Power Sums Related to Semigroups S (d1, d2, d3)
Leonid G. Fel† and Boris Y. Rubinstein‡
†
Department of Civil and Environmental Engineering,
Technion, Haifa 32000, Israel
and
‡
Department of Mathematics, University of California, Davis,
One Shields Ave., Davis, CA 95616, U.S.A.
September 26, 2005
Abstract
The explicit formulas for the sums of positive powers of the integers si unrepresentable
by the triple of integers d1 , d2 , d3 ∈ N, gcd(d1 , d2 , d3 ) = 1, are derived.
Let S (d1 , . . . , dm ) be the semigroup generated by a set of integers {d1 , . . . , dm } such that
1 < d1 < . . . < d m ,
gcd(d1 , . . . , dm ) = 1 .
(1)
For short we denote the tuple (d1 , . . . , dm ) by dm and consider the generating function Φ (dm ; z)
X
Φ (dm ; z) =
z si = z + z s2 + . . . + z sG(dm ) ,
(2)
si ∈ ∆(dm )
for the set ∆ (dm ) of the integers s which are unrepresentable as s =
∆ (dm ) = {s1 , s2 , . . . , sG(dm ) } ,
s1 = 1 .
Pm
i=1
xi di , xi ∈ N ∪ {0}
(3)
The integer G (dm ) is known as the genus for semigroup S (dm ). Recall the relation of Φ (dm ; z)
¤
£
with the Hilbert series H(dm ; z) of a graded monomial subring k z d1 , . . . , z dm [1]
X
Q(dm ; z)
1
,
(4)
, where H(dm ; z) =
z s = Qm
H(dm ; z) + Φ(dm ; z) =
dj
1−z
j=1 (1 − z )
m
s ∈ S(d )
1
and Q(dm ; z) is a polynomial in z. The calculation of the power sums
X
gn (dm ) =
si ∈
sni
(5)
∆(dm )
was performed in [2] for m = 2
¡
gn d
2
¢
n+1 n+1−k
X
X µn + 2¶µn + 2 − k ¶
Bn+1
1
,
Bk Bl d1n+1−k dn+1−l
−
=
2
(n + 1)(n + 2) k=0 l=0
n+1
k
l
(6)
where Bk are the Bernoulli numbers. The formula (6) generalizes the known Sylvester’s expression
[3] for G (d2 ) = g0 (d2 ) and further results for n = 1 [4] and n = 2, 3 [5]
¡ ¢
1
g0 d 2 =
(d1 − 1)(d2 − 1) ,
2
¡ ¢
g0 (d2 )
g1 d 2 =
(2d1 d2 − d1 − d2 − 1) ,
6
¡ ¢
g0 (d2 )
g2 d 2 =
d1 d2 (d1 d2 − d1 − d2 ) ,
6
¡
¡ ¢
¢
¤
g0 (d2 ) £
(1 + d1 )(1 + d2 ) 1 + d21 + d22 + 6d21 d22 − 15d1 d2 (d1 + d2 ) .
g3 d 2 =
60
As for higher dimensions, m ≥ 3, the first two sums, g0 (d3 ) and g1 (d3 ), were found in [5]. The
use was made of the explicit expression for the Hilbert series H(d3 ; z) of a graded subring for
semigroups S (d3 ) which was recently established [5].
In this paper we derive the formula for the power sum gn (d3 ) related to the symmetric and
non-symmetric semigroups S (d3 ). This will be done by applying to the relation (4) an approach
based on the generating function of the Bernoulli polynomials of higher order.
Following Johnson [6] recall the definition of the minimal relations for given d 3 = (d1 , d2 , d3 )
a11 d1 = a12 d2 + a13 d3 , a22 d2 = a21 d1 + a23 d3 ,
a33 d3 = a31 d1 + a32 d2 ,
where
(7)
a11 = min {v11 | v11 ≥ 2, v11 d1 = v12 d2 + v13 d3 , v12 , v13 ∈ N ∪ {0}} ,
a22 = min {v22 | v22 ≥ 2, v22 d2 = v21 d1 + v23 d3 , v21 , v23 ∈ N ∪ {0}} ,
(8)
a33 = min {v33 | v33 ≥ 2, v33 d3 = v31 d1 + v32 d2 , v31 , v32 ∈ N ∪ {0}} .
The auxiliary invariants aij are uniquely defined by (8) and
gcd(a11 , a12 , a13 ) = 1 , gcd(a21 , a22 , a23 ) = 1 ,
2
gcd(a31 , a32 , a33 ) = 1 .
(9)
According to [5] the numerator of the Hilbert series for non-symmetric semigroups S (d 3 ) reads
Q(d3 ; z) = 1 −
3
X
3
3
z aii di + z 1/2[ha,di−J (d )] + z 1/2[ha,di+J (d )] , where
(10)
i=1
3
X
¡ ¢
aii ajj di dj + 4d1 d2 d3 ,
J 2 d3 = ha, di2 − 4
ha, di =
3
X
aii di .
(11)
i=1
i>j
The case of symmetric semigroups S (d3 ) is much simpler. Here two off-diagonal elements in one
column of the matrix aij vanish, e.g. a13 = a23 = 0 that results in a11 d1 = a22 d2 . The numerator
of the Hilbert series for symmetric semigroups S (d3 ) with above symmetry is given by
Qs (d3 ; z) = (1 − z a22 d2 )(1 − z a33 d3 ) .
(12)
Move on to calculation of the power sums gn (d3 ). First, denote z = et and represent (4) as follows
X
si ∈
e si t =
∆(d3 )
1
Q(d3 ; et )
−
,
1 − et (1 − ed1 t ) (1 − ed2 t ) (1 − ed3 t )
(13)
and apply the sequence of identities
X
si ∈
e
si t
∆(d3 )
=
X
si ∈
∆(d3 )
∞
X
sn t n
i
k=0
n!
=
∞
X
tn
k=0
n!
X
si ∈
∆(d3 )
sni
=
∞
X
¡ ¢ tn
.
gn d 3
n!
n=0
(14)
Thus, gn (d3 ) could be found by expanding the right hand side of (13) in the power series.
To present the result in compact form we use a definition [7] of the Bernoulli polynomials of
higher order
∞
ext
1 X tn−m (m)
Bn (x|dm ) ,
=
d
t
i
πd n=0 n!
− 1)
i=1 (e
Qm
πd =
m
Y
di ,
(15)
i=1
satisfying the recursion relation
n µ ¶
³x´
X
n p
(m−1)
n
m
m−1
(1)
(m)
), Bn (x|d) = d Bn
,
Bn (x|d ) =
d Bp Bn−p (x|d
d
p m
p=0
(16)
where Bn (x) denotes the regular Bernoulli polynomial. Because each term in the right hand side
of (13) has the form of the left hand side of (15), it is easy to write the answer as sum of the
Bernoulli polynomials of higher order.
First, contribution of the term 1/(1 − z) to gn (d3 ) is found trivially
∞ n−1
∞
X
X
t
tn−1 Bn
1
(1)
=
−
B
(0|1)
=
−
,
n
1 − et
n!
(n
−
1)!
n
n=0
n=0
3
(17)
so that the corresponding term in gn (d3 ) is −Bn+1 /(n + 1). Consider a general term for n = 3
∞
etx
1 X tn−3 (3)
=
−
B (x|d3 ),
(1 − ed1 t )(1 − ed2 t )(1 − ed3 t )
d1 d2 d3 n=0 n! n
(18)
thus its contribution to gn (d3 ) reads:
g(x; d3 ) = −
n!
(3)
B (x|d3 ).
(n + 3)! d1 d2 d3 n+3
(19)
(3)
The Bernoulli polynomial of higher order Bn+3 (x|d3 ) can be expanded into the triple sum over
the Bernoulli numbers
(3)
Bn+3 (x|d3 )
¶µ ¶µ ¶
j
k µ
n+3 X
X
X
k k−l j−k n+3−j
j
n+3
d1 d2 d3
Bk−l Bj−k Bn+3−j xl ,
=
l
k
j
j=0 k=0 l=0
(20)
which for x = 0 reduces to the double sum
(3)
Bn+3 (0|d3 )
=
¶µ ¶
j µ
n+3 X
X
n+3
j
j=0 k=0
j
k
n+3−j
dk1 dj−k
Bk Bj−k Bn+3−j .
2 d3
(21)
The expression for gn (d3 ) for the non-symmetric semigroups has the form
h
Bn+1
n!
(3)
(3)
(3)
=−
Bn+3 (0|d3 ) − Bn+3 (a11 d1 |d3 ) − Bn+3 (a22 d2 |d3 )−
+
n + 1 (n + 3)! d1 d2 d3
i
£
¡ ¢¤
£
¡ ¢¤
(3)
(3)
(3)
Bn+3 (a33 d3 |d3 ) + Bn+3 (1/2 ha, di − J d3 |d3 ) + Bn+3 (1/2 ha, di + J d3 |d3 ) .
gn(n) (d3 )
(22)
In case of the symmetric semigroups we obtain
gn(s) (d3 )
h
Bn+1
n!
(3)
(3)
(3)
=−
+
Bn+3 (0|d3 ) − Bn+3 (a22 d2 |d3 ) − Bn+3 (a33 d3 |d3 )+
n + 1 (n + 3)! d1 d2 d3
i
(3)
Bn+3 (a22 d2 + a33 d3 |d3 ) .
(23)
Thus, for the symmetric semigroup S(d3 ) the first three sums read
(s)
2g0 (d3 )
g
= 1 − sd + ha,
di ,
sd =
3
X
i=1
di ,
g
ha,
di = ha, di − a11 d1 ,
³
´³
´
g
g
= sd − ha, di sd − 2ha, di + (d1 d2 + d1 d3 + d2 d3 ) − πd − 1 ,
´
³
(s)
g
g
g
12g2 (d3 ) = (sd − ha,
di) sd ha,
di − ha,
di2 − (d1 d2 + d1 d3 + d2 d3 ) + πd .
(s)
12g1 (d3 )
4
(24)
(25)
(26)
In the non-symmetric case we obtain:
(n)
2g0 (d3 ) = 1 − sd − πa + ha, di ,
πa = a11 a22 a33 ,
(n)
12g1 (d3 ) = ha, di(2ha, di − 3sd − 2πa ) + sd (sd + 3πa ) −
(27)
3
X
aii ajj di dj +
i6=j
(d1 d2 + d1 d3 + d2 d3 ) + πd − 1 ,
3
h
i
X
πa
πd
(n)
12g2 (d3 ) =
Ai (2Ai + 1)d3i − (Ai + 2)d2i + πd di + (2Ai + 1) +
2
2
i=1
3 h
X
Cij d2i dj −
i,j=1
πa
πd i
Bij di dj − Fij ,
2
2
(28)
(29)
where
Ai = aii − 1 , Bij = Ai Aj − Ai − Aj , Cij = Ai (Ai Aj − Ai − 1) , Fij = Ai (2Aj + 1) .
We finish the paper with a compact version of (6) through the Bernoulli polynomials of higher
orders
gn (d2 ) = −
Bn+1
1
(2)
(2)
−
[Bn+2 (0|d2 ) − Bn+2 (d1 d2 |d2 )] ,
n + 1 d1 d2 (n + 1)(n + 2)
(30)
¶µ ¶
j µ
n+2 X
X
n+2
j j−k n+2−j
=
Bj−k Bn+2−j xk .
d1 d2
j
k
j=0 k=0
(31)
where
(2)
Bn+2 (x|d2 )
References
[1] R. P. Stanley, Combinatorics and Commutative Algebra,
Birkhäuser Boston, 2nd ed., (1996)
[2] Ö. J. Rödseth, A note on Brown and Shiue’s paper on a Remark Related to the Frobenius
Problem, Fibonacci Quarterly, 32, 407 (1994)
[3] J. J. Sylvester, Mathematical Questions with Their Solutions,
Educational Times, 41, 171 (1884)
[4] T. C. Brown and P. J. Shiue, A Remark Related to the Frobenius Problem,
Fibonacci Quarterly, 31, 32 (1993)
5
[5] L. G. Fel, Frobenius Problem for Semigroups S (d1 , d2 , d3 ),
preprint, (2004), [http://arXiv.org/abs/math.NT/0409331]
[6] S. M. Johnson, A Linear Diophantine Problem,
Canad. J. Math., 12, 390 (1960)
[7] H. Bateman and A. Erdelýi, Higher Transcendental Functions, Vol.1, Ch.1,
McGraw-Hill Book Co., NY (1953)
6