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RESEARCH ARTICLE

h-Type indices, partial sums and the
majorization order

Leo Egghe1 and Ronald Rousseau2,3

1University of Hasselt, Belgien
2University of Antwerp, Faculty of Social Sciences, B-2020 Antwerp, Belgien
3KU Leuven, MSI, Facultair Onderzoekscentrum ECOOM, Naamsestraat 61, Leuven B-3000, Belgien

Schlüsselwörter: partial sums of an array, h-index, g-index, R-index, Gini index, Lorenz curve

ABSTRAKT

We study the array of partial sums, PX, of a given array X in terms of its h-type indices.
Concretely, we show that h(PX) can be described in terms of the Lorenz curve of the array X
and obtain a relation between the sum of the components of PX and the Gini index of X.
Darüber hinaus, we obtain sharp lower and upper bounds for h-type indices of PX.

EINFÜHRUNG

h-type indices such as the h-index itself and the g-index have interesting mathematical prop-
erties as shown, Zum Beispiel, in Egghe and Rousseau (2019A), although they are only probably
approximately correct (PAC) in research evaluation exercises (Bouyssou & Marchant, 2011;
Waltman & van Eck, 2012; Rousseau, 2016). In this investigation we continue our theoretical
investigation of the mechanism leading to h-type indices. Concretely, we study properties re-
lated to h-type indices of the array of partial sums of a given array X. We recall that these
partial sums form the basis of the Lorenz curve and the related Gini index. As a consequence,
we also obtain relations with the Gini index and the Lorenz curve of the original array X. Wir
will further derive sharp lower and upper bounds for h-type indices of PX.

In the following section we recall the definitions we will use in this investigation.

2. DEFINITIONS
Let (R+)N be the set of all arrays of length N with nonnegative real values. An array X = (xr)r=1,2,
…,N in (R+)N is said to be decreasing if, for all r = 1, 2, …, N, xr
≥ xr+1. The set (R+)N has a
natural partial order defined by X ≤ Y if, for all r = 1, 2, …, N, xr ≤ yr. Equality between X
and Y only occurs if xr = yr for all r. We denote the set of all decreasing arrays in (R+)N with
at least one component larger than or equal to 1 by ΦN.

Next we recall the definition of some h-type indices for arrays in (R+)N.

2.1. The h-index (Hirsch, 2005)

2 ΦN. The h-index of X, denoted h(X), is the largest natural number such
Let X = (xr)r=1,2,…,N
that the first h coordinates have each at least a value h. If all components of a decreasing array
X are strictly smaller than 1, then h(X) = 0. Such arrays are not considered further on because
we will work with arrays in ΦN. If xN, the last element in X, is larger than or equal to N, then h
(X) = N.

Keine offenen Zugänge

Tagebuch

Zitat: Egghe, L., & Rousseau, R.
(2020). h-Type indices, partial sums
and the majorization order. Quantitative
Science Studies, 1(1), 320–330. https://
doi.org/10.1162/qss_a_00005

DOI:
https://doi.org/10.1162/qss_a_00005

Erhalten: 22 Januar 2019
Akzeptiert: 27 Juli 2019

Korrespondierender Autor:
Ronald Rousseau
ronald.rousseau@uantwerpen.be

Handling-Editor:
Vincent Larivière

Urheberrechte ©: © 2019 Leo Egghe and
Ronald Rousseau. Published under a
Creative Commons Attribution 4.0
International (CC BY 4.0) Lizenz.

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h-Type indices, partial sums, and the majorization order

Recall that an h-index (and similarly for the other h-type indices defined further on) can
≥ r; umgekehrt, if for all r ≤

only be defined for decreasing arrays. Darüber hinaus, for r ≤ h(X), xr
≥ n, then n ≤ h(X). Weiter, if r > h(X), then xr < h(X) + 1. n, xr 2.2. The g-index (Egghe, 2006a, b) 2 ΦN. The g-index X, denoted g(X), is defined as the highest natural number Let X = (xr)r=1,2,…,N g such that the sum of the first g coordinates is at least equal to g2. If the sum of all coordinates of X is strictly larger than N2, then we extend the array X with coordinates equal to zero, making it into an array in ΦM, M > N, until it is possible to apply the definition.

2.3. The R-index (Jin et al., 2007)

Let X = (xr)r=1,2,…,N 2 ΦN. The R-index of X is defined as the square root of the sum of all
coordinates up to and including the one with index h(X). Omitting the square root yields
the R2-index. As it is easier to work with R2 than with R if their properties are for our purposes
the same, all concrete examples will be given for R2.

2.4. Kosmulski’s h(2) Index (Kosmulski, 2006)

Let X = (xr)r=1,2,…,N
number h(2) such that the first h(2) coordinates have each at least a value (H(2))2.

2 ΦN. The h(2) or Kosmulski index of X, denoted h(2)(X), is the largest natural

2.5. The Majorization Order (Hardy et al., 1934)

Let X, Y 2 ΦN, where N is any finite number in N = {1, 2, 3, …}. The array X is majorized by Y, or X
is smaller than or equal to Y in the majorization order, denoted as X ≤ Y if for all i = 1, …, N

>>>>< >>>>:

xi ¼

yi and

i¼1

xj ≤

yj; ∀i ¼ 1; …; N:

j¼1

We note that this definition is also valid for arrays in which all values are between zero (enthalten)
Und 1 (not included).

3. AN INEQUALITY RELATED TO THE g-INDEX AND THE MAJORIZATION ORDER
Theorem 1. ∀ N 2 N: X, Y 2 ΦN: X − < Y ⇒ g(X) ≤ g(Y) (*) Proof. Although this theorem is implied in Egghe (2009, p. 487) we present here two short l D o w n o a d e d f r o m h t t p : / / d i r e c t . m i t . / e d u q s s / a r t i c e - p d l f / / / / 1 1 3 2 0 1 7 6 0 8 0 7 q s s _ a _ 0 0 0 0 5 p d . / f b y g u e s t t o n 0 7 S e p e m b e r 2 0 2 3 proofs. First proof: For each i ≤ g(X), P i Second proof: For each i > G(Y),

G(X) ≤ g(Y).

Comments

j¼1 yj ≥
P

ich

j¼1 xj ≥ i2. This implies that g(Y) ≥ g(X).

ich

j¼1 xj ≤

ich
j¼1 yj < i2. Also this inequality implies that A. B. This theorem proves that g is an order-preserving mapping from (R+)N with the major- ization order to the positive real numbers with their natural order. The converse of inequality (*) does not hold. Consider for instance X = (5, 2, 2) and Y = (4, 4, 1). Then g(X) = g(Y) = 3 but neither X ≤ Y nor Y ≤ X holds. Quantitative Science Studies 321 h-Type indices, partial sums, and the majorization order C. D. E. F. The inequality (*) can be strict. Indeed, take X = (2, 1, 1) and Y = (2, 2, 0). Then X ≤ Y, but g(X) = 1 and g(Y) = 2. Yet, inequality (*) cannot be strict for N = 2. Indeed, consider X = (x1, x2) and Y = (y1, y2), ≤ y1 (hence g(X), g(Y) ≥ 1) and x1 + x2 = y1 + y2. As N = 2, this sum with X ≤ Y. Then 1 ≤ x1 completely determines the value of the g-index. Hence this value must be equal for X and Y. We note that even here there is no upper bound to the value of g(X) = g(Y). If it were allowed that x1 < 1 then the previous Comment D is not valid. Indeed, take X = (½, ½) and Y = (1, 0) then X ≤ Y, and g(X) = 0 and g(Y) = 1. Inequality (*) does not hold for the h- or the R2-index. Consider X = (3, 3, 3) and Y = (5, 2, 2). Then X ≤ Y but h(X) = 3 > H(Y) = 2. Darüber hinaus, R2(X) = 9 > R2(Y) = 7.

For small N we even have the opposite relation for the h-index. This is shown in the next

proposition.

Proposition 1.

(A)
(B)

For X, Y 2 Φ2 and X ≤ Y, H(X) ≥ h(Y)
For X, Y 2 Φ3, X ≤ Y and if the components of X and Y are strictly positive natural
Zahlen, then h(X) ≥ h(Y).

Proof.

(A). N = 2, X ≤ Y then x1
y1 and 2 > x2
2, it follows that h(Y) ≤ 2 = h(X).

≤ y1 and x1 + x2 = y1 + y2. Hence x2

≥ y2. If now h(X) = 1, Dann 1 ≤
≥ y2. This implies that h(Y) = 1 = h(X). The case h(X) = 2 is trivial: As N =

(B). N = 3 and X ≤ Y, then x1
implies that x3

≤ y1; x1 + x2

≤ y1 + y2 and x1 + x2 + x3 = y1 + y2 + y3. This already

≥ y3. We now consider three cases: H(Y) = 3, H(Y) = 2 and h(Y) = 1.

Assume first that h(Y) = 3. Then y3

≥ 3. Hence we see that x1

≥ x2

≥ x3

≥ y3

≥ 3, from which

we derive that h(X) = 3 = h(Y).

Assume next that h(Y) = 2. Then y3 < 3, hence y3 = 2 or y3 = 1, and y1 ≥ 2. We first consider the case y3 = 1. We know already that x3 is at least equal to 1. So, if x3 is equal to 1, ≤ x2. Now h(Y) = 2 leads then x1 + x2 = y1 + y2. As x1 to 2 ≤ y2 ≤ x2, or h(X) ≥ 2 = h(Y). Still with y3 = 1 we now consider the case that x3 > 1. Dann
≥ x2
x1

≤ y1 the previous equality implies that y2

≥ 2, leading to h(X) ≥ 2 = h(Y).

≥ y2

≥ x3

Next we consider the case y3 = 2. Then x1

≥ x2

≥ x3

≥ y3 = 2, which implies that h(X) ≥ 2 = h(Y).

Endlich, as components are assumed to be strictly positive natural numbers, H(X) and h(Y)

are at least equal to 1. Hence h(Y) = 1 implies h(X) ≥ h(Y).

Comments

Proposition 1(B) is not valid if some of the components are zero. This is illustrated as
follows. Let X = (2, 1, 1) and Y = (2, 2, 0). Then X ≤ Y, but h(X) = 1 and h(Y) = 2.
Proposition 1(B) is also not valid if some of the components are not natural numbers.
In der Tat, let X = (2, 1.5, 1.5) and Y = (2, 2, 1). Then X ≤ Y, but h(X) = 1 and h(Y) = 2.
Propositions 1(A) Und 1(B) are not valid for R2.

(A) N = 2. Consider X = (1, 1) and Y = (2, 0). Then X ≤ Y, H(X) = h(Y) = 1; R2(X) = 1 Und

R2(Y) = 2.

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h-Type indices, partial sums, and the majorization order

(B) N = 3. Consider X = (2, 2, 2) and Y = (3, 2, 1). Then X ≤ Y, H(X) = 2 = h(Y) and R2(X)

= 4 < R2(Y) = 5. D. Proposition 1 with N ≥ 4 is not valid for the h-index h. Consider X = (6, 5, 2, 2) and Y = (6, 5, 3, 1). Then X ≤ Y, h(X) = 2 < h(Y) = 3. We further remark that R2(X) = 11 < R2(Y) = 14. 4. INTRODUCING THE ARRAY OF PARTIAL SUMS Now we come to the main part of this article. First we introduce some notation. Let X = (xr)r=1,2, …,N 2 ΦN and consider the partial sums: X i j¼1 xj; i ¼ 1; …; N: Ranking these partial sums again in decreasing order leads to the array PX. The ith component of PX, denoted as yi, is equal to xj. An example: Let X = (4, 3, 2, 1). Then PX = (10, 9, 7, 4) and h(PX) = 4. N−iþ1 j¼1 P Remarks P N j¼1 xj; (PX)N = x1. (PX)1 = A = If X ends with p zeros, then PX starts with p + 1 As. 1. 2. 3. Clearly X ≤ PX, as (PX)N = x1. Hence h(X) ≤ h(PX), g(X) ≤ g(PX) and R(X) ≤ R(PX). (Egghe & Rousseau, 2019a; Proposition 2). If X denotes the number of received citations of an author’s publications, then the indicator value h(PX) shows how many of the less cited publications can be removed so that the total number of the remaining items of X is higher than the rank of this total in the array PX. This is another way of describing the impact of the most cited publications. Contrary to the case of h(X), h(PX) may increase if a publication in X’s h-core, not necessarily the most cited one, increases its number of citations. We provide an example: let X = (4, 2, 0, 0, 0, 0, 0, 0). Then h(X) = 2; PX = (6, 6, 6, 6, 6, 6, 6, 4) and h(PX) = 6. Consider now X = (4, 3, 0, 0, 0, 0, 0, 0). Then h(X ) = 2; PX0 = (7, 7, 7, 7, 7, 7, 7, 4) and h(PX0) = 7. 0 0 5. A RELATION WITH THE GINI INDEX We recall (Rousseau et al., 2018, formula (4.19)) that the Gini concentration index of a de- creasing array X of nonnegative real numbers, (xj)j=1,…,N is obtained as G Xð Þ ¼ (cid:2) 1 N N þ 1− 2 A X N j¼1 j xj (cid:3) ; (1) P where A = S(X), is N j¼1 xj . Consider now PX. The sum of all components of PX, denoted as XN xj þ XN−1 xj þ … þ X1 xj ¼ XN ð N−j þ 1 j¼1 j¼1 j¼1 j¼1 ð Þxj ¼ N þ 1 ÞA− XN jxj: j¼1 Quantitative Science Studies 323 l D o w n o a d e d f r o m h t t p : / / d i r e c t . m i t . / e d u q s s / a r t i c e - p d l f / / / / 1 1 3 2 0 1 7 6 0 8 0 7 q s s _ a _ 0 0 0 0 5 p d / . f b y g u e s t t o n 0 7 S e p e m b e r 2 0 2 3 h-Type indices, partial sums, and the majorization order From this result we obtain a relation between G(X) and S(X): (cid:2) 1 N 1 NA 2 N þ 1− ð ð N þ 1 Þ ÞA−S Xð Þ A Þ: Þ ð 2S Xð Þ−A N þ 1 ð (cid:3) ¼ 1 N (cid:2) ð N þ 1−2 N þ 1 Þ þ (cid:3) 2S Xð Þ A G Xð Þ ¼ ¼ Conversely S Xð Þ ¼ A ð ð 2 N 1 þ G Xð Þ Þ þ 1 Þ: (2) (3) An example: If X = (a, a, a, a), a > 0, then G(X) = 0 (by definition), N = 4, A = 4a and
S(X) = 10a. Now we check formula 3 and find that, In der Tat, 10a = 4a

2 (4 + 1).

6. A GEOMETRIC INTERPRETATION OF H(PX) IN TERMS OF THE LORENZ CURVE LX

For the decreasing array X of nonnegative real numbers, (xj)j=1,…,N and for aj =

Lorenz curve of X, denoted as LX, connects points with coordinates
average of array X is denoted as (cid:2)x ¼ 1
N

N
i¼1 xi.

Now h(PX) is equal to the largest natural number i such that yi =

equal to (N + 1) minus the smallest natural number i such that yN−i+1 =
Dividing by the sum of all elements in X this yields

X
ich

j¼1 aj ≥ N−i þ 1
(cid:2)X
N *

(cid:2)

1
(cid:2)X

1− s þ

(cid:3)

xjP

(cid:4)
s ¼ i
N

;

N
k¼1 xk
P
ich
j¼1 aj

= xj
N(cid:2)X , Die
(cid:5)

. Der

N−iþ1
j¼1

xj ≥ i, was auch so ist
P
j¼1 xj ≥ N − i + 1.
ich

Folglich, H(PX) is equal to N (1 − the smallest s such that LX(S) ≥ 1
(cid:2)X

(cid:6)

(cid:7)

1−s þ 1
N

An illustration: If X = (3, 2, 1, 0), N = 4, (cid:2)x = 3

(cid:6)
3 1− 2

(cid:7)
4 þ 1
4

6 ≥ 2

2, PX = (6, 6, 5, 3) and h(PX) = 3. Now LX

6 < 4. Hence, the smallest s is equal to 2/4 and h(PX) = = 2 2 (cid:7) (cid:6) = 5 3 1− 1 = 2 4(1 − 2/4) + 1 = 2 + 1 = 3. 3; but LX 4 þ 1 (cid:6) (cid:7) 2 4 4 ) + 1. (cid:6) (cid:7) 1 4 = 3 In Egghe and Rousseau (2019b) we studied h(PX) and its relation with the Lorenz curve in a continuous context. This led to a new geometric interpretation of the h-index. 7. BOUNDS ON h-TYPE INDICES In the next sections we derive bounds for h-type indices of PX. This is of importance for the following reason. A function relates an input to a unique output. In this way the standard h-index is a function which maps an array to a natural number. Yet it is not an explicit func- tion, such as the function that maps the real number x to x2 + 4x + 7 or the function which maps a finite array to its sum. Finding an h-index needs a procedure and hence it is not pos- sible to study properties in an analytical way (e.g., using integrals). The bounds obtained in this article are explicit functions which can be studied using analytical methods. We denote by ⌊a⌋, the floor function of a (i.e., the largest integer smaller than or equal to a). We note that a ≥ ⌊a⌋ > a − 1. Using the notation just introduced we come to the following
interesting theorem.

Quantitative Science Studies

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h-Type indices, partial sums, and the majorization order

Theorem 2. Let X = (xr)r=1,2,…,N 2 ΦN, Dann

min N; Að

Þ ¼ min N; N(cid:2)X
D

Þ ≥h PXð

(cid:8)
D
Þ
Þ≥ N þ 1

(cid:9)
;

(cid:2)X
(cid:2)x þ 1

(4)

Before proving Theorem 2 we make three remarks:

1. The first inequality, namely min(N, A) ≥ h(PX) is easy to see because, on the one hand,
an h-index can never be larger than the length of the array and on the other 1 ≤ h(PX) ≤
P

Þþ1

N−h PXð
j¼1

xj ≤ A.

2. H(PX) = N if and only if x1
3. H(PX) = 0 can never occur in our context. In der Tat, this may only occur if all components

≥ N.

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are strictly smaller than 1, which is excluded. Noch, in Egghe and Rousseau (2019C)
we showed that formula 4 is also correct in cases for which h(PX) = 0.

Proof of Theorem 2. We only have to show the second inequality. By definition we know

that h(PX) is equal to the largest index i such that yi =
xj =
(N − i + 1).(the average of (x1, x2, …, xN−i+1)) ≥ (N − i + 1). (cid:2)X (as the array X is ranked in decreasing
Befehl).

xj ≥ i. We know that yi =

N−iþ1
j¼1

Jetzt, Wenn (N − i + 1). (cid:2)x ≥ i then certainly yi

≥ i. Solving this inequality for i leads to

i ≤ N þ 1

(cid:2)X
(cid:2)x þ 1

As the index i is a natural number it follows that h(PX) ≥ N þ 1

k
. This proves this theorem.

Þ (cid:2)X
(cid:2)xþ1

In order to make these bounds more concrete we provide a table (Tisch 1) for some values
of N and (cid:2)X (or A), showing how sharp these bounds often are. Largest differences occur when
the average number of items is one.

The next theorem shows that the second inequality in Theorem 2 becomes an equality

for the array

(cid:2)
X = ((cid:2)X, (cid:2)X, …, (cid:2)X) 2 (R+)N.

Theorem 3. Let X = (xr)r=1,2,…,N

2 ΦN, Dann

min N; Að

Þ ¼ min N; N(cid:2)X
D

D
Þ≥h P(cid:2)
X

(cid:8)
D
Þ ¼ N þ 1

(cid:9)

(cid:2)X
(cid:2)x þ 1

;

(5)

Tisch 1. Some specific bounds for h(PX) according to Eq. 4

N
10

100

200

0.5
0.1
1 ≥ h(PX) ≥ 1
5 ≥ h(PX) ≥ 3
3 ≥ h(PX) ≥ 2
15 ≥ h(PX) ≥ 10
10 ≥ h(PX) ≥9
50 ≥ h(PX) ≥ 33
20 ≥ h(PX) ≥ 18 100 ≥ h(PX) ≥ 67

1
10 ≥ h(PX) ≥ 5
30 ≥ h(PX) ≥ 15
100 ≥ h(PX) ≥ 50
200 ≥ h(PX) ≥ 100

(cid:2)X

2
10 ≥ h(PX) ≥ 7
30 ≥ h(PX) ≥ 20
100 ≥ h(PX) ≥ 67
200 ≥ h(PX) ≥ 134

5
10 ≥ h(PX) ≥ 9
30 ≥ h(PX) ≥ 25
100 ≥ h(PX) ≥ 84
200 ≥ h(PX) ≥ 167

10
10 ≥ h(PX) ≥ 10
30 ≥ h(PX) ≥ 28
100 ≥ h(PX) ≥ 91
200 ≥ h(PX) ≥ 182

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325

h-Type indices, partial sums, and the majorization order

Proof. We see that P(cid:2)

X = (N(cid:2)X, (N − 1) (cid:2)X, …, 2(cid:2)X, (cid:2)X). Then h(P(cid:2)

X) is the largest natural number i
such that (N − i + 1) (cid:2)x ≥ i. We observe that then h(P(cid:2)X) is equal to the largest natural number i
J
X) = N þ 1
D
such that i ≤ (N + 1)

. This proves Theorem 3.

(cid:2)X
(cid:2)xþ1 and hence h(P(cid:2)

Þ (cid:2)X
(cid:2)xþ1

We next present some examples, illustrating different aspects of the previous results.
Example 1. Returning to the example introduced before, we have X = (4, 3, 2, 1), mit (cid:2)x =
J
D
= ⌊3.571⌋ = 3. Das

2.5 and PX = (10, 9, 7, 4). Now N = h(PX) = 4 > N þ 1

= 5 (cid:2) 2:5
2:5þ1

Þ (cid:2)X
(cid:2)xþ1

illustrates that the second inequality in Theorem 2 can be strict. Continuing now with
that N = 4 > h(P(cid:2)X) = h(10, 7.5, 5, 2.5) = 3 = 5 (cid:2) 2:5

3:5

(cid:2)
X we see

Example 2. Consider X = (4, 2, 1, 1) mit (cid:2)x = 2 and PX = (8, 7, 6, 4). Now N = h(PX) = 4 >
Þ (cid:2)X
N þ 1
D
X) = h(8, 6, 4, 2) =
(cid:2)xþ1
k
J
3 = 5 (cid:2) 2
3

. This example illustrates that the floor function is really needed, Weil 3 < = 5 (cid:2) 2 2þ1 j k = 10 3 (cid:2) X we see that N = 4 > H(P(cid:2)

= 3. Continuing with

j k
= 10
3

10/3.

(cid:11)

(cid:10)
5 (cid:2) 1
2

Example 3. Let X = (4, 0, 0, 0) mit (cid:2)x = 1 and PX = (4, 4, 4, 4). Now N = h(PX) = 4 > 5 (cid:2) 1
1þ1

=
= ⌊2.5⌋ = 2. This is another example that the second inequality in Theorem 2 can be
= ⌊2.5⌋. Das ist
strict. Continuing with
not only another example that the floor function is really needed, but it also illustrates that the
first inequality in Theorem 3, and hence also in Theorem 2, can be strict.

(cid:2)
X we see that N = A = 4 > h(P(cid:2)

X) = h(4, 3, 2, 1) = 2 = 5 (cid:2) 1

(cid:10)

(cid:11)

Example 4. In the previous examples h(PX) = N. Next we present an example where
H(PX) < N. Let X = (3, 2, 1, 0). Then (cid:2)x = 3/2 and PX = (6, 6, 5, 3). Now N = 4 > H(PX) =
(cid:2)
3 ≥ 5 (cid:2) 3=2
X we see that N = 4 > h(P(cid:2)X) = h(6, 4.5,

= ⌊3⌋ = 3. Continuing with

3=2ð

J
= 5 (cid:2) 3
5
k

Þþ1
J

3, 1.5) = 3 = 5 (cid:2) 3
5

= 3.

Example 5. Endlich, we present an example where min(N, A) = A < N. Let X = (2, 0, 0, 0). j Then (cid:2)x = ½ and PX = (2, 2, 2, 2). Now A = 2 = min(N, A) = h(PX) = 2 ≥ 5 (cid:2) 1=2 ð 1=2 Þþ1 (cid:2) X we see that A = 2 = min(N, A) > H(P(cid:2)

X) = h(2, 3/2, 1, 1/2) = 1 = 5 (cid:2) 1

J
= 5 (cid:2) 1
3
k
J

= 1.

Continuing with

Corollaries

Wenn (cid:2)x ≥ N, then h(PX) = N.

Proof. As limt ⇒ ∞

T
tþ1 = 1, there exist a number t0 such that for all t > t0.

Þ
N ≤ N þ 1
D

T
t þ 1

< N þ 1: This double inequality clearly holds if we take t0 = N. With (cid:2)x in the role of t we see that in these circumstances N þ 1 = N and thus by Theorem 2, Corollary A is proved. k j ð Þ (cid:2)x (cid:2)xþ1 B. lim(cid:2)x ⇒ ∞ h(PX) = N This follows immediately from Corollary A. Quantitative Science Studies 326 l D o w n o a d e d f r o m h t t p : / / d i r e c t . m i t . / e d u q s s / a r t i c e - p d l f / / / / 1 1 3 2 0 1 7 6 0 8 0 7 q s s _ a _ 0 0 0 0 5 p d / . f b y g u e s t t o n 0 7 S e p e m b e r 2 0 2 3 h-Type indices, partial sums, and the majorization order Remark When applied to publications, corollaries A and B show that for large (cid:2)x we only need those publications in X with the highest citations to determine h(PX). This is in accordance with the principle and meaning of an h-index. 8. PARTIAL SUMS AND THE G-INDEX Using the same notations as before, we next prove the analogue of Theorem 2 for the g-index. We recall that the g-index has no upper limit. Theorem 4. Let X = (xr)r=1,2,…,N 2 ΦN. Then (cid:8) 8 >>>>< >>>>:

g PXð

Þ≥

2N þ 1
D
R(cid:8)

(cid:2)X
(cid:2)x þ 2
ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
(cid:2)X
N N þ 1
2

yj < N2 6að Þ j¼1 XN yj ≥N2 6bð Þ j¼1 l D o w n o a d e d f r o m h t t p : / / d i r e c t . m i t . (cid:9) (cid:9) if if o Proof. 8 < : g PXð Þ ¼ P n max i 2 N; n max i 2 N; j¼1 yj ≥ i2 i P j¼1 yj ≥ i2 N o if if P j¼1 yj < N2 N P j¼1 yj ≥N2 N P P (cid:4) P (cid:5) P P i j¼1 yj ¼ N−1 s¼1 xs þ … þ xk Now, (cid:4) … + (N − i + 1) (cid:2)x (because X is ordered decreasingly) = (cid:2)x × N Nþ1 s¼1 xs þ N−jþ1 k¼1 i j¼1 ¼ N Þ ð 2 P N−iþ1 s¼1 ð − N−i xs ≥ N(cid:2)x + (N − 1) (cid:2)x + (cid:5) Þ ð Þ N−iþ1 2 = (cid:2)x 2 i(2N − i + 1). Now we require that this expression is larger than or equal to i2. This leads to: i ≤ ð 2N− i þ 1 2 Þ(cid:2)x : (cid:2)x (cid:2)xþ2. Taking into account that g(PX) is an integer, we obtain that if k Solving for i yields: i ≤ (2N + 1) P j¼1 yj < N2 then g(PX) ≥ j N ð 2N þ 1 Þ (cid:2)x (cid:2)xþ2 . N j¼1 yj ≥ N2 then we have to study P P P If j¼1 yj ≥ (cid:2)x. N Nþ1 (cid:4) N 2 ð (cid:5) Þ ≥ i2 (is all we need). Hence i ≤ N j¼1 yj ≥ i2. In the same way as above we find that qj ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ k (cid:2)x Þ ð 2 N N þ 1 q ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ (cid:2)x Þ ð 2 N N þ 1 or imax = , where imax denotes the maximal value the index i can take here. Consequently, if P N j¼1 yj ≥ N2 then g(PX) ≥ qj ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ k (cid:2)x Þ ð 2 N N þ 1 . Similar to the theory for the h-index, the next theorem shows that inequality in Theorem 4 becomes an equality for the array (cid:2) X. Quantitative Science Studies 327 / e d u q s s / a r t i c e - p d l f / / / / 1 1 3 2 0 1 7 6 0 8 0 7 q s s _ a _ 0 0 0 0 5 p d . / f b y g u e s t t o n 0 7 S e p e m b e r 2 0 2 3 h-Type indices, partial sums, and the majorization order Theorem 5. Let X = (xr)r=1,2,…,N 2 ΦN, then (cid:8) 8 >>< >>:

D
g P(cid:2)
X

Þ ¼

(cid:9)

D
2N þ 1
R(cid:8)

(cid:2)X
Þ
(cid:2)x þ 2
ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
(cid:9)
(cid:2)X
N N þ 1
2

Wenn

(cid:2)X < (cid:2)x ≥ 2N N þ 1 2N N þ 1 Þ 7að 7bð Þ P N Proof. Now: (cid:4) … + 1. (cid:2)x = (cid:2)x. N Nþ1 j¼1 yj ¼ (cid:5) Þ . ð 2 P N j¼1 (cid:4) P (cid:5) (cid:2)x ¼ P N−jþ1 k¼1 N s¼1 (cid:2)x þ P N−1 s¼1 (cid:2)x þ … þ P1 s¼1 (cid:2)x = N(cid:2)x + (N − 1) (cid:2)x + P Hence, Similarly, N ð j¼1 N − j þ 1 P Þ(cid:2)x < N2⇔(cid:2)x < 2N Nþ1. N ð j¼1 N−j þ 1 Þ(cid:2)x≥N2⇔(cid:2)x≥ 2N Nþ1. Comment. Also here we can make the remark that lower bounds for g(PX) and g(P(cid:2)X) depend only on N and (cid:2)x. Examples Example 1. Take X = (4, 4, 4, 4), (cid:2)x = 4 and PX = (16, 12, 8, 4). Then g(PX) = 6 (as 40 > 62 Und
40 < 72). As this is a case where 40 > 42 we have to check formula 6b. This formula states that
6 = g(PX) ≥

= 6. Thanks to the use of the floor function we obtain an

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
4(cid:2)4(cid:2)5
2

ﬃﬃﬃﬃﬃ
(cid:11)
40

P(cid:10)

equality.

Example 2. Take X = (4, 3, 2, 1), (cid:2)x = 2.5 and PX = (10, 9, 7, 4). Then g(PX) = 5 (als 30 > 52 Und

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2:5(cid:2)4(cid:2)5
2

P(cid:10)

(cid:11)
ﬃﬃﬃﬃﬃ
25

30 < 62). Also here we have to check formula 6b. We see that 5 = g(PX) ≥ 5. This is an example where the floor function is not necessary. Example 3. For X = (4, 0, 0, 0), (cid:2)x = 1 and PX = (4, 4, 4, 4). Here the sum, namely 16, is larger than or equal to N2 = 42; hence we have to check formula 6b. This leads to 4 = g(PX) ≥ qj = 2. This is another case where we have strict inequality. ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ k 12 (cid:2) 1 2 = (cid:11) p(cid:10) ﬃﬃﬃ 6 Example 4. For X = (2, 0, 0, 0), (cid:2)x = 0.5 and PX = (2, 2, 2, 2). Here the sum, namely 8 < 42, = 1. Also here we have k hence we have to check formula (6a). This leads to 2 = g(PX) ≥ 9(cid:3)0:5 strict inequality. 2:5 j e) Finally we consider a case for which N ≠ 4. Let X = (5, 4, 3, 2, 1), (cid:2)x = 3 and PX = (15, 14, 12, 9, 5). Here the sum namely 55 > 52; hence we check formula 6b. We first note that g(PX) =
7 (55 > 72 Und 55 < 82). Now 7 = g(PX) ≥ = 6. This is again a case with a ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 3(cid:2)5(cid:2)6 2 ﬃﬃﬃﬃﬃ (cid:11) 45 qj p(cid:10) = k strict inequality. Corollary lim(cid:2)x ⇒ ∞g PXð Þ ¼ ∞ Quantitative Science Studies 328 l D o w n o a d e d f r o m h t t p : / / d i r e c t . m i t . / e d u q s s / a r t i c e - p d l f / / / / 1 1 3 2 0 1 7 6 0 8 0 7 q s s _ a _ 0 0 0 0 5 p d / . f b y g u e s t t o n 0 7 S e p e m b e r 2 0 2 3 h-Type indices, partial sums, and the majorization order Proof. If (cid:2)x is large, then we have to consider formula 6b. Then the right-hand side of formula 6b becomes unlimited large and hence this also holds for g(PX). This result confirms the fact that the g-index has no upper limit. 9. PARTIAL SUMS, THE R(R2)-INDEX AND KOSMULSKI’S INDEX H(2)(X) In the previous sections we studied the h-index and the g-index. As a final case we mention the R2-index and Kosmulski’s h(2)-index. For proofs of the results we refer the reader to Egghe and Rousseau (2019c). Theorem 6. Let X = (xr)r=1,2,…,N 2 ΦN. Then (cid:6) (cid:2)x (cid:2) N þ 1 ð Þ2 (cid:2) N (cid:2) (cid:2)x 2 (cid:2)x þ 1 ð Þ 2 R PXð Þ >

2 þ 2N (cid:2) (cid:2)x −(cid:2)x− 2

(cid:7)
:

Theorem 7. Wenn (N + 1)

(cid:2)X
(cid:2)xþ1 is a natural number and X = (xr)r=1,2,…,N 2 ΦN, Dann

D
P(cid:2)
X

Þ ¼

(cid:2)xð Þ2

Þ N(cid:2)x þ 2N þ 1
N þ 1
D
Þ
D
Þ2
2 (cid:2)x þ 1
D

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Endlich, we extend our results to the case of Kosmulski’s index, denoted as h(2), referring to
Egghe and Rousseau (2019C) for proofs.

Theorem 8. Let X = (xr)r=1,2,…,N

P(cid:4)

min N;

(cid:5)

ﬃﬃﬃﬃ
A

2 ΦN. Dann
(cid:8)

2ð Þ

≥h

PXð

Þ≥

1
2

(cid:2)

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
(cid:2)xð Þ2 þ 4 N þ 1
Þ(cid:2)X
D

−(cid:2)X

(cid:3)

(cid:9)

Similarly to Theorem 3 we have

Theorem 9. Let X = (xr)r=1,2,…,N

P(cid:4)

min N;

(cid:5)

ﬃﬃﬃﬃ
A

≥h

2 ΦN. Dann
(cid:8)
1
2

D
P(cid:2)
X

Þ ¼

2ð Þ

(cid:2)

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ
(cid:2)xð Þ2 þ 4 N þ 1
Þ(cid:2)X
D

(cid:3)

(cid:9)

−(cid:2)X

(10)

(11)

10. DISCUSSION AND CONCLUSION

In this article we studied arrays of partial sums, PX, of a given array X in terms of their h-type
indices. We showed that h(PX) can be described in terms of the Lorenz curve of the array X.
Darüber hinaus, we obtained sharp lower and upper bounds for these h-type indices. We found
bounds that only depend on N, the length of the array, and the average of array X, or equiv-
alently, on the length of the array and the total sum of all items in the array.

As h(PX) is an h-index it is not surprising that it is not strictly independent in the sense of
Bouyssou and Marchant (2011). This means that if h(PX) < h(PY) and if one adds to X and Y the same items (X becomes X ) then it is possible that h(PX0) > H(PY0). An exam-
Bitte: Let X = (2, 0, 0, 0, 0) and Y = (1, 1, 1, 1, 1). Then PX = (2, 2, 2, 2, 2) with h(PX) = 2, and PY =
(5, 4, 3, 2, 1) with h(PY) = 3, hence h(PX) < h(PY). Adding 5 times 1 to each of them yields X = 0 (2, 1, 1, 1, 1, 1, 0, 0, 0, 0), PX0 = (7, 7, 7, 7, 7, 6, 5, 4, 3, 2) with h(PX0) = 6, and Y = (1, 1, 1, 1, 1, 1, 1, 1, 1, 1), PY0 = (10, 9, 8, 7, 6, 5, 4, 3, 2, 1) with h(PY0) = 5, hence h(PX0) > H(PY0).

0
, and Y becomes Y

Quantitative Science Studies

329

h-Type indices, partial sums, and the majorization order

A reviewer asked if h(PX) can be described in terms of Vannucci’s (2010) dominance dimen-
ich
,

sion. Es kann: Using Vannucci’s notation we see that h(PX) = dom projn
N

Nþ1−n
j¼1

j n1ð

H
(cid:4)

(cid:5)

(cid:4)

where X = (xn)n=1,…,N is an array of length N.

Our investigation illustrated the rich mathematical structure hidden in the mechanism lead-
ing to h-type indices (see also Egghe & Rousseau, 2019D). In this article we considered the
discrete case, requiring the floor function in order to get the correct results. In further research
we intend to study the continuous case, where by definition no floor function will be needed.
Dann, bounds will be differentiable and integrable functions.

ACKNOWLEDGMENTS

The authors thank anonymous reviewers for useful suggestions to improve the presentation of
dieser Artikel.

BEITRÄGE DES AUTORS
Leo Egghe: conceptualization; formal analysis; investigation; methodology; writing—original
Entwurf; writing—review and editing. Ronald Rousseau: validation; writing—review and editing.

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COMPETING INTERESTS

The authors have no competing interests.

FUNDING INFORMATION

No funding has been received.

DATA AVAILABILITY

Not applicable.

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Quantitative Science Studies

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