. In this paper Carlitz introduced the polynomials ßm(X) defined by
A comparison with (6) shows that
(18)
Aim,
f) = ß,-(ßjf.
Department of Mathematics and Statistics
Case Western Reserve University
Cleveland, Ohio 44106
1. L. CARLITZ, "A degenerate Staudt-Clausen theorem," Arch. Math., v. 7, 1956, pp. 28-33.
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