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Journal of Integer Sequences, Vol. 17 (2014),
Article 14.5.3
23 11
Iterative Procedure for Hypersums
of Powers of Integers
José Luis Cereceda
Distrito Telefónica, Edificio Este 1
28050 Madrid
Spain
jl.cereceda@movistar.es
Abstract
Relying on a recurrence relation for the hypersums of powers of integers put forward
recently, we develop an iterative procedure which allows us to express a hypersum of
arbitrary order in terms of ordinary (zeroth order) power sums. Then, we derive the
coefficients of the hypersum polynomial as a function of the Bernoulli numbers and the
Stirling numbers of the first kind.
1
Introduction
For every integer m,Pm ≥ 1, the hypersums of powers of integers are defined recursively as
(m)
(m−1)
(0)
follows: Pk (n) = nj=1 Pk
(j), where Pk (n) is the sum of the first n positive integers
(0)
each raised to the integer power k ≥ 0, Pk (n) = 1k + 2k + · · · + nk [1, 2]. The latter is given
(m)
by a polynomial in n of degree k + 1 with zero constant term. Hence, inductively Pk (n) is
given by a polynomial in n of degree k + m + 1 with zero constant term:
(m)
Pk (n) =
k+m+1
X
crk,m nr .
(1)
r=1
An explicit formula for the coefficients crk,m involving the Stirling numbers of the first
and second kinds has been given by the author [3]. In this paper (Section 2), by an iterative
(m)
procedure, we obtain a new representation of the m-th order hypersum Pk (n) in terms of
1
(m)
ordinary (zeroth order) power sums. Specifically, we will show that Pk (n) can be expressed
(0)
(0)
(0)
as a linear combination of Pk (n), Pk+1 (n), . . . , Pk+m (n), as follows:
(m)
Pk (n)
=
m
X
(−1)i qm,i (n)
m!
i=0
(0)
Pk+i (n),
(2)
where qm,i (n) is a polynomial in n of degree m − i. (Note that formula (2) holds for m = 0 if
we set q0,0 (n) = 1.) In Section 3, we determine the explicit form of the coefficients of qm,i (n).
Then, using (2), we obtain the coefficients crk,m in terms of the Bernoulli numbers Bk and
the (unsigned) Stirling numbers of the first kind nk (see A008275 in [4]). In particular, we
proved that
m
1 X
1
i m+1
ck,m =
Bk+i ,
(3)
(−1)
i+1
m!
i=0
in accordance with the result obtained by Inaba [2, Proposition 1]. (Please note that,
throughout this paper, we use the convention that B1 = 12 .)
For later reference, we recall that the recurrence relation defining the numbers nk is
given by [5, p. 214]:
m
m
m+1
,
(4)
+
=m
i
i+1
i+1
with 00 = 1, and n0 = n0 = 0 for n ≥ 1.
(m)
Formula (2) is noteworthy since it neatly shows how the hypersum Pk (n) is constructed
(0)
out of the building blocks Pk+i (n), i = 0, 1, . . . , m. Moreover, we point out that the polynomials qm,i (n) are interesting in their own right. Indeed, for fixed m, the coefficients corresponding to the set of polynomials {qm,i (n)}m
i=0 may be arranged in a Pascal-like triangular
array with a specific rule of formation.
2
Iterative procedure for the hypersum
(m)
The basic tool we use to obtain Pk (n) in terms of ordinary power sums is the following
recurrence relation, a proof of which has been given by the author [3, Theorem 8]:
(j)
(j−1)
Theorem 1. The hypersums Pk (n), Pk
(j)
Pk (n) =
(j−1)
(n), and Pk+1 (n) satisfy the recurrence relation
1 (j−1)
n + j (j−1)
Pk
(n) − Pk+1 (n),
j
j
(m)
k ≥ 0, j ≥ 1.
(5)
To obtain Pk (n), we repeatedly apply the recurrence (5) to successive values of j =
2
1, 2, . . . , up to j = m. Proceeding in this way, it is easy to see that, for example,
1
(0)
(3)
Pk (n) = (n + 1)(n + 2)(n + 3)Pk (n)
6
(0)
1
− (n + 1)(n + 2) + (n + 1)(n + 3) + (n + 2)(n + 3) Pk+1 (n)
6
(0)
1
1 (0)
+ (n + 1) + (n + 2) + (n + 3) Pk+2 (n) − Pk+3 (n),
6
6
which is of the form (2) with q3,0 (n) = (n + 1)(n + 2)(n + 3), q3,1 (n) = (n + 1)(n + 2) +
(n + 1)(n + 3) + (n + 2)(n + 3), q3,2 (n) = (n + 1) + (n + 2) + (n + 3), and q3,3 (n) = 1.
From this procedure, the general form of the polynomial qm,i (n) is argued to be
qm,i (n) =
m−i
Y
X
(n + st ),
i = 0, 1, . . . , m,
(6)
1≤s1 <s2 <···<sm−i ≤m t=1
where st , t = 1,Q2, . . . , m − i, is an integer. Note the special cases qm,m−1 (n) =
and qm,0 (n) = m
i=1 (n + i). Furthermore,
qm,i (0) =
m−i
Y
X
st = σm−i (1, 2, . . . , m),
Pm
i=1 (n
+ i)
(7)
1≤s1 <s2 <···<sm−i ≤m t=1
where σm−i (1, 2, . . . , m) is the (m − i)-th elementary symmetric polynomial evaluated on the
first m integers {1, 2, . . . , m} [6, Chapter 6].
Lemma 2. For m ≥ 1, the polynomials qm,i (n) satisfy the recurrence relation
(n + m)qm−1,i (n) = qm,i (n) − qm−1,i−1 (n),
i = 0, 1, . . . , m − 1,
(8)
where it is understood that qm−1,i−1 (n) = 0 for i = 0.
Proof. Relation (8) follows directly from the definition of qm,i (n). Hence, from (6), we obtain
(n + m)qm−1,i (n) =
X
1≤s1 <···<sm−i−1 ≤m−1
m−i−1
Y
(n + st )(n + m).
(9)
t=1
On the other hand, we have
qm−1,i−1 (n) =
X
m−i
Y
(n + st ).
1≤s1 <···<sm−i ≤m−1 t=1
Clearly, the sum of the right-hand side of (9) and (10) is identical to (6).
3
(10)
(m)
Now we show by induction on m that Pk (n) have the form (2) with qm,i (n) given by
(6). This statement is readily verified for the base cases m = 0, 1, 2, and 3. Assuming the
(m−1)
inductive hypothesis holds for Pk
(n) (with m ≥ 1), Equation (5) yields
"
#
m−1
m−1
X
X
1
(0)
(0)
(m)
(n + m)
(−1)i qm−1,i (n)Pk+i (n) −
(−1)i qm−1,i (n)Pk+i+1 (n) .
Pk (n) =
m!
i=0
i=0
Using (8), it follows that
(m)
Pk (n)
"m−1
1 X
(0)
(−1)i qm,i (n)Pk+i (n)
=
m! i=0
−
m−1
X
(−1)
i
(0)
qm−1,i−1 (n)Pk+i (n)
i=1
+
m
X
(−1)
i
(0)
qm−1,i−1 (n)Pk+i (n)
i=1
#
#
"m−1
1 X
(0)
(0)
=
(−1)i qm,i (n)Pk+i (n) + (−1)m qm−1,m−1 (n)Pk+m (n)
m! i=0
m
=
1 X
(0)
(−1)i qm,i (n)Pk+i (n),
m! i=0
where we used that qm−1,m−1 (n) = qm,m (n) = 1 to justify the last equation. This completes
the inductive step and the proof of the above statement. We formally state this result in the
following theorem.
(m)
Theorem 3. The hypersum Pk (n) admits a representation of the form (2) with qm,i (n)
given by (6).
3
The coefficients of the hypersum polynomial
In this section, we provide an explicit expression for the coefficients crk,m in terms of the
Bernoulli numbers and the Stirling numbers
Pm−i ofs thes first kind. To this end, we first put
qm,i (n) in polynomial form as qm,i (n) = s=0 qm,i n . On the other hand, according to the
(0)
well-known Bernoulli formula, Pk+i (n) can be written as [7, Equation 9]
k+i+1
X k + i + 1 1
(0)
Pk+i (n) =
Bk+i+1−t nt .
k + i + 1 t=1
t
(Remember that we are taking B1 = 12 in the above formula.) Then, substituting the
(0)
aforementioned expressions for qm,i (n) and Pk+i (n) into (2) and comparing the resulting
polynomial with (1), gives
m
crk,m
1 X
(−1)i Qrk,m,i ,
=
m! i=0
4
r = 1, 2, . . . , k + m + 1,
(11)
where
Qrk,m,i
r−1
X
1
k+i+1
h
=
Bk+i+h+1−r .
q
r−h
k + i + 1 h=0 m,i
(12)
In particular, from (11) and (12), we quickly obtain
m
c1k,m
1 X
0
=
(−1)i qm,i
Bk+i ,
m! i=0
k, m ≥ 0.
(13)
s
Now let us address the question of the nature of the coefficients qm,i
, s = 0, 1, . . . , m − i,
0
of qm,i (n). Let us first look at the constant term qm,i . This is the value of qm,i (n) at n = 0.
0
have qm,i
= σm−i (1, 2, . . . , m). On the other hand, the Stirling numbers
Hence, from (7),we
n
all the permutations of size n with k cycles. It turns out that
of the first kind k enumerate
n+1
σk (1, 2, . . . , n) = n+1−k [5, pp. 213–214], and then
m+1
.
σm−i (1, 2, . . . , m) =
i+1
(14)
0
. Putting this in (13), we obtain formula (3).
Thus, we have qm,i
= m+1
i+1
s
In order to systematically derive the coefficients qm,i
, it is useful to note that
m−i
Y
t=1
(n + st ) =
m−i
X
σm−i−s (s1 , s2 , . . . , sm−i )ns ,
s=0
where σm−i−s (s1 , s2 , . . . , sm−i ) is the (m − i − s)-th elementary symmetric polynomial on the
variables s1 , s2 , . . . , sm−i . Substituting this expression into (6), we deduce that
X
s
qm,i
=
σm−i−s (s1 , s2 , . . . , sm−i ).
(15)
1≤s1 <s2 <···<sm−i ≤m
Clearly, the right-hand side of (15) is a symmetric function on {s1 , s2 , . . . , sm−i }. This
function is asum of products
of m − i − s distinct integers chosen from {1, 2, . . . , m}, with
a total of mi times m−i
terms.
On
s
the other hand, the elementary symmetric polynomial
m
σm−i−s (1, 2, . . . , m) is a sum of i+s terms, each of which is a product of m − i − s distinct
m
, we conclude that
= i+s
integers chosen from {1, 2, . . . , m}. Therefore, since mi m−i
i+s
s
s
i+s
the right-hand side of (15) is necessarily s times σm−i−s (1, 2, . . . , m). Hence, using (14),
we find that
i+s
m+1
s
, s = 0, 1, . . . , m − i.
(16)
qm,i =
i+s+1
s
m−i
Note, in particular, that qm,i
= mi . From (16), we also deduce the symmetry property
s
i
qm,i
= qm,s
. As a concrete example, Table 1 displays the coefficients of the polynomials
q8,i (n), i = 0, 1, . . . , 8, where we use [ns ] to denote the coefficient of ns . Note that the
5
q8,8 (n)
q8,7 (n)
q8,6 (n)
q8,5 (n)
q8,4 (n)
q8,3 (n)
q8,2 (n)
q8,1 (n)
q8,0 (n)
[n0 ]
1
36
546
4536
22449
67284
118124
109584
40320
[n1 ]
–
8
252
3276
22680
89796
201852
236248
109584
[n2 ]
–
–
28
756
8190
45360
134694
201852
118124
[n3 ]
–
–
–
56
1260
10920
45360
89796
67284
[n4 ]
–
–
–
–
70
1260
8190
22680
22449
[n5 ]
–
–
–
–
–
56
756
3276
4536
[n6 ]
–
–
–
–
–
–
28
252
546
[n7 ]
–
–
–
–
–
–
–
8
36
[n8 ]
–
–
–
–
–
–
–
–
1
Table 1: The coefficients of the polynomials q8,i (n), i = 0, 1, . . . , 8.
symmetry property implies that the table of coefficients is symmetric about a 45◦ diagonal.
4
2
For example, we have q8,2
= q8,4
= 8190.
Finally, combining the Equations (11), (12), and (16), we obtain
crk,m
m
r−1 1 X (−1)i X i + h k + i + 1
m+1
B
,
=
i + h + 1 k+i+h+1−r
h
r−h
m! i=0 k + i + 1 h=0
which constitutes the generalization of Inaba’s formula (3) to arbitrary r = 1, 2, . . . , k+m+1,
with k, m ≥ 0, and B1 = 21 .
On the other hand, from (8), we immediately derive the following recurrence relation for
s
the coefficients qm,i
:
s−1
s
s
s
qm,i
= mqm−1,i
+ qm−1,i−1
+ qm−1,i
.
(17)
0
0
0
= mqm−1,i
+qm−1,i−1
. Therefore, comparing this relation
For s = 0, relation (17) becomes qm,i
1
0
0
. For s = 1 we
= m+1
with (4) and noting that q0,0 = 1 = 1 , we retrieve the result qm,i
i+1
1
1
1
0
1
0
have qm,i = mqm−1,i + qm−1,i−1 + qm−1,i , which is satisfied when we set qm,i = (i + 1)qm,i+1
=
m+1
(i + 1) i+2 . In general, the solution of the recurrence (17) is given by
s
so that qm,i
=
i+s
s
0
qm,i+s
1
s−1
s
qm,i
= (i + 1)qm,i+1
s
1 1
s−2
=
(i + 1)(i + 2)qm,i+2
ss−1
..
.
1
0
,
= (i + 1)(i + 2) · · · (i + s)qm,i+s
s!
m+1 , in accordance with (16).
= i+s
i+s+1
s
6
s−1
s
Thus, Table 1 is generated by the rule qm,i
= 1s (i + 1)qm,i+1
, s ≥ 1, which enables one to
s−1
s
determine the element qm,i in row m − i and column s from the preceding element qm,i+1
in
row m − i −
1
and
column
s
−
1,
the
elements
of
the
starting
0-th
column
being
given
by
0
.
qm,i
= m+1
i+1
We conclude with three brief remarks.
(m)
(m)
. Then, letting
Remark 4. For k = 0 the hypersum Pk (n) is equal to P0 (n) = n+m
m+1
Pm (−1)i qm,i (n) (0)
(0)
. Solving for Pm (n), we get
Pi (n) = n+m
k = 0 in (2), we will have i=0
m+1
m!
(−1)
m
Pm(0) (n)
m−1
X
n+m
(0)
= m!
+
(−1)i+1 qm,i (n)Pi (n),
m+1
i=0
(0)
m ≥ 1,
(0)
(0)
which allows us to compute recursively Pm (n) from the power sums P0 (n), P1 (n), . . . ,
(0)
Pm−1 (n), and the polynomials qm,i (n), i = 0, 1, . . . , m − 1.
Remark 5. The leading coefficient of the hypersum polynomial (1) has been given by the
k+m+1
k!
. On the other hand, the leading coefficients of qm,i (n) and
author [3]: ck,m
= (k+m+1)!
(0)
m
m−i
1
Pk+i (n) are given by qm,i = i and ck+i+1
k+i,0 = k+i+1 , respectively. Therefore, equating the
terms of maximum degree on the two sides of (2) yields the combinatorial identity
m
X
i=0
k! m!
m
(−1)i
,
=
k+i+1 i
(k + m + 1)!
k, m ≥ 0.
Remark 6. From formula (3), we deduce an identity relating the harmonic number Hm =
the Stirling numbers of the first kind. Indeed,
1 + 21 + · · · + m1 to the Bernoulli numbers and
m+1
1
from c0,m = 1/(m + 1) [2, 3], recalling that 2 = m!Hm , and from (3) we obtain
⌊m/2⌋ 2 X m+1
2m
B2j .
+
Hm =
m + 1 m! j=1 2j + 1
4
Acknowledgement
The author would like to thank an anonymous referee for giving an extensive list of comments
and suggestions that led to an improvement of an earlier version of this paper.
References
[1] D. E. Knuth, Johann Faulhaber and sums of powers, Math. Comp. 61 (1993), 277–294.
[2] Y. Inaba, Hyper-sums of powers of integers and the Akiyama-Tanigawa matrix, J. Integer
Sequences 8 (2005), Article 05.2.7.
7
[3] J. L. Cereceda, Generalized Akiyama-Tanigawa algorithm for hypersums of powers of
integers, J. Integer Sequences 16 (2013), Article 13.3.2.
[4] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, Published electronically
at http://oeis.org.
[5] L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, D. Reidel Publishing Company, 1974.
[6] W. Fulton, Young Tableaux: With Applications to Representation Theory and Geometry,
Cambridge University Press, 1997.
[7] T. M. Apostol, Bernoulli’s power-sum formulas revisited, Math. Gaz. 90 (2006), 276–279.
2010 Mathematics Subject Classification: Primary 11B57; Secondary 11C08, 11Y55.
Keywords: hypersum polynomial, iterative procedure, elementary symmetric polynomial,
Bernoulli number, Stirling number of the first kind.
(Concerned with sequence A008275.)
Received October 14 2013; revised versions received January 19 2014; March 17 2014. Published in Journal of Integer Sequences, March 23 2014.
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