(J,
...
A SYMMETRY PROPERTY OF THE GENOCCHI NUMBERS
by
Dominique Dumont
D~partement de Math~mtique
Universit~ de Stpasbourg
....
'
Dominique Foata
Department of Statistias
University of North Carolina. at ChapeZ Hill
Institute of Statsstics Mimeo Series No. B8Z
January, 1975
.,~
A SYMMETRY PROPERTY OF
THE GENOCCHI NUMBERS
by
Dominique Dumont* and Dominique Foata i :*
ABSTRACT
A refinement of the Genocchi numbers into a sum of symmetric
coefficients is obtained, together with a combinatorial interpretation
for this refinement.
*Universite de Strasbourg, France
**University of North Carolina, Chapel Hill,U.S.A.
~2-
INTRODUCTION.
1.
The Genoaohi numbePB
G2n en > 0) are positive integers, which may
be defined by means of their exponential generating function (or rather
that of the coefficients
(_l)n G )
2n
2u/(eu + 1)
(1)
=u
l
+
(u
2n
n~l
/(2n)!) (~l)n G •
2n
They are the closest integers to the BepnouZZi numbePB
the following sense.
(2)
u
u/(e - 1)
2n (n
Let the latter numbers be defined by
=1
u/2 +
-
B
> 0)
in
(u 2n/(2n)l) (_l)n+l B .
2n
I
n~l
Then, the following identity is a simple consequence of von Staudt-Clausen
theorem (see e.g. Carlitz [2], or Hardy and Wright [6] p. 91)
a(a
~)
for any integer a.
a
=2
2n
=0
- 1) B2n
mod
1
The least non-trivial integer to be considered is
and this gives precisely the Genocchi numbers
~)
(n > 0) •
The first values of these two sequences of numbers are shown in table 1.
TABLE 1
n
1
2
G
2n
1
1
1/6
1/30
3
1/42
6
7
155
2073
38227
5/66
691/2730
7/6
4
5
17
1/30
-3.-
The purpose of this paper is to prove the following two theorems
2.
1a and
lao
Theorem. Let
three variabZes
(Fn(x,y,x))n>O
x, y, z defined by the recurrenoe reZation
FI (x,y,z)
(5)
and for n
=1
~ 2
= (x+z)
F (x,y,z)
n
(6)
The~
be the sequenoe of polynomiaZs in
for any
(y~z)
Fn- l(x,y,z+l) - z
n
to the set of the three variables x, y,
Z.
Fn- l(x,y,z) •
is symmetrio
the polynomial F (x,y,z)
n > 0
2
~ith
respeot
Moreover
(n > 0) •
(7)
Clearly, the coefficients of the polynomials
numbers into symmetric coefficients.
F
n
(x,y, z) =
~
l..
l~i,j,k
More precisely, let
a.. k x
n,~,J,
i-I
y
j-1 zk-J.
A straightforward algebraic argument shows that
lent to the recurrence relation on the
(9)
and for
alII
, , ,1
n
~
2,
are non-
Theorem 1 then gives a refinement of the Genocchi
negative integers.
(8)
Fn(x,y,z)
=1
1
~
,
a l ,~,J
. . ,k = 0
i, j, k
~
n
(5)
(n > 0)
and
(6)
a n,J.,J,
. . k's
if
(i,j,k) f (1,1,1)
are equiva-
-4-
(10)
= k-l~i~n-l
l
an,i,j,k
[i-I]
k-2
[i-I]
k-l an-1,i-l,j-l,i +
[i-I] an-l,i,j,i·
(an-l,i-l,j,i + an-l,i,j-l,i) + k-3
We then have another way of stating theorem 1
lb.
Theorem.
Let an,1,J,
. . k (n
integers defined by
(i, j, k)
and any
and
(9)
1
1,
( 1" ,
~
i, j, k
~
n) be the sequenae of
Then:lfor any sequenae of integers
(10).
pe~tation
(11)
~
J·v , k')
of
(i, j, k)
an, i' , J., , k'
Moreover
(12)
G2n+2
=
l
l~i,j,k~n
a n,1,J,
. . k •.
A table of the first values of the coefficients
a n,1,J,
. . k
appears in
table 2.
Theorem la is proved in section 2.
It is a consequence of an earlier
result due to Carlitz [2], Riordan and Stein [7].
The main result of our paper is stated in theorem 2, which provides
with a aombinatoriaZ interpretation for the sequence
map
f
of the interval
[2n] = 0, 2,
exaeedant, if the inequality x
each n > 0
let
onto the subset
to
An
and
x
An
f(x)
2n}
[2n].
... ,
2n}
holds for every
x
of the even integers.
The ordered pair
A
into itself is said to be
in
[2n]
For
denote the set of all exceedant surjections of
{2, 4, 6,
to
~
... ,
(Fn(x,y'Z))n>O.
(x, f(x))
Let
f
[2n]
belong
is said to be an
-5-
upper reaord of f
for all
y with
= 1,
if either x
1
~
Y ~ x,
maximum oaaurrenae of f
or
1 < x
a fixed point
1 ~ x ~ 2n-l
if
of
and
~
2n
f
and
if
fey) < f(x)
f(x)
= 2n
f(x)
= x and a
The number
of upper records (resp. fixed points, maximum occurrences) of f
denoted by
2.
I(f)
Theorem.
(resp.
For eaah
J(f),
is
K(f)).
the trivariate generating funation
n > 0
(13)
of the veator
(I, J,
K)
over An is equaZ to
It follows from theorems
(14)
G2n+ 2
1 and
= card
xyz Fn(x,y,z) .
2 that
(n > 0) •
An
In fact, this result was already established by the first author
[4],
when he gave the first combinatorial interpretations for the Genocchi
It also follows from theorem 1 that the distribution of the
numbers.
vector
is symmetria.
(I, J, K)
'This property of symmetry can be proved
directly, as ShOMl in sections 3 and 4.
elements
f
points, and
in An
K(f)
having
=k
3 it is proved that
prove the
role of
s~netry
i
and
I(f)
=i
Let
upper records,
maximum occurrences
a n,l.,],
.. k
(1
~
(10)
be the number of
J(f)
=j
i,j,k,n).
satisfies recurrence
it suffices to establish
k.
a n,1,],
.. k
fixed
In section
relation
(10) •
after permuting the
More precisely, it suffices to prove that
To
-6-
r
an,i,j,k = ].• 1 0
1 [.e.-IJ
i-I an-l,.e.,j-l,k-l
~-l.~n-
(15)
.e.-I)
[ i-2 (an-l,.e.,j,k-l
This is done in section 4.
+
an-l,.e.,j-l,k~
[.e.-I)
i-3 an-l,.e.,j,k·
As the distribution of the vector
is symmetric, there exist invoZutions
An
+
+
f
+
f9
and
f
+
fll
(1, J, K)
of the set
with the following properties
1 (f t ) = J(f)
, J(f' )
l(f) ,
K(f' ) = K(f)
J(f ll ) = K(f) ,
K(fll) = J(f)
=
on the one hand, and
l(fll) = l(f) ,
on the other hand.
Such involutions are described in section 5.
They
are based upon the combinatorial constructions given in section 3 and 4.
Numerical tables are added at the end of the paper.
The authors thank Professor Carlitz for his interest in the present
results, and also his giving them an interesting note [3] with .explicit
formulas for
2.
THE GANDHI CONJECTURE.
Gandhi
bers.
Fn(x,y,z) •
Let
[5]
proposed the following conjecture about Genocchi num-
(P (z)) >0
n
n-
be the sequence of polynomials in one variable
defined by the recurrence relation
z
-7-
Po(z)
=1
Pn(z)
= z2
(16)
P _1 (z+1) - (z_1)2 P _ (z)
n
n 1
for
n ~ 1 •
Then, Gandhi conjectured that
(17)
G2n +2 = Pn (1)
held for any
by Carlitz
n
[2]
~
o.
The conjecture was immediately proved to be correct
and Riordan and Stein
The change of variables
[7] .
~
(n
(18)
(~(z))n>O
yields a sequence of polynomials
1)
defined by the recurrence
relation
Ql (z) = 1
(19)
~(z) = (Z+1)2 ~_l(z+l) - z2 ~(z)
relation
G2n + 2 = ~ (1)
(20)
~~en
comparing
(6)
with
(19)
implies
(17)
(n > 0) •
it is readily
(21)
Hence
(n > 1)
see~
that
(n > 0) •
(7)
is a consequence of
(20)
and
(21).
will be completed if the symmetry of the polynomials
is established.
This can be done as follows.
symmetric with respect to the pair
{x,y}.
The proof of theorem 1
Fn(x,y,z)
Relation
(6)
(n > 0)
is clearly
It then suffices to show
-8-
that
(6)
also holds when x and
z are permuted.
In other words, it
suffices to establish the following recurrence relation
(22)
Fn(X,y,z) = (x+y)(x+z) Fn_l(x+l,y,z)
First
Fl(x,y,z) = 1 and
(6)
let m > 2 and assume that
x
2
Fn- l(x,y,z)
(n > 1) •
imply that
(22)
F (x,y,z) = xy + yz + zx. Now
2
has been established for any n < m •
Then
F (x,y,z)
m
= (x+z)(y+z)
F l(x,y,z+l) - z2 F l(x,y,z)
mm-
= (x+z)(y+z)((x+y)(x+z)
2
- Z
by induction on m.
F _ (x+l,y,z+1) - x 2 F _ (x,y,z+1))
m2
m2
((x+y)(x+z) F _2 (x+l,y,z) - x 2 F _ (x,y,z))
m
m2
By associating the first with the third
second with the fourth one, and using
(6)
we clearly obtain
te~m,
(22)
and the
(with
m replacing n).
3.
EXCEEDANT SURJECTIONS.
For any n, i, j, k > 0 denote by An,l,J,
. . k the set of all exceedant
surjections f of [2n] onto {2, 4, 6, ... , 2n} which have I(f) = i
upper records,
J(f) = j
fixed points, and
K(f)
=k
maximum occurrences.
Also let
A
n,.,.,k
=
U
., I
1,J~
A
n,i,j,k
and A ,
=
n,l,~,.
The purpose of this section is to prove that
U
j,k~l
A
an,l,J,
. . k
• • k •
n,l,J,
= card
An,l,J,
. . k
-9-
satisfies recurrence relation
(10).
As was noticed in the introduction,
this is equivalent to proving theorem 2.
Let
If
= card
an, • , . , k
(23)
(x,
f(x))
An, • , • , k and
k, R.
>
f(x)
0 denote by
(c , c 2 '
l
... ,
= card
An , ].,
. •, •
is an upper record (resp. fixed point, maximum occurrence)
of an exceedant surjection f,
the index and
a
n ,.]., • ,.
c _ )
k l
it will be convenient to say that x
the vaZue of the ordered pair
(x ,
f(x)) .
is
For any
([R.+1] ) the set of all the (R.+l) increasing sequences
k-1
k-l
with length k - 1 satisfying the inequalities
(24)
If k
=1
(25)
the set
([~:~])
Construction of a bijection
iP :
of An,.,.,k onto
For each f
valued map
Clearly g
t l , t z,
in A
n
f + ((c , c ' ••• , c _ ) , g)
Z
k 1
l
[R.+l]) x A
U
k-l~R.~n-1
n-1,.,.,R. for'
( k-l
n > I
.
with n > 1 define the following integral-
g by
g(x)
(26)
Let
only contains the empty sequence.
= min
(Zn-2 , f(x))
for
is an exceedant surjection of
... ,
1 s x s 2n-2 .
[Zn-2]
onto
{Z, 4, 6, .•. , 2n-2}
t.e be the indices of the R. maximum occurrences of g,
labeled from left to right, so that
1
~
t
l
<
tz
< ••• <
tR. = Zn-3.
Also
-10-
t l +1 = 2n-2. The equality f(x) = 2n
equal to 2n, 2n-1 or one of the indices
let
cannot hold, unless
x
is
have
t , ,t 2 , ••• , t + . Let f
1
l 1
k maximum occurrences, whose indices are denoted by sl' 52' ••• , sk '
with
1
S
sl
<
s2
< ••• <
sk
= 2n-1
.
As
occur in the
(t 1 , t 2 , ••• , t + 1 ) , the map g belongs to An-l,.,.,-{..b with
l
k - 1 s l. Also l S n - 1 , because every exceedant surjection in
sequence
An _l
cannot have more than
the sequence
subscripts
n - 1 maximum occurrences.
(c l , c 2 ' ... , c _ )
k l
c such that
If
k > 1 define
as being the increasing sequence of the
(27)
By construction
1 S c
(28)
If k
(2n-l
=1
1
< C
z<
••• <
c _ S l + 1 •
k 1
let
(c l , c ' ••• , c _1 ) be the empty sequence.
k
2
f(2n-l)) is the only maximum occurrence of f and
the restriction of
f
to the interval
f
f(x)
g(x)
(c 1 '
=1
=6
=6
3 4
5
6
7
8
2 4 8
6
8
8
8
2 4 6
6· 6
2
= k = 3 , K(g) = l = 3
... , ck_1) = (2, 4) .
K(f)
is simply
((c , c ' ... , c _l ) , g)
k
l
2
defined by:
x
Here
g
[2n-2].
In the following example we determine the pair
associated with
In this case
, (t ,
1
.... ,
t l + 1)
= (1,
4, 5, 6)
and
-11-
Clearly
is injective.
~
To prove that
is also bijective and
~
give the aonstruation of its inverse
~-l we proceed as follows.
g belong to A
$
with
0
n-l,.,.,~
satisfy
denoted by
f
= ~-l
The map
(Z8)
t l , t z,
f(x)
= Zn
g has
,
((c , c ,
l
z
,
t,e'
=tc
if x
Clearly
(Z6)
if
g(x)
holds.
With
1
$
l
x
t
$
n - I
and let
t maximum occurrences that will be
c _ ) , g)
k l
(29)
=
k - I
= 2n-Z
tt+l
define
by
, t C , ..• , t c _ , Zn-l , Zn
z
k l
$
2n-Z
and
x
The inequality k - 1
t c , t C , ... , t c _ .
k l
l
z
~
$
together with
t
that the set
[t+l]\{c , c ' ..• , c _l } is non-empty.
2
k
l
ment of this set. Then (29) inplies that
proves that
f
is also an exceedant surjection.
is the increasing sequence of the
c.1 's
3.
Proposition. Let;
f
((C , c ' ... , c _ ) , g).
I
J(g)
2
= J(f)
Proof.
and
From
k l
or J(f) - 1)
(29)
...,
(c l , c z,
beZong to
Then
)
c.
1
I (f)
j
(20) imply
be any eleThis
2n-2 .
(c l , c ' •.• , c k _l )
2
= 2n
and
J(f)
are
.
n,.,.,K1 and ~(f) =
leg) = I(f) or l(f) - 1
A
aaaording as
it follows that
c _ )
k l
Let
Finally
f(t
with
The next proposition shows how the numbers
related with the sequence
Let
f(x)
I
(resp.
= g(x)
t + 1)
if x
g have the same upper records within the interval
(resp.
does
o~
does
Hence
- 1].
f
If
-12-
c1
=1
then
(t 1 ' g(t 1 )) (resp. (t 1 ' f(t 1))
Hence leg) = l(f)
(resp. f)
.
record of g
.
is the rightmost upper
If
1 < cl
and
1 < k ,
(t , 2n-2) is an upper record for both f and g • It is
l
also the rightmost upper record for g
The pair (t , 2n) is also
cl
an upper record, but for f only. As there are no upper records of f
the pair
between
Finally, if
, 2n), we conclude that
c1
the map g is the restriction of
k = 1 ,
the relation
[1, 2n-3] .
In the interval
g(2n-2)
= 2n-2
f
and
Moreover
[2n-2, 2n]
< f(2n-2)
(2n, 2n)
(i)
= 2n
(2n, 2n)
since
(2n-2 , 2n-2)
In case
(ii) g(n-2) = f(2n-2) = 2n-2
= J(f)
is a fixed point of f.
is a fixed point of both
f
and
g.
f.
Hence
In case
Hence
J(g)
c _ = t + 1 and (ii)
k l
(c l , c ' ••• , c _ l ) .
2
k
corresponds to the case where
t
and
there are two possibilities to consider
J(g)
case where
[2n-2]
f.
is a fixed point of g, .. but not of
4.
to
is always a fixed point of
(2n-2, 2n-2)
(i)
f
- 1 •
g have the same fixed points within the
(i)
Now
= I(f)
leg)
leg) = ref) - 1 is evident.
Clearly, the maps
interval
(t
and
(t l , 2n-2)
1 does not occur in
+
Corollary. The sequenae
(a
. . k)
n, 113,
satisfies reaurrenae
(ii)
= J(f)
to the
Q.E.D.
(9), (10) .
since Al only contains the element
1,1,1,1 = 1 ,
f = f(l) f(2) = 2 2. For n > 1 it follows from proposition 3 that the
Proof.
set
First
a
A . . k is mapped under
n,1,J,
~
- 1 .
onto the union of the following sets
-13-
1, l+li{cl""~_l}
g)
U{((c 1 ,···,C k_l )
~
c _
k l
=l+l ,
g€An- 1 ,1. 1,J,
. l}
= c1
~
c _
k 1
<
l+l
g€An- 1 ,1,). . 1 , l}
= cl
~
c _
k l
=l+l
g€An- 1 ,1,J,
. . l} ,
U{((cl,···,c k_l ) , g)
1
<
U{((c 1,·· .,c k_l )
g)
1
U{((cl,···,c k _l )
g)
1
where the range of l
g€An- 1 ,1. 1 ,J. 1, l}
cl
is the interval
[k-l, n-l] •
Clearly, the cardi-
nalities of these four unions are the four summations occurring in the right
hand side member of relation
(10).
Q.E.D.
4.
TIED UPPER RECORDS.
A new combinatorial construction is used to prove identity
will be described in this section.
notion of tied upper reaord.
Let
upper records, whose indices are
(15)
and
The keyrole will be played by the
f
belong to
An (n>l)
51' s2' ..• , si
and have
i
labeled from right to
left, so that
1 = si
Clearly,
convention
sl
<
si_l
is the smallest integer
So
=
2n + 1.
sl
< ••• <
x
2n •
such that
The upper record
called tied, if there exists an integer
<
f(x)
(sk' f(sk))
= 2n
By
of f
is
x with the property that
(30)
Otherwise, the upper record is said to be untied.
As
f(2n-1)
= f(2n) = 2n
,
-14-
the map
f
always has a tied upper record.
the tied upper record of
Let
) f(sp))
denote
Put
P(f)
=P
the upper record:: "(5.1 )f(s.))
].
is
(5
P
with least value" f(sp) .
f
so that
=i
I(f)
For any x
= 3)
4) ••• ) 2n
= P(f)
P
~
1 •
define
hex) = f(x)
(31)
~
if x;t
S.
1-
1) .•• ) sp+1
= f(sk+1) if x = sk and i
In particular) when
P(f)
already tied) and h
=P =i
)
> k ~ P
is simply the restriction of f
to the interval
[3) 2n] .
5.
Lemma.
If f
is an exceedant surjection of
no fixed point of f
Proof.
since
6.
If f(x)
f
= x)
Let
then
x
But then
f
is even) say 2k.
(2k, f(2k))
Then
{2) 4) 6) .•. ) 2n:
f(2k-1)
[3) 2n]
onto
~
2k ,
cannot be an upper record.
Q.E.D.
belong to An and h defined by
h is an exceedant surjection of
Moreover h
onto
can be an upper record of f.
is exceedant.
Propos it i on.
[2n]
(31) •
Then
{4, 6) 8, ... , 2n} •
has the same fixed points as the restriction of f to
[3, 2n] •
-15-
Proof.
Let
3
hex) = f(x)
x
2n .
~
h(sk)
= f(sk+1) = f(sk- 1)
then
sk
record
and
val
sk+1
(sk+l' f(sk+1»
then less than
[2j-2]
= sk
= sk
(i) sk+1
= 2j
When x ~ si_l' ••• , sp+1 ' sp'
When x
x.
~
considered:
~
with
1
>
i
<
P
~
k
two cases are to be
(ii) sk+1 < sk - 1.
In case
Then h(sk) ~ sk'
sk - 1.
= 2j-l
with
= (2j-1
, 2j)
j
The map
> 1.
Each
f(x)
2j • Hence, the restriction of
f
is itself an exceedant surjection.
= f(sk+1)
(ii) h(sk)
is an untied upper record.
sk = 2j
with
j
because
(sk+1
of f
1.
On the one hand,
f(sk+1»
is untied.
>
f
to
[3, 2n]
is
latter set.
f
x = 1 ,
As
with x < 2j-l
3
~
If
x = sk+1
~
3
= s.1 = 1
As
If
with
Finally, if x = s
h(z) = f(z),
again
f(sk+1)
>
i < k
P
f(2j-l)
f(2j-1)
(31)
~
2j
and
Let
1
f(l) , f(2)
x
~
~
P
s.1- l'
=y
f(s.)
... ,
again h(z) = y .
~
4 > 2
s.1 = 1
Also
h(s.1- 2)
there is an integer
sk
y be an integer of the
then
p+1
>
It remains to show that
is equal to
5
= sk
that h and the restriction
is surjective, there is an integer x with
= f(s.)
= Y . If x = 2 ,
1
and 2 = f(s.)
< f(s.1- 1) = y
1
Suppose now x
(sk+1' f(sk+1)
In both cases the strict inequality h(sk)
h(s.1- 1)
s.1- 2
=
On the other hand,
{4, 6, 8, ••• , 2n}.
then x
is
Hence h(sk) > sk .
If h(sk) = sk
have the same fixed points.
at least one of the values
and
has the upper
because
= h(sk)
sk
'
Accordingly, it has a tied
sk'
It then follows from lemma 5 and
the range of h
If
~
h(sk)
= sk
to the non-empty inter-
f(sk- l ) ~ sk - 1,
Thus
leading to a contradiction.
holds.
>
(i)
If h(sk)
upper record and this contradict the definition of P(f).
In case
then
, S
P
and
z with
= f(2),
2
because
Then
Si_I
f(x) = Y
s.l- 1
~
=x =2
,
= f(s.1- 1) = y
then hex)
3
•
= f(x)
= y
f(sk+1) = y
and
f(Sp)
Q.E.D.
= fez)
.
-16-
Let us keep the same notations as in
i
>
k
~ P
3 ~ sk.
and assume that
for any x < sk+l'
since
(sk+l' f(sk+l))
each
j > k,
is untied.
As
we conclude that
3
with
x < sp
~
sk.
(31).
sk+l < x < sk'
hex) < h(sk)
>
>
integer
(s
p
f(s))
P
for
for each
3
~
x
~
sk.
for
Hence
1, ..• , P + 1 , P
k = i
(x, hex))
(sp_l' h(sp_l))' •.. , (sl ' h(sl))
is a tied upper record of
x with the property that
hex)
because
are the only upper records of h with indices at least equal to
Finally, as
~
f(x)
f(Sj+l) = h(Sj)
Of course, there may be other upper records
On the other hand,
Let
is an upper record of f.
h(sk) = f(sk+1)
(sk ' h(sk)) .is an upper record of h
when
and
Then h(sk) = f(Sk+l)
The same inequality holds for any x with
(sk+l ' f(sk+l))
(30)
p- l '
there is an
f
s
s
< x < s 1 and f(s )=f(x).
Let
p
pp
x be the smallest integer with this property. Then (x , hex)) = (x , f(x))
is the upper record of h with the
p-th
For each n > 1 denote by AIn- 1
of
[3 , 2n]
onto
indices of the
the set of all exceedant surjections
{4, 6, 8 , .•. , 2n} .
have obvious meanings.
uppe~
Put
I(h) = i
records of h
greatest value.
Notations such as
A'n- 1.
,1,.,. , ...
and denote by
t l , t 2, ... , t
the
i
labeled from right to left, so that
(32)
Also let
t + = 2.
i l
sl' s2' ••• , si_l
From the above remarks it follows that the indices
sense to define the sequence
(33)
(t l , t 2 , ... , t i + l ) . It makes
(d , d , ... , d _ ) by the equation
i l
2
l
occur in the sequence
.•. , t d
)
i-I
-17-
If
(d I , d , ... , d _ )
2
i I
i = 1 the sequence
is empty.
Furthermore
(34)
with
(d , d , ... , dp _I ) = (1, 2, ... , P - 1)
I
2
(35)
and
p + 1 ~ d
(36)
When f(I) # f(Z),
P
< ••• <
d. 1 s l
1-
+
1 •
the upper record of h with the
value is
p-th
greatest
Hence
h(tp )
(37)
= f(s p )
.
When f(I) = feZ) ,. the map h is the restriction of f
.2
(38)
= f(s p )
to
[3,2n]
.
Finally
hEAln- 1 ,~,.,.
0
with
(39)
Note that
(35)
occurring in
(40)
and
i-I
mean that
(36)
~
l S n - 1 .
p is the smallest integer not
(d , d 2 , ... , di _l ) .
I
Construation of a bijeation
~
of An,i,.,.
: f
onto
+
((d I , d Z' ... , d i _I ) , h)
i-lJSn.-i
([f:i1) x A~_I,l,.,.
(n>I).
and so
-18-
With f
and
=P
P(f)
..., d.
and the sequence
(31)
7.
in A.
n,1,.,.
1-
the map
1)
h id
defined by
(33) •
by
Example. Consider the following map f
x
f(x)
Here
=1
=!
2 8
5 6
6, 10
= i = 6.
I(f)
lined.
2 3 4
7
8
6 8
12
9 10
10
12
11
12
13
14
15
16
12
14
16
14
16
16
Indices and values of upper records of
fare under-
Thus
(si_1' .•• , 52' 51) = (1, 3, 5. 8, 12, 13) •
The pair
with least value.
hex)
=
=
I (h)
= £. = 7
x
and
A5
4 6 8
... ,
(5.1- l'
in
K(h)
Hence
•
6
8
Then
52' s 1)
.
=
is bijeative.
= K(f)
or
= P = 3.
, 12)
is the tied upper record
According to
(31)
8
9
10
11
12
13
14
15
16
10
10
12
12
14
16 14
16
16
(t£.+l' • •• , t 2, t 1 ) = (2, 3, 4. 5. 8. 10. 12. 13) •
(3. 5, 8. 12, 13) , the (d • d 2 , ... , d.1- 1) =
1
Proposition. The mapping
(40)
= (8
' f(S3))
P(f)
3 4 5 6 7
(1, 2, 4. 5. 7)
8.
= (s3
(sp' f(sp))
~:
f
FUrthe~ore
((d1, d2•
J(h)
= J(f)
K(f) - 1) aaaording as £.
+
1
• ••• d. 1) • h)
l-
or
(resp.
J (f) - 1
1)
defined
(resp.
does or
-19-
does not
~
of
in
(d 1 , d 2, ••• , di _1) • FinaZZy, the inverse
is defined by the foZZowing aZgorithm (i) - (vi) .
OCJCUP
~
-1
((d 1 , d 2, ••. , di _1) , h) satisfy (34) and (39), and
denote by t 1, t , ••• , t l the indices of the l upper records of h,
2
so that (32) holds. The algorithm for ~-1 is the following:
Let
(i)
(ii)
(iii)
determine p by
=1
put
s.~
by
(33) ;
f(l)
, t +
l 1
(v)
(vi)
= h(sk)
f(sk+1)
if sk
= hex) if x ~
f(sp) = 2 or h(t p )
f(x)
If d. 1
~-
=l
f(l)
=2
either i
< f(2).
= 1,
+ 1,
= (2
(5.1- 1 ' f(5.1- 1»
and define
(51' 52' ••• , s..1- 1)
~
~
3 and
3 and
x
~
is obvious.
=td
~.
, f(2»
1-
p
~(f)
Let
1
2
= f(2)
~
(41)
d.1- 1
=l
.
+ lor not.
= ((d l ,
d , ••• , d _ ) ,
2
i 1
= to 1 = 2 • The pair
~+
In particular,
is an,upper record of f.
, then si_1 = t d . 1
1-
(d , d , ... , d _ ) ,
2
i 1
l
If i = 1 the map f
= 2n
has only one upper record and necessarily f(l)
2n > f(l)
P ;
=i =1
i > land d.1- 1 < l + 1 •
i > 1 and d. 1 < l + 1
~-
(resp.
sk_1' ••• , sp+1
according as
then s·_l
~
i > k
When l + 1 does not occur in
or
d.~- 1
(d!, d 2, ••• , di _1);
Proof. The injectivity of
h).
=2
(36) ;
f(2)) = 2 if l + 1 is equal to
(resp.
does not occur in
(iv)
and
(35)
~
tl
, f(2)
-=
3 •
=2
In this case
Thus
+ 1 if f(l)
= 2 < f(2) ,
and
s.1- 1
• If
=2
,
-20-
and
l + 1 does not occur in
(42)
if f(l)
~
2
= f(2)
(d , d , ••• , d _ )
1 2
i 1
and s.1- 1 ~ 3 •
Taking into accout proposition 6 the number J(h)
h is equal to
J(f)
or J(f) - 1,
of fixed points of
according as l + 1 does or does
not occur in
(d , d , ••• , d _1)
i
1 2
In the same manner, if d1 = 1, then (35) and (36) imply
that 1 < P = P(f). The maps f and h have the same maximum occur-
rences.
P(f)
If 1 does not occur in
= P = 1.
(d , d , •.• , d _1), then
i
1 2
In this case h(sl) = f(s2) < 2n = f(sl) . As f(x)
for every x > s1'
we conclude that
K(h)
= K(f)
- 1.
= hex)
This proves the
first part of the proposition.
Conversely, let
and f
f
defined by
((d , d 2,
1
(i) - (vi) •
... , d.
1-
1) , h)
satisfy
(34)
(39) ,
It is straightforward to verify that
is an exceedant surjection, and its image under
~
is the above
ordered pair.
9.
and
Q.E.D.
Corollary. The sequence
(a
. . k)
n, 1,),
satisfies
rea~enae
(9) ,
(15) •
be in A . . k and ~(f) = ((d 1 , d 2 , ••. , d1._ 1) , h) •
n,l,J,
According to proposition 8 the map h belongs to A'n-1,-\..,J-1,
o·
k-1 '
Proof.
Let
f
-21-
Atn- 1 ,~,J,
0 • k 1 ' At 1 0 • 1 k
or AIn- 1,~,J,
0 • k
n- ,~,J- ,
, according as neither 1
nor .f.. + 1 occur in (d , d , ••• , d _ ) , 1 < d S d _ = .f.. + 1 ,
1 2
l
i 1
i l
1 = d S d _ < .f.. + 1 , 1 = d S d _ = t + 1 •
1
i 1
1
i 1
/
Q.E.D.
5.
INVOLUTIONS.
Let
1 S c
l
<
c
2
< ••• <
ck_l S
t
The mapping
+ 1.
where
is an involution of the set
the single element of Al
and
f
is in A
n,.,.,k
~
(43)
f-+ ((c , c ,
z
l
(44)
f-+ ((d ' d ,
1
2
~
.
[[.f..+l 1)
Let f' = fll
k-l
When n > 1 , 1 S k S n
A
)
n,i,.,.
(resp.
and
(resp.
... ,
d _ ) , h) -+ ((di' di, ... , d!1- 1) , h")
i 1
~
f
1
S
z,
••• t
k_1 )
c
~-l
, g t) -+
iy-l
-+
i
S
n)
f'
f" ,
of A 1 and h -+ h" of At
nn-l
Note that (43) and (44) involve the
g -+ g'
defined in
is
form the sequences
c _ ) , g) -+ ((ci, c
k 1
have been defined by induction.
~
when
... ,
assuming that the involutions
bijections
=f
(25)
and
(40)
respectively.
-22-
10.
Proposition. For eaah n > 0 the transfoPmations f
defined by
f + fil
A
n
and
(43)
(44)
+
f'
and
respeative Zy are invo Zutions of
with the foZZowing properties
l(f')
= J(f)
, J(f')
= l(f)
, K(f')
= K(f)
and
= ref)
, J(f")
By construction 1
(resp.
l(f")
Proof.
(ci, c
z, ... , ck_1)
occur in
sequ~nce
and
f
, K(f")
= J(f)
•
does or does not occur in
l + 1)
if and only if l + 1
(resp.
1)
does or does not
(c l , c 2' ••• , ck_l ). Proposition 10 is then a simple conof propositions 3 and 8. For instance if 1 = c 1 ~ c k _l < l + 1
is in An, 1,),
•. k '
g'
is in A
' 1 ,1,.(..
. h
n- I,Jin An,J,1,
.. k •
11.
= K(f)
Example.
•
then
As
Let us determine
element of A •
4
g belongs to A
' . 1 ,.(..h • By induction
n- I,1,J1 < ci ~ c _1 = l + 1, the map f' is
k
Q.E.D.
f'
and
fll
when
f
is the following
-23-
x =
1 234 5 6 7 8
323
f(x) =
4 2 6 8 6 8 8 8
223
g(x) =
c 1,···,ck _1
£+1
42666 6
2,4
4
1 2 2
4 244
2,3
3
111
2 2
1
2
I J K
k_
ci,···,c 1
2
2
1 1 1
2 2
212
2 444
1,2
3
223
266 466
1,3
4
23 3
f' (x) =
2 8 6 4 8 6 8 8
x ='
1 2 3 456 7 8
323
f(x) =
4 2686888
3 1 2
hex) =
d 1 ,···,di _1
£+1
P
4 6 6 8 8 8
2,3
4
1
2 1 2
6 8 8 8
1,3
3
2
1 1 1
8 8
2
2
1
1
2
2
I J K
di,.··,di_l
1 1 1
8 8
221
668 8
1,3
3
2
321
4 6 668 8
2,3
4
1
332
f"(x)
=
4 2 6 8 6 6 8 8
-24-
6.
TABLES.
Table 2 gives the first values of the coefficients
an,1.,J,
° ° k •
Because of the symmetry only the values of an,1.,J,
. ° k with
are shown.
TABLE
n
i j k
an. 1..
°i •k
1
111
1
2
1 2 2
1
3
123
1
223
2
123
4
5
i j k
1
S
j S k
2
° i °k
an. 1..
i,j ,k
an.i.i,k
,
133
1
222
2
1
124
2
133
4
134
22;)
3
144
1
222
2
8
224
6
233
12
234
3
333
6
123
124
8
125
6
133
3
16·
134
24
135
11
144
22
145
6
155
1
222
6
223
32
224
48
225
22
233
84
2 3 4·
74
235
18
244
36
245
4
333
126
334
60
335
6
344
12
-25-
REFERENCES
1.
L. Car1itz, The Staudt-Clausen Theorem, Math. Magazine
131-146.
2.
L. Car1itz, A conjecture concerning Genocchi numbers, K. norske
Vidensk. SeZsk. Sk. 9 (1972), 1-4.
3.
L. Car1izt, Private communication (1974).
4.
D. Dumont, Interpr~tations combinatoires des nombres de Genocchi,
Duke Math. J. 41 (1974), 305-318.
5.
J.M. Gandhi, A conjectured representation of Genocchi numbers,
Amer. Math. MonthZy 77 (1970), 505-506.
6.
G.H; Hardy and E.M. Wright, An Introduation to the Theory of Nwnbers,
Oxford University Press, London, 1938.
7.
J. Riordan and P. Stein, Proof of a conjecture on Genocchi numbers,
Disarete Math. 5 (1973), 381-388.
35 (1961),
Present address of the two authors:
de Math~matique
de Strasbourg
D~partement
Universit~
~rue Ren~-Descartes
67084 Strasbourg
FRANCE
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