Fibonacci polynomials and central factorial numbers
Johann Cigler
Abstract
I consider some bases of the vector space of polynomials which are defined by Fibonacci and Lucas polynomials and compute the matrices of corresponding basis transformations. It turns out that these matrices are intimately related to Stirling numbers and central factorial numbers and also to Bernoulli numbers, Genocchi numbers and tangent numbers. There is also a close connection with the Akiyama-Tanigawa algorithm. Since such numbers have been extensively studied it would be no surprise if some of these results are already known, but hidden in the literature. I would therefore be very grateful for hints to relevant papers or books or for other comments.
1. Introduction
We consider (a variant of) the Fibonacci polynomials defined by
⎢⎥n−1 ⎣⎦⎢⎥2 ⎛⎞nk−−1 k Fsn ()= ∑ ⎜⎟ s . (1.1) k=0 ⎝⎠k
They satisfy the recursion Fsnn()=+ F−−12 () s sFs n () with initial values Fs0 ()= 0 and Fs1()= 1.
The first terms are
0,1,1,1 , 1 2 , 1 3 ,1 4 3 ,1 5 6 , 1 6 10 4 ,… .
The corresponding Lucas polynomials are defined by
⎢⎥n ⎣⎦⎢⎥2 n ⎛⎞nk− k Ln ()ss= ∑ ⎜⎟ (1.2) k=0 nk− ⎝⎠k
and satisfy the same recurrence as the Fibonacci polynomials, but with initial values Ls0 ()= 2 and Ls1()= 1. The first terms of this sequence are
2,1,1 2 , 1 3 , 1 4 2 ,1 5 5 ,1 6 9 2 ,… .
Let
1
114++s α = (1.3) 2
and
114−+s β = (1.4) 2
be the roots of the equation zzs2 −−=0.
1 Then for s ≠− the well-known Binet formulae give 4
α nn− β Fs()= (1.5) n α − β
and
nn Lsn ()=+α β . (1.6)
1 For s =− it is easily verified that 4
⎛⎞1 n Fn ⎜⎟−=n−1 . (1.7) ⎝⎠42
It is also well known that Lsnn()=+ F+−11 () s sFs n (). This follows immediately from s =−αβ and α nn++11−−βαβαβ() nn −− 11 − = ααββαβ n() −+− n( ).
Since deg()degFs21nnnn++==== Fs 22 ()deg()deg() Ls 2 Lsn 21 + each of the sets
{F21nnnn++(),sFsLsLs} { 22 (),} { 2 (),} { 21 + ()} is a basis for the vector space of polynomials in s.
I am interested in the matrices which transform one basis into another. Their entries are related to Bernoulli numbers, Genocchi numbers and tangent numbers. Furthermore there are interesting connections with Stirling numbers and central factorial numbers.
The Genocchi numbers G = 1,1,3,17,155,2073,38227,929569," (cf. OEIS A110501 ) ( 2n )n≥1 ( ) and their relatives (gnn )≥1 =−( 1, 1,0,1,0, − 3,0,17,0, − 155,") (cf. OEIS A036968) are defined by their exponential generating function
zn2 n 21zez− n z zz=+zz =∑∑ gnn =+ z(1) − G2 , (1.8) 11++eennn≥≥11 ! (2)! n
2 and the tangent numbers (T21n+ ) = (1,2,16,272,7936,353792,22368256,") (cf. OEIS A000182) are defined by
ez221zk−1 + (1)kT . 2z =−∑ 21k+ (1.9) ek++1(21)!k≥0
Note that by comparing coefficients we get
gGT (1)−==k −1 22kkk+++ 22 21 . (1.10) 22222kk++21k +
nn2 It is also well-known that GB22nn=−(1)2(12 − ) , where (Bn ) is the sequence of Bernoulli numbers defined by
n ⎛⎞n Bnk= ∑⎜⎟B (1.11) k=0 ⎝⎠k
for n > 1 with B0 = 1. The list of Bernoulli numbers begins with
1, , ,0, ,0, ,0, ,0, ,0, ,0, ,0, ,…
For later uses let us also recall the generating function of the Bernoulli numbers
zzn ∑Bn = z . (1.12) n≥0 ne!1−
2. Connection constants
The following theorem gives an explicit computation of some basis transformations.
Theorem 2.1
The bases ()F22n+ and (F21n+ ) are connected by
n nk− G222nk−+⎛⎞22n + Fs22nk++()=−∑ (1)⎜⎟ F 21 , (2.1) k=0 21k + ⎝⎠2k and
n ⎛⎞21n + B22nk− Fs21nk++()= ∑⎜⎟ Fs 22 (). (2.2) k=0 ⎝⎠21k + k +1
3
The bases consisting of Lucas polynomials are connected by
nnTG⎛⎞21nn++ ⎛⎞ 21 L sLsLsnk−−221nk−+ nk 222 nk −+ (2.3) 21nk+ ()=−∑∑ (1)221nk−+⎜⎟ 2 () =− (1) ⎜⎟ 2 k () kk==002222⎝⎠22kknk−+ ⎝⎠ and
n ⎛⎞2n B22nj− L221nj()sLs= 2∑⎜⎟ + (). (2.4) j=0 ⎝⎠2 j 21j +
Proof
1) Since αβ+=1 we have
eeαβzz−= e(1−− β ) z − e (1 α ) z =− eee zz( −− α − β z). (2.5)
This gives
Fs() Fs () ∑∑nnzenz= −−() z n (2.6) nn≥≥00nn!! or
1+ ez−+zn z21 n ∑∑Fsnn()= F21+ () s (2.7) 2!nnnn (21)!+
This reduces our task to finding the matrix which corresponds to the operator "multiplication by 1 + e− z " of the corresponding exponential generating functions. 2
Comparing coefficients we see that for xx= ()nn≥0 and yy= ()nn≥0
1+ ez−zn z n ∑∑xynn= (2.8) 2!!nnnn
is equivalent with yMx=
nk− 1 ⎛⎞n where M = (mi(, j )) with mnk(,)=− (1) ⎜⎟ for kn< , mnn(,)= 1 and mnk(,)= 0 for 2 ⎝⎠k kn> .
For example
4
100000 10000 1 1 0 0 0 M = . 6 100 23 210 5 5 1 The inverse of (2.8) is
zzzzzgnnjkn2()− n ⎛⎞n x yg y(1)nk− nk−+1 y, thus ∑∑∑∑∑∑nnjk==− z +1 =−⎜⎟ k nnjknkne!1++ n ! ( jjkn 1)! ! !=0 nk −+ 1⎝⎠k
n nk− gnk−+1 ⎛⎞n xnk=−∑(1)⎜⎟y . (2.9) k=0 nk−+1⎝⎠k
For example
1 0 0 000 1 0 000 0 1 1 000 M −1 = . 6 0 100 0 10 210 0 0 1
Thus (2.7) implies
n−1 gnk+−22 ⎛⎞n Fsnk()=−∑ (1) ⎜⎟ F21− nk+−22⎝⎠21k − and we get as special case
n gG222nk−+⎛⎞22nn++nk− 222 nk −+ ⎛⎞ 22 Fs22nkk+++()=−∑∑⎜⎟ F 21 = (1) − ⎜⎟ F 21 , kk≥=00222nk−+⎝⎠21kk+ 21 k + ⎝⎠ 2 i.e. (2.1).
From
eeeαβz +=zzzzzz(1−− β ) + e (1 α ) = eee( −− α + β) (2.10) we get in the same way as before
5
1+ ez−zn z2 n ∑∑Lsnn()= L2 () s . (2.11) 2!(2)!nnnn
In this case (2.9) implies (2.3).
2) Another consequence of (2.5) is
Fs() F () s ()12.−=ez− z ∑∑nnnn2 z2 (2.12) nn≥≥00nn!(2)!
Now
knn−1 n −−−znjzz1 ⎛⎞n z (1−=−=−exk )∑∑∑ ( ) ( 1)⎜⎟ xjyn ( ) ∑ ( 1) is equivalent with knjkn!!=≥01⎝⎠j n n !
n nj− ⎛⎞n +1 yn()=−∑ (1)⎜⎟ x (). j (2.13) j=0 ⎝⎠j
The inverse is
zzbzzknkj1 −1 xk() yn ( 1)k y ( j 1) ∑∑∑∑=−=−−z knnnke!(1)− n ! k ! j !
1 where bn()= B for n ≠ 1 and b(1)= . This follows from n 2 zzzzeznn z bBzz . ∑∑nn=+=+==zz− z nnnnee!!−−− 111 e
This implies
n ⎛⎞n bnj− x()nyj= ∑⎜⎟ (). (2.14) j=0 ⎝⎠j j +1
From (2.12) we see that with x()nFs= n () and yn(2− 1)== 2 Fsyn2n ( ), (2 ) 0, we get
21nn+ n ⎛⎞21nn++bb21n+− j ⎛⎞ 21 22 nj − ⎛⎞ 21 n + b 22 nj − Fs21n+ ()==+=∑∑⎜⎟ yj () ⎜⎟ yj (2 1) ∑ ⎜⎟ F22j+ , jj==00⎝⎠jjjj++1221 ⎝⎠21++ j = 0 ⎝⎠ 21 j j + i.e. (2.2).
Ls() L () s (2.10) implies ()12.−=ez−+zn∑∑nn21+ z21 n nn≥≥00nn!(21)!+ 6
Therefore we get from (2.12) with x()nLs= n () and yn(2 )= 2 L21n+ ( syn ), (2+= 1) 0,
nn ⎛⎞22nnbB22nj−− ⎛⎞ 22 nj L2n ()syjLs==∑∑⎜⎟ (2) ⎜⎟ 221j+ (), jj==00⎝⎠22jj21jj++ ⎝⎠ 21 i.e. (2.4).
The matrix corresponding to "multiplication by (1− e− z ) " of the generating functions is
∞ ⎛⎞ij− ⎛⎞i +1 Cji=−⎜⎟(1)[ ≤ ]⎜⎟ . (2.15) ⎝⎠j ⎝⎠ij,0= Its inverse is given by
∞ −1 ⎛⎞⎛⎞i bi()− j Cji=≤⎜⎟[] . (2.16) ⎜⎟j j +1 ⎝⎠⎝⎠ ij,0=
−1 For example for n = 5 the corresponding finite parts C5 and C5 are
100 00 1 2 0 0 0 C5 = 1 330 0 14 640 1 510 105 and
10000 000 −1 00 C5 = 0 0 0
−1 If we apply C to the column vector y with entries yn(2+ 1)== 2 Fn+2 ( syn ), (2 ) 0, then the −1 entries x()n of x = Cy are x()nFs= n ().
n ⎛⎞22nnb221nj−− ⎛⎞F2n For even n this is trivial because in x(2nyjF )=+==∑⎜⎟ (2 1) ⎜⎟ 2n j=0 ⎝⎠21jn+−22jn+ ⎝⎠ 21 2
7 the only non-vanishing term occurs for jn= −1. For odd n we get the identities (2.2).
−1 If we apply C to the column vector y with entries yn(2 )= 2 Lsynn ( ), (2+= 1) 0, then the −1 entries x()n of x = Cy are x()nLs= n (). For even n we get (2.4). For odd n the corresponding identity is again trivial.
3. Stirling numbers and central factorial numbers
The matrices (2.15) and (2.16) are intimately connected with Stirling numbers.
In order to make the paper self-contained I shall first state some well-known facts about Stirling numbers and some generalizations (cf. [4]).
Let wwn= () be an increasing sequence of positive numbers. Define the w− Stirling ( )n≥0 numbers of the second kind Snkw (,) by
SnkSnww(,)=−−+ ( 1, k 1) wkSn () w ( − 1,) k (3.1)
with initial values Skkw (0, )== [ 0] and Snwn(,0)= w (0)and the w− Stirling numbers of the first kind s w (,)nk by
s ww(nk , )=−−−− s ( n 1, k 1) wn ( 1) s w ( n − 1, k ) (3.2)
n−1 with skkw (0, )== [ 0] and snwn(,0)(1)=− ∏ wj (). j=0
This is equivalent with
n ∑snkxwk( , )=− ( xw (0))( xw − (1))" ( xwn − ( − 1)), (3.3) k=0
nk−1 ∑∏Snkwn(,) ( xwj− ())= x (3.4) kj==00 and
xk ∑Snkxwn(,)= . (3.5) nk≥ (1−−wx (0) )(1 wx (1) )" (1 − wkx ( ) )
Let
8
n−1 Snww()= Sij (,) (3.6) ( )ij,0= and
n−1 snww()= sij (,) . (3.7) ( )ij,0=
Then it is clear that
−1 snww()= ( Sn ()) . (3.8)
The same holds for the infinite matrices SSijww= (, ) and ssijww= (, ) . ( )ij,0≥ ( )ij,0≥
For later applications we note the trivial
Proposition 3.1
If an nn×−matrix M n has different eigenvalues rj() and if the eigenvector corresponding to n−1 rj() is the column vector Sijw (, ) , then M = SnDnsnww() () (), where D()n is the ( )i=0 n diagonal matrix with entries rrn(0)," , (− 1).
For wn()= n we get the Stirling numbers snk(,) of the first kind and Snk(,) of the second kind.
Recall the first values of the Stirling numbers:
1000000 0100000 0110000 6 Si(, j ) = , ()ij,0= 0131000 0176100 0 1 15 25 10 1 0 0 1 31 90 65 15 1 1000000 0100000 0 11 0 0 00 6 si(, j ) = ()ij,0= 02 31 0 00 0 611 61 00 024 5035 1010 0 120 274 225 85 15 1
9
Note that Si(, j ) and si(, j ) are the same as Sijw (, ) and sijw (, ) ( )ij,1≥ ()ij,1≥ ( )ij,0≥ ( )ij,0≥ respectively for wn()=+ n 1.
It is well-known that the generating function of the Stirling numbers of the second kind is (1)ezzk− n Fzk ()==∑ Snk (,) . kn!!n
It can be characterized by
Proposition 3.2
zn The formal power series Fzk ()= ∑ Snk (,) is the uniquely determined solution of the n n! − z ′ differential equation (1−=eFzkFz )kk ( ) ( ) with Skk(,)= 1.
Proof
(1)ez − k It is clear that Fz()= satisfies the differential equation. k k!
zn If on the other hand Fz()= ∑ xn () satisfies the differential equation then n≥0 n!
zzij z n ∑∑(1)−=i−1 xj () k ∑ xn () . ij≥≥00ij!! n ≥ 0 n !
Comparing coefficients we see that for nk< all x()n vanish and that for nk= we get an arbitrary constant which can be chosen to be xk()= 1.
ij− ⎛⎞i +1 Consider the infinite matrix Ccij= ()(, ) with ci(, j )=− ( 1) ⎜⎟ for ji≤ and ci(, j )= 0 ij,0≥ ⎝⎠j for ji> , which has been defined in (2.15). The eigenvalues are the numbers 1, 2, 3," .
A vector xxnkkn= (())≥0 is an eigenvector of C with respect to the eigenvalue k if Cxkk= kx or equivalently
zznn zn if (1−=−exnkxn−z ) ( ) ( 1) . This means that Fz()=− xn ( 1) satisfies ∑∑kk kk∑ nnnn!! n n! − z ′ (1−=eFzkFz )kk ( ) ( ).
Therefore by Proposition 3.2 xk ()nSnk=+ ( 1,)is a Stirling number of the second kind. 10
This leads to
Theorem 3.3
The matrices C and C−1 can be factored in the following way:
CSij=++( 1, 1)∞∞∞ ( i += 1)([ ij ] sij ( ++ 1, 1) ()()()ij,0= ij ,0== ij ,0 (3.9) and
∞ −1 ∞∞⎛⎞1 CSij=++(1,1) ([] ijsjk = ( ++ 1,1). (3.10) ()ij,0= ⎜⎟() jk ,0= ⎝⎠i +1 ij,0=
This theorem could also be derived from the following simple identities.
The identity
⎛⎞⎛⎞⎛⎞ii ⎛+1 ⎞ Cji⎜⎟⎜⎟[]≤=≤ [] ji (3.11) ⎜⎟jj ⎜ ⎟ ⎝⎠⎝⎠⎝⎠ij,0≥≥ ⎝ ⎠ ij ,0
is another formulation of the trivial result
nn+1 nj−−⎛⎞⎛⎞⎛⎞njn++11 nj ⎛⎞⎛⎞⎛⎞ njn ++ 11 ∑∑(1)−=+−=⎜⎟⎜⎟⎜⎟ (1) ⎜⎟⎜⎟⎜⎟ jj==00⎝⎠⎝⎠⎝⎠j kk ⎝⎠⎝⎠⎝⎠ jkk for kn≤ .
The identity
⎛⎞⎛⎞i ⎜⎟[]ji≤=++()( Sij (,) Si (1,1) j ) (3.12) ⎜⎟j ij,0≥≥ ij ,0 ⎝⎠⎝⎠ij,0≥ is a well-known property of the Stirling numbers and easily proved by induction:
nn⎛⎞nn−−11 ⎛⎞−−11 n ⎛⎞ n n − 1 ⎛⎞ n − 1 ∑∑⎜⎟Sjk(,)=+=+++ ⎜⎟ Sjk (,) ∑ ⎜⎟ Sjk (,) Snk (, 1) ∑ ⎜⎟ Sj ( 1,) k jj==00⎝⎠jjj ⎝⎠ j = 0 ⎝⎠−1 j = 0 ⎝⎠ j nn−−11⎛⎞nn−−11 ⎛⎞ =++Snk(, 1)∑∑⎜⎟ S (, jk −+ 1) k ⎜⎟ S (,) jk jj==00⎝⎠jj ⎝⎠ =+++Snk(, 1) Snk (,) kSnk (, +=++ 1) Sn ( 1, k 1).
Furthermore we need
11
⎛⎞⎛⎞i +1 ⎜⎟[]ji≤=+++()( Sij (,) Si (1,1)(1). j j ) (3.13) ⎜⎟j ij,0≥≥ ij ,0 ⎝⎠⎝⎠ij,0≥
This also is easily proved by induction:
nnn⎛nnn+1 ⎞ ⎛⎞ ⎛ ⎞ n−1 ⎛⎞ n ∑∑∑⎜ ⎟Sjk(,)=+ ⎜⎟ Sjk (,) ⎜ ⎟ Sjk (,) =++++ Sn ( 1, k 1) ∑ ⎜⎟ Sj ( 1,) k jjj===000⎝jjj ⎠ ⎝⎠ ⎝−1 ⎠ j = 0 ⎝⎠ j nn−−11⎛⎞nn ⎛⎞ =+++Sn(1,1) k∑∑⎜⎟ S (,1) jk −+ k ⎜⎟ S (,) jk jj==00⎝⎠jj ⎝⎠ =++++−Sn( 1, k 1)() Sn ( 1, k ) Snk ( , −+ 1) k() Sn ( ++− 1, k 1) S (nk,) =+(kSnk 1) ( + 1, ++ 1)() SnkSnkkSnkkSnk ( + 1, ) − ( , −− 1) ( , ) =+ ( 1) ( + 1, + 1).
To derive (3.9) observe that (3.11) gives
−1 ⎛⎞⎛⎞⎛⎞ii+1 ⎛⎞ ⎛⎞ Cji=≤⎜⎟⎜⎟[]⎜⎟ [] ji ≤ , ⎜⎟jj⎜⎟ ⎜⎟ ⎝⎠⎝⎠⎝⎠ij,0≥≥⎝⎠ ⎝⎠ ij ,0
(3.13) gives
⎛⎞⎛⎞i +1 −1 ⎜⎟[]j≤=+++ i()() Si (1,1)(1)(,) j j Si j ⎜⎟j ij,0≥≥() ij ,0 ⎝⎠⎝⎠ij,0≥ and (3.12) gives
−1 ⎛⎞⎛⎞⎛⎞i −1 ⎜⎟⎜⎟[]ji≤=()( Sij (,) Si (1,1). ++ j ) ⎜⎟⎜⎟j ij,0≥≥() ij ,0 ⎝⎠⎝⎠⎝⎠ij,0≥
Combining these identities gives (3.9).
Another consequence is
−1 ⎛⎞⎛⎞⎛⎞⎛⎞ii ⎛+1 ⎞ ⎜⎟⎜⎟⎜⎟[ji≤≤=+= ] [ ji ]()( Sij ( , ) ( i 1)([ i j ] )() sij ( , ) . (3.14) ⎜⎟⎜⎟jj ⎜ ⎟ ij,0≥≥≥ ij ,0 ij ,0 ⎝⎠⎝⎠⎝⎠⎝⎠ij,0≥≥ ⎝ ⎠ ij ,0
If we choose wn()= n2 we get the central factorial numbers tnk(,)= sw (,) nk of the first kind and the central factorial numbers Tnk(,)= Sw (,) nk of the second kind respectively.
12
These numbers have been introduced in [12] with a different notation. Further results can be found in [14], Exercise 5.8.
The following tables show the upper part of the matrices of central factorial numbers. See also OEIS A036969.
10 0 0 0 0 0 01 0 0 0 0 0 01 1 0 0 0 0 6 Tij(, ) = ()ij,0= 01 5 1 0 0 0 0 1 21 14 1 0 0 0 1 85 147 30 1 0 0 1 341 1408 627 55 1 10 0 0 000 01 0 0 000 0 11 0 0 00 6 ti(, j ) = . ()ij,0= 04 51 000 0 36 49 14 1 0 0 0 576 820 273 30 1 0 0 14400 21076 7645 1023 55 1
By (3.5) the generating function of Tnk(,) is
xk Tnkx(,)n . ∑ = 22 n (1−−x )(1 2xkx )" (1 − )
The following partial fraction expansion is easily verified:
(2kx )!2k 2k ⎛⎞2k 1 (1)j 22222=−∑ ⎜⎟ (1−−x )(1 2xkx )" (1 − )j=0 ⎝⎠j 1 −− ( kjx )
This implies that
2k 1 j ⎛⎞2k 2n Tnk(,)=−∑ (1)⎜⎟ ( j − k ) . (3.15) (2k )! j=0 ⎝⎠j
From this we imply the known exponential generating function (cf. [12])
2k ze22nkzzk− (1− e ) 1 ⎛⎞ z Tzk ( )===∑ Tnk ( , )⎜⎟ 2sinh( ) . (3.16) (2nk )! (2 )! (2 k )!⎝⎠ 2
In the same way as above we get
13
Proposition 3.4
z2n The formal power series Tzk ()= ∑ Tnk (,) is the uniquely determined solution of the n (2n )! ez −1 differential equation Tz′()= kTz ()with Tkk(,)= 1. ez +1 kk
For wn()=+ nn ( 1) we get the Legendre-Stirling numbers LS(,) n k and ls(,) n k studied in [2] and [8] (cf. OEIS A071951 and A129467).
10 0 0 0 00 01 0 0 0 00 02 1 0 0 00 6 LS(, i j ) = ()ij,0= 04 8 1 0 00 0 8 52 20 1 0 0 0 16 320 292 40 1 0 0 32 1936 3824 1092 70 1
1000000 0100000 0 21 0 0 00 6 ls(, i j ) = . ()ij,0= 012 81 0 00 0 144 108 20 1 0 0 0 2880 2304 508 40 1 0 0 86400 72000 17544 1708 70 1
There are some interesting relations between central factorial numbers and Legendre-Stirling numbers which are analogous to the corresponding results about Stirling numbers.
Theorem 3.6
n ⎛⎞2nj− LS(,) j k= T ( n++ 1, k 1) (3.17) ∑⎜⎟j j=0 ⎝⎠ and
n ⎛⎞21nj+− ∑⎜⎟LS(,) j k= ( k+++ 1)( T n 1, k 1) (3.18) j=0 ⎝⎠j
14
Proof
For nk<+1 all terms vanish. For nk=+1 we have
k ⎛⎞21kj+− ⎛⎞ 21 kk +− ∑⎜⎟LS(,) j k= ⎜⎟=+=+ ( k 1)( k 1)( T k + 1, k + 1) j=0 ⎝⎠jk ⎝⎠ and
k ⎛⎞2kj− ⎛⎞ k ∑⎜⎟LS(,) j k===++ ⎜⎟ 1 T ( k 1, k 1). j=0 ⎝⎠jk ⎝⎠
Now assume that (3.17) and (3.18) are already known up to n − 1. Then
nn⎛⎞22121nj−−−−−⎛⎞ ⎛⎞⎛⎞ njnj ∑∑⎜⎟LS(,) j k=+⎜⎟ ⎜⎟⎜⎟ LS (,) j k jj==00⎝⎠jjj⎝⎠ ⎝⎠⎝⎠−1 nn−−11⎛⎞2(nj−− 1) ⎛ 2( n −+− 1) 1 j ⎞ =++LS(1,) j k LS (,) j k ∑∑⎜⎟ ⎜ ⎟ jj==00⎝⎠jj ⎝ ⎠ nn−−11⎛⎞2(nj−− 1) ⎛⎞ 2( nj −− 1) n−1 ⎛⎞2(nj−+− 1) 1 =−++∑∑⎜⎟LS(, j k 1) k ( k 1) ⎜⎟ LS ( j,)kLSjk+ ∑⎜⎟ (,) jj==00⎝⎠jj ⎝⎠j=0 ⎝⎠j =++++++=++Tnk( , ) kk ( 1) Tnk ( , 1) ( k 1) Tnk ( , 1) Tn ( 1, k 1).
In the same way we get
nn⎛21n+− j ⎞⎛⎞ ⎛⎞⎛⎞ 2 nj − 2 nj − ∑∑⎜ ⎟LS(,) j k=+⎜⎟ ⎜⎟⎜⎟ LS (,) j k jj==00⎝jjj ⎠⎝⎠ ⎝⎠⎝⎠−1 nn−1 ⎛⎞2(nj−+− 1) 1 ⎛⎞ 2 nj − =++∑∑⎜⎟LS(1,) j k ⎜⎟ LS (,) j k jj==00⎝⎠jj ⎝⎠ nn−−11⎛⎞2(nj−+− 1) 1 ⎛⎞ 2( nj −+− 1) 1 =−+++∑∑⎜⎟LS(, j k 1) k ( k 1) ⎜⎟ LS (,) j k Tn(1,1)++ k jj==00⎝⎠jj ⎝⎠ =++++++=+++kTnkkkTnkTnk(,) ( 1)(,2 1) ( 1, 1)( kTnk 1)( 1, 1).
Corollary 3.7
⎛⎞⎛⎞2ij− ⎜⎟[]ji≤=++⎜⎟() LSij() ,() Ti (1,1) j (3.19) ⎝⎠⎝⎠j and
15
⎛⎞⎛⎞21ij+− ⎜⎟[]ji≤=+++⎜⎟() LSij() ,() (1)(1,1). j Ti j (3.20) ⎝⎠⎝⎠j
For example
10 0 0 0 0 10 0 0 00 10 0 0 00 11 0 0 0 0 01 0 0 00 11 0 0 00 13 1 0 0 0 02 1 0 00 15 1 0 00 . = . 15 6 1 0 0 04 8 1 00 121141 00 1715101 0 0 8 52 20 1 0 1 85 147 30 1 0 192835151 016320292401 1 341 1408 627 55 1 and
10 0 0 00 10 0 0 00 10 0 0 00 12 0 0 00 01 0 0 00 12 0 0 00 14 3 0 00 02 1 0 00 1103 0 0 0 . = . 16104 00 04 8 1 00 142424 0 0 18212050 0 8 52 20 1 0 1 170 441 120 5 0 1103656356 0 16 320 292 40 1 1 682 4224 2508 275 6
Corollary 3.8
−1 21ij+− 2 ij − ⎛ ⎛ ⎞⎞⎛⎛ ⎛ ⎞⎞⎞ −1 ⎜[]j≤≤=++=+++ i ⎜ ⎟⎟⎜⎜ [] j i ⎜ ⎟⎟⎟ ()()() Ti (1,1)[](1)(1,1). j i j j() Ti j (3.21) ⎝ ⎝jj ⎠⎠⎝⎝ ⎝ ⎠⎠⎠ and
−1 ⎛⎛ ⎛221ij−+− ⎞⎞⎞ ⎛ ⎛ i j ⎞⎞ −1 ⎜⎜[]ji≤≤ ⎜ ⎟⎟⎟ ⎜ [] ji ⎜ ⎟⎟ ==+()()()LS(, i j ) [ i j ]( j 1)() LS (, i j ) . (3.22) ⎝⎝ ⎝jj ⎠⎠⎠ ⎝ ⎝ ⎠⎠
Proof
−1 ⎛⎞⎛⎞⎛⎞21ij+−⎛⎞ ⎛⎞ 2 ij − []ji≤≤ [] ji ⎜⎟⎜⎟⎜⎟⎜⎟ ⎜⎟ ⎝⎠⎝⎠⎝⎠jj⎝⎠ ⎝⎠ −11− =++=+()()()T(1,1)[ i j i j ](1)(,) j() LS i j()() LS (,) i j() T (1,1) i ++ j
In the same way we get (3.22).
Another interesting result is
16
Theorem 3.9
The following identities hold:
n ⎛⎞n +1 ∑⎜⎟Tj(1,1)(1,1)+ k+= LSn + k + (3.23) j=0 ⎝⎠22nj− and
n ⎛⎞n +1 ∑⎜⎟Tj(1,1)(1)(1,1).+ k+= k + LSn + k + (3.24) j=0 ⎝⎠221nj−+
Equivalently this means
⎛⎞⎛⎞i +1 ⎜⎟[]ji≤++=++()() Ti (1,1) j LSi (1,1) j (3.25) ⎜⎟22ij− ij,0≥≥ ij ,0 ⎝⎠⎝⎠ij,0≥ and
⎛⎞⎛⎞i +1 ⎜⎟[ji≤ ]()( Ti (1,1) ++=+++ j ( j 1)(1,1) LSi j ) ⎜⎟22ij−+ 1 ij,0≥≥ ij ,0 ⎝⎠⎝⎠ij,0≥ (3.26) =++LS(1,1)[ i j i =+ j ](1). j ()()ij,0≥≥ ij ,0
Proof
For nk< both sides vanish. For nk= we get 1 and k + 1 respectively.
nnn⎛⎞nnn+1 ⎛⎞ ⎛ ⎞ ∑∑∑⎜⎟Tj(1,1)++= k ⎜⎟ Tj (1,1) +++ k ⎜ ⎟ Tj (1,1) ++ k jjj===000⎝⎠22nj−−−− ⎝⎠ 22 nj ⎝ 221 nj ⎠ nn⎛⎞nn ⎛⎞ =+++++∑∑⎜⎟Tj(1,1) k ⎜⎟ Tj (1,1) k jj==10⎝⎠2(nj−+− 1) 2 2 ⎝⎠ 2( nj −+− 1) 1 2 n−1 ⎛⎞n =++++∑⎜⎟Tj(2,1)(1)(, k k LSnk +1) j=0 ⎝⎠2(nj−− 1) 2 nn−−11 ⎛⎞nn2 ⎛⎞ =∑∑⎜⎟Tj(1,)(1) + k ++ k ⎜⎟ Tj (1,1)(1)(,1) + k +++ k LSnk + jj==00⎝⎠2(nj−− 1) 2 ⎝⎠ 2( nj −− 1) 2 =+++++=++LS( n , k ) (( k 1)2 ( k 1)) LS ( n , k 1) LS ( n 1, k 1).
For the second sum we get 17
nnn⎛⎞nnn+1 ⎛⎞ ⎛⎞ ∑∑∑⎜⎟Tj(1,1)++= k ⎜⎟ Tj (1,1) +++ k ⎜⎟ Tj (1,1) ++ k jjj===000⎝⎠221nj−+ ⎝⎠ 221 nj −+ ⎝⎠ 22 nj − nn⎛⎞nn ⎛⎞ =+++++∑∑⎜⎟Tj(1,1) k ⎜⎟ Tj (1,1) k jj==11⎝⎠2(nj−+− 1) 3 2 ⎝⎠ 2( nj −+− 1) 2 2 n−1 ⎛⎞nnn−1 ⎛⎞ =+++∑⎜⎟Tj(2,1) k ∑⎜⎟Tj(2,1)++ k j=0 ⎝⎠2(nj−+− 1) 1 2j=0 ⎝⎠ 2( n − 1)2− j
nn−−11 ⎛⎞nn2 ⎛⎞ =+++++∑∑⎜⎟Tj(1,)(1) k k ⎜⎟ Tj (1,1) k jj==00⎝⎠2(nj−+− 1) 1 2 ⎝⎠ 2( nj −+− 1) 1 2 nn−−11 ⎛⎞nn2 ⎛⎞ ++++++∑∑⎜⎟Tj(1,)(1) k k ⎜⎟ Tj (1,1) k jj==00⎝⎠2(nj−− 1) 2 ⎝⎠ 2( nj −− 1) 2 =++++++++kLSnk(,)(1)(1)(,1)(,)(1)(, k22 k LSnk LSnk k LSnk 1) =+(1)(,)(1)(2)(,1)(1)(1,1).k() LSnk ++ k k + LSnk + =+ k LSn + k +
Corollary 3.10
−1 ⎛⎞⎛⎛ii++11 ⎞⎞⎛⎛ ⎞⎞ ⎜⎟⎜⎜[]ji≤≤ ⎟⎟⎜⎜ [] ji ⎟⎟ 22ij−−+ 22 ij 1 ⎝⎠⎝⎝ ⎠⎠⎝⎝ij,0≥≥ ⎠⎠ ij ,0 (3.27) −1 = Ti(1,1)[++ j i = j ](1) j + Ti (1,1) ++ j ()()()ij,0≥≥ ij ,0() ij ,0 ≥ and
−1 ⎛ ⎛ii++11 ⎞⎞⎛⎞ ⎛ ⎛ ⎞⎞ ⎜[]ji≤≤ ⎜ ⎟⎟⎜⎟ ⎜ [] ji ⎜ ⎟⎟ 22ij−+ 1 22 ij − ⎝ ⎝ ⎠⎠ij,0≥≥⎝⎠ ⎝ ⎝ ⎠⎠ ij ,0 (3.28) −1 = LS(1,1)[ i++ j i = j ](1) j + LS (1,1) i ++ j . ()()()ij,0≥≥ ij ,0() ij ,0 ≥
Proof
The left-hand side of (3.27) equals
−11− LSij(1,1)[++ ijj = ](1) + Tij (1,1) ++ Tij (1,1) ++ LSij (1,1) ++ . ()()()ij,0≥≥ ij ,0( ij ,0 ≥) ()() ij ,0 ≥() ij ,0 ≥
In the same way the left-hand side of (3.28) equals
18
−1 −1 Ti(1,1)++ j LSi (1,1) ++ j LSi (1,1)[ ++ j i = j ](1) j + Ti (1,1) ++ j . ()()ij,0≥≥() ij ,0()()() ij ,0 ≥≥≥ ij ,0() ij ,0
Comparing (3.27) with (3.21) we see that
−1 −1 ⎛ ⎛21ij+− ⎞⎞⎛⎛ ⎛ 2 ij − ⎞⎞⎞ ⎛⎞⎛⎛ii++11 ⎞⎞⎛⎛ ⎞⎞ ⎜[]ji≤≤= ⎜ ⎟⎟⎜⎜ [] ji ⎜ ⎟⎟⎟ ⎜⎟⎜⎜[]ji≤≤ ⎟⎟⎜⎜ [] ji ⎟⎟. (3.29) jj 22ij−−+ 22 ij 1 ⎝ ⎝ ⎠⎠⎝⎝ ⎝ ⎠⎠⎠ ⎝⎠⎝⎝ ⎠⎠⎝⎝ij,0≥≥ ⎠⎠ ij ,0
This implies
⎛⎞⎛ ⎛ii++11 ⎞⎞⎛⎛21ij+− ⎞⎞ ⎛ ⎛ ⎞⎞ ⎛⎛⎛⎞⎞⎞ 2 ij − ⎜⎟⎜[]ji≤≤ ⎜ ⎟⎟⎜⎜[]ji≤= ⎟⎟ ⎜ [] ji ⎜ ⎟⎟ ⎜⎜⎜⎟⎟⎟ [] ji ≤ 22ij−−+jj 22 ij 1 ⎝⎠⎝ ⎝ ⎠⎠ij,0≥ ⎝⎝ ⎠⎠ ⎝ ⎝ ⎠⎠ij,0≥ ⎝⎝⎝⎠⎠⎠ (3.30) −1 =++LS(1,1)[ i j i =+ j ](1) j LS (,) i j . ()()()ij,0≥≥≥ ij ,0() ij ,0
Remark
Michael Schlosser has shown me a simple direct proof of the first identity in (3.30).
It suffices to show that
nn⎛⎞⎛⎞⎛njkn++−+121 1 ⎞⎛⎞ 2 jk − ∑∑⎜⎟⎜⎟⎜= ⎟⎜⎟. jk==⎝⎠⎝⎠⎝22nj−−+ k jk 221 nj ⎠⎝⎠ k
This is equivalent with
2n j+1 ⎛⎞⎛⎞njk++−11 ∑ (1)−=⎜⎟⎜⎟ 0. jk=−21 ⎝⎠⎝⎠2nj− k
This follows from the Chu-Vandermonde formula:
2n j+1 ⎛⎞⎛⎞njk++−11 ∑ (1)− ⎜⎟⎜⎟ jk=−21 ⎝⎠⎝⎠212nj−+− j k
2nk+− 12 212 +− k i ⎛⎞⎛⎞⎛⎞⎛⎞⎛⎞nikn++111 +−−− k nk =−∑∑(1)⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟ = = = 0. ii==00⎝⎠⎝⎠⎝⎠⎝⎠⎝⎠212nkii+−− 212 nkii +−− 221 nk −+
We later need the following result.
19
Lemma 3.11
Let w(0)= 1 and wnˆ()= wn (+ 1). Then
SnkSnwwˆ (,)= (++− 1, k 1) Snk w (, + 1) (3.31) and
k swwˆ (,)nk=−∑ s ( n + 1,). j (3.32) j=0
If
nn ww ww Fn1(,)j ==+∑∑Sn (,)AA F() s (,) Aj FSnFs (0) (,) AA () (,) Aj (3.33) AA==01 for some function F()A then
nk wwˆˆ ∑∑Sn(,)AA F()+= 1 s (,) A k ( FnjFn11 (,) −+ ( 1,). j ) (3.34) A==00j
Proof
The first assertions follow from
xxxk (1− ) k+1 ∑Snkxwnˆ (,) == n (1−−wx (1))(1 w (2))(1 x"" −+ wkx ( 1)) x (1 − w (0))(1 x − wx (1))(1 −+ wkx ( 1)) =−(1xSnkx )∑∑wnw ( , + 1)−1 =() Snk ( + 1, +− 1) Snkx w ( , + 1) n nn and
n n−1 ∑∑swk(nk , ) x=− ( x w (0))" ( x − wn ( − 1)) =− ( x 1) s wˆ ( n − 1, k ) x k k=0 k=0 or
nn1 +1 ∑∑swkˆ (,)nk x=− s w ( n + 1,) k x k . kk==001− x
This implies
20
nn k ∑∑Snwwwˆˆ(,)AA F()+=−++−+++ 1 s (,) A k() Sn ( 1, A 1) Sn w (, A 1) F() A 1 ∑ s w ( A 1,) j AA==00j = 0 kn kn−1 =−+++++∑∑Snwwww(1,1)1(1,)AA F() s A j ∑∑ Sn (,1)1(1,) AA +++ F () s A j AA jj==00 == 00 kn+1 k n =−∑∑∑∑Snwwww( + 1,)AA F() s (,) A j + Sn (,) AA F () s (,) A j jj==01AA == 01 k =−+∑()Fnj11(,) Fn ( 1,). j j=0
4. Some interesting matrices
We know already that
Fs() Fs () ∑∑nnzenz=−(). − z n (4.1) nn≥≥00nn!!
This can also be written in the form
Fs() ez − 1 F () s 221nnzz221nn= + + . (4.2) ∑∑(2ne )!z ++ 1 (2 n 1)!
If we define a linear functional λ on the vector space of polynomials in s by
λ (Fs21n+ ()) = [ n= 0], (4.3) we get
⎡⎤ze2nz−1 λ(())Fs2n = ⎢⎥ zz . (4.4) ⎣⎦(2ne )!+ 1
Using (1.8) this gives (cf. [5],[3])
n−1 λ(())(1).Fs22nn=− G (4.5)
By comparing coefficients we get again (cf. [3])
n Fs22nk++()= ∑ ankFs (, ) 21 () (4.6) k=0 with
21
nk− 1 ⎛⎞22n + ank(,)=− (1)⎜⎟ G222nk− + . (4.7) 21k + ⎝⎠2k
n−1 We call the infinite matrix Aank= (,) and the finite parts Aaij= (, ) Genocchi ()nk,0≥ n ()ij,0= matrices.
E.g.
⎛⎞10000 ⎜⎟ ⎜⎟−12 000 A5 = ⎜⎟35300− . (4.8) ⎜⎟ ⎜⎟−−17 28 14 4 0 ⎜⎟ ⎝⎠155−− 255 126 30 5
ez −1 It is clear that A is the matrix version of "multiplication with " of a certain kind of ez +1 generating functions. More precisely we have
Proposition 4.1
⎛⎞x0 ⎛⎞y0 ⎜⎟x ⎜⎟y ⎜⎟1 ⎜⎟2
Let x = ⎜⎟x2 , y = ⎜⎟y2 and Aank= ()(,) . ⎜⎟ ⎜⎟ nk,0≥ ⎜⎟# ⎜⎟# ⎜⎟ ⎜⎟ ⎝⎠# ⎝⎠#
Then yAx= is equivalent with
ze22nz+ −1 x yzn 21n+ . ∑∑n = z (4.9) nn≥≥00(2ne+++ 2)! 1 (2 n 1)!
By considering the highest powers of s we see from (1.1) that akk(,)= k+ 1. Therefore the eigenvalues of A are 1, 2, 3," .
For later uses we note that
n ∑ank(,)= 1. (4.10) k=0
This follows from (4.6) for s = 0. 22
For special values of s (4.6) gives some interesting identities. Consider for example s =−1.
Here (Fsnn ( ))≥0 =−−( 0,1,1, 0, 1, 1,") is periodic with period 6.
nn This gives for example ∑∑an(3+= 2,3 k ) an (3 ++ 2,3 k 2). kk==00
1 ⎛⎞1 n For s =− we get Fn ⎜⎟−=n−1 . Therefore 4 ⎝⎠42
n ∑4(21)(,)nk− kankn+=+ 1. k=0
By comparing coefficients of sk we get
n ⎛⎞⎛⎞221j −+−knk ∑an(,) j⎜⎟⎜⎟= for nk≥ . jk= ⎝⎠⎝⎠kk
This is equivalent with the matrix identity
−1 ⎛⎞⎛⎞⎛⎞⎛⎞21ij+− 2 ij − Aji=≤⎜⎟⎜⎟[]⎜⎟⎜⎟ [] ji ≤ . (4.11) ⎝⎠⎝⎠⎝⎠⎝⎠jj
⎛⎞⎛⎞1 F1 ⎛⎞⎛⎞1 F2 ⎜⎟⎜⎟sF ⎜⎟⎜⎟sF ⎜⎟⎜⎟3 ⎜⎟⎜⎟4 ⎛⎞⎛⎞2ij− 2 ⎛⎞⎛⎞21ij+− 2 If we recall that ⎜⎟[]ji≤=⎜⎟⎜⎟⎜⎟sF5 and ⎜⎟[]ji≤=⎜⎟⎜⎟⎜⎟sF6 ⎝⎠j ⎜⎟⎜⎟3 ⎝⎠j ⎜⎟⎜⎟3 ⎝⎠ij,0≥ sF ⎝⎠ij,0≥ sF ⎜⎟⎜⎟7 ⎜⎟⎜⎟8 ⎜⎟⎜⎟ ⎜⎟⎜⎟ ⎝⎠⎝⎠## ⎝⎠⎝⎠## we see that (4.11) is the same as (4.6).
By (3.21) we get the following representation
−1 A =++=+()()()Ti(1,1)[](1)(1,1). j i j j() Ti ++ j (4.12)
By (3.29) we also have
−1 ⎛⎞⎛⎞⎛⎞⎛⎞ii++11 ⎛ ⎞ A = ⎜⎟[]ji≤≤ [] ji . (4.13) ⎜⎟⎜⎟⎜⎟⎜⎟22ij−−+ ⎜ 22 ij 1 ⎟ ⎝⎠⎝⎠⎝⎠⎝⎠ij,0≥≥ ⎝ ⎠ ij ,0
23
Thus we get
Theorem 4.2
The Genocchi matrix A has the following representations
−1 ⎛⎞⎛⎞⎛⎞⎛⎞21ij+− 2 ij − Aji=≤⎜⎟⎜⎟[]⎜⎟⎜⎟ [] ji ≤ , (4.14) ⎝⎠⎝⎠⎝⎠⎝⎠jj
−1 ⎛⎞⎛⎞⎛⎞⎛⎞ii++11 ⎛ ⎞ A = ⎜⎟[]ji≤≤ [] ji (4.15) ⎜⎟⎜⎟⎜⎟⎜⎟22ij−−+ ⎜ 22 ij 1 ⎟ ⎝⎠⎝⎠⎝⎠⎝⎠ij,0≥≥ ⎝ ⎠ ij ,0 and
ATij=++(1,1)[ iji =+ ](1) tij (1,1), ++ (4.16) ( )ij,0≥≥≥( ) ij ,0( ) ij ,0 where ti(, j ) are the central factorial numbers of the first kind and Tij(, )the central factorial numbers of the second kind.
Note the analogy with (3.9) and (3.11).
⎛⎞⎛⎞⎛⎞⎛⎞ii++11 ⎛ ⎞ (4.15) implies ⎜⎟[]ji≤≤A = [] ji . ⎜⎟⎜⎟⎜⎟⎜⎟22ij−−+ ⎜ 22 ij 1 ⎟ ⎝⎠⎝⎠⎝⎠⎝⎠ij,0≥≥ ⎝ ⎠ ij ,0
By considering the first column of these matrices we get
n ⎛⎞n +1 j ∑⎜⎟(1)−==Gn22j+ [] 0. j=0 ⎝⎠22nj−
Slightly reformulated this is
Seidel's identity for Genocchi numbers ([13])
n ⎛⎞n k ∑⎜⎟(1)− Gn22nk− == [ 1]. (4.17) k=0 ⎝⎠2k
−1 ⎛⎞⎛⎞2ij− The matrix ⎜⎟[]ji≤ ⎜⎟ has been evaluated by Dumont and Zeng [6] . We don't need this ⎝⎠⎝⎠j result. We are only interested in the first column. Let H 1,1, 2, 8, 56, 608," be the ( 21n+ )n≥0 = ( ) median Genocchi numbers (cf. OEIS A005439).
24
Lemma 4.3
−1 ⎛⎞⎛⎞2ij− The elements of the first column of []ji≤ are the numbers (1)− n H . ⎜⎟⎜⎟ 21n+ ⎝⎠⎝⎠j
The first two columns of
−1 ⎛⎞⎛⎞⎛⎞221ij−+− ⎛ i j ⎞ ⎜⎟⎜⎟[]ji≤≤⎜⎟ [] ji ⎜ ⎟ ⎝⎠⎝⎠⎝⎠jj ⎝ ⎠
⎛⎞10 ⎜⎟ ⎜⎟02 ⎜⎟02− are ⎜⎟. The numbers of the second column are ⎜⎟08 ⎜⎟ ⎝⎠## rH(0)=−13 + 1 = 0, rHrHrHrH (1) = + 1 = 2, (2) =− 579 =− 2, (3) = = 8, (4) =− =− 56,"and in nn−1 general for n ≥ 2 rn()=−λ ( s ) = (1) − H21n+ .
−1 ⎛⎞⎛⎞2ij− The left upper part of the matrix ⎜⎟[]ji≤ ⎜⎟ is ⎝⎠⎝⎠j
100000 1 1 0 0 0 0 2 31 000 8 13 6 1 0 0 56 92 45 10 1 0 608 1000 493 115 15 1
−1 ⎛⎞⎛⎞⎛⎞221ij−+− ⎛ i j ⎞ and of the matrix ⎜⎟⎜⎟[]ji≤≤⎜⎟ [] ji ⎜ ⎟is ⎝⎠⎝⎠⎝⎠jj ⎝ ⎠
1000 10 0 0 10 00 1 1 0 0 12 0 0 02 00 . = . 2 310 14 3 0 0 23 0 8 13 6 1 16104 08 84
To prove this lemma we need the fact (cf. [5]) – which can serve as definition of the median Genocchi numbers – that
25
n nj− ⎛⎞21nj+− GH221nj=−∑(1)⎜⎟+ . (4.18) j=0 ⎝⎠j
n+1 ⎛⎞21nj+− jn Since ∑⎜⎟λλ()sFsG==−()22nn++ () (1) 22 by (4.5) we see that j=0 ⎝⎠j
nn λ()(1)sH=− 21n+ . (4.19)
By (4.6) we get
⎛⎞⎛⎞Fs1() 1 −1 ⎛⎞2()ij− ⎜⎟⎜⎟ Fs s ⎛⎞⎜⎟⎜⎟3 ⎜⎟[]ji≤=⎜⎟ 2 (4.20) ⎝⎠⎝⎠j ⎜⎟⎜⎟Fs5 () s ⎜⎟⎜⎟ ⎝⎠⎝⎠## and therefore by applying λ
⎛⎞11 ⎛ ⎞ −1 20()ij− ⎜⎟ ⎜ s ⎟ ⎛⎞⎛⎞⎜⎟ ⎜λ ⎟ ⎜⎟[]ji≤=⎜⎟ 2 . ⎝⎠⎝⎠js⎜⎟0() ⎜λ ⎟ ⎜⎟ ⎜ ⎟ ⎝⎠## ⎝ ⎠
To prove the second assertion we note that F21n+ (0)= 1and therefore (4.20) gives the first column.
n ⎛⎞2nj− j To obtain the second column we recall that ∑⎜⎟(1)− HFs21jn++==λ () 21 () 0 j=0 ⎝⎠j
n ⎛⎞2nj− and therefore ∑⎜⎟rj()=+ 1 (2 n − 1)2. = n j=0 ⎝⎠j
(3.22) gives
−1 ⎛⎛ ⎛221ij−+− ⎞⎞⎞ ⎛ ⎛ i j ⎞⎞ −1 ⎜⎜[]ji≤≤ ⎜ ⎟⎟⎟ ⎜ [] ji ⎜ ⎟⎟ ==+()()()LS(, i j ) [ i j ]( j 1)() LS (, i j ) . ⎝⎝ ⎝jj ⎠⎠⎠ ⎝ ⎝ ⎠⎠
If we delete the first row and first column we get the matrix
−1 LS(1,1)[ i++ j i = j ](2) j + LS (1,1) i ++ j . (4.21) ()()()ij,0≥≥ ij ,0(()ij,0≥ )
26
The left-upper part begins with
20000 2 3 0 0 0 8 8400 . 5656 2050 608 608 216 40 6
Theorem 4.2 implies that the eigenvectors of Aaij= (, ) with respect to the eigenvalue k ( )ij,0≥ are
⎛⎞Tk(1, ) ⎜⎟ ⎜⎟Tk(2, ) xk()= ⎜⎟Tk(3, ) . (4.22) ⎜⎟ ⎜⎟Tk(4, ) ⎜⎟ ⎝⎠#
This means that ∑an(,)( jT j+= 1,) k kTn ( + 1,). k j≥0
This is equivalent with
z eTnkTnk−1(,)21nn− (,) 2 z ∑∑zk= z. (4.23) en+−1nn≥≥11 (2 1)! (2 n )!
Let
Tnk(,)2n Tzk ()= ∑ z . (4.24) n≥0 (2n )!
Then (4.23) can be written as
ez −1 Tz′()= kTz (). (4.25) ez +1 kk
This result can also be obtained from Proposition 3.4.
27
Corollary 4.4
Define linear functionals ϕ by ϕ F ()sTnk= (, )for k ≥ 1. Then ϕ F ()skTnk= (, ). k kn( 21− ) kn( 2 ) Proof
This follows from (4.2).
An explicit expression for these functionals gives
Theorem 4.5
The following formulae hold:
n ϕk+1()s = LS (,), n k (4.26)
n−1 ⎛⎞22nj−− ϕkn+−121()Fs()==+∑⎜⎟ LSjkTnk (, ) (, 1) (4.27) j=0 ⎝⎠j
n−1 ⎛⎞21nj−− ϕkn+12()Fs()==++∑⎜⎟ LSjk (, ) ( k 1)(, Tnk 1). (4.28) j=0 ⎝⎠j
Proof
This is an immediate consequence of Theorem 3.6.
For our next results we need the square of A.
For example
10000 3 4 0 0 0 2 A5 = 17 25 9 0 0 . 155 238 98 16 0 2073 3255 1428 270 25
Theorem 4.6
The square of An is given by
Aaaijji2 =≤(, )[ ]n−1 n ()ij,0= (4.29) with 28
nk− ⎛⎞22n + nk++2 aa(,) n k=− (1)⎜⎟ G224nk− + . (4.30) ⎝⎠2k (2kn++− 1)( 2 k )
Proof
x Let fz()= ∑ n z21n+ . n≥0 (2n + 1)!
22nz+ ze−1 2 Since yAx= is equivalent with ∑ yfzn = z () we see that zAyAx== is n≥0 (2ne++ 2)! 1 equivalent with
ze221nz−−−111 z n− ee z⎛⎞ z ′ ∑∑zynn−−11==zzz⎜⎟ fz() nn≥≥11(2ne )!+−++ 1 (2 n 1)! e 1⎝⎠ e 1 (4.31) 2 ⎛⎞eeezzz−−−111 ⎛⎞′ =+fz′() fz () ⎜⎟eeezzz+++111 ⎜⎟ ⎝⎠ ⎝⎠ Since
2 ⎛⎞eGzzn−1 2 n+1 22n+ (4.32) ⎜⎟z =−∑(1) ⎝⎠enn++11(2)!n≥1 and
eezz−−11⎛⎞⎛⎞′′′ e z − 1 Gz23 n− (1)n−1 2n , (4.33) −==−zz⎜⎟⎜⎟ z ∑ ee++11⎝⎠⎝⎠ e + 1n≥1 2(23)! nn − we get
2 ⎛⎞ezzn−1 2 n ⎛⎞2n G f ′()zxk (1)nk− 224nk−+ () (4.34) ⎜⎟z =−∑∑⎜⎟ ⎝⎠en+−+1(2)!2nk=1 ⎝⎠22k − nk and
⎛⎞ezzn−1 ′′ 2 n ⎛⎞2n G nk− 224nk−+ (4.35) −=⎜⎟z f ()zxk∑∑⎜⎟ (1) − (). ⎝⎠en+−+1(2)!224nk=0 ⎝⎠21k − nk
From (4.34) and (4.35) and using (4.31) we get immediately (4.30).
It only remains to prove (4.32).
29
This follows from
2 ⎛⎞eezz−−11 ⎛⎞′ (21) nz +2 n =−12 =− 12 (1) −n G ⎜⎟eezz++1 ⎜⎟ 1∑ 22n+ (2 nnn )!(2 ++ 1)(2 2) ⎝⎠ ⎝⎠ n≥0 Gz2n =−∑(1)n−1 22n+ . n≥0 nn+1(2)!
2 Theorem 4.7 (Further properties of An )
aa(,0) n= −+ a ( n 1,0) (4.36) and for 11≤≤−kn
aa(,) n k= a (, n k−− 1) a ( n + 1,) k (4.37)
Proof.
This follows from (4.30).
Intimately connected with the linear functional λ is the linear functional λ* defined by
* λλ(())Fsnn=− ( sFs ()). (4.38)
Since λ(sFsnn ( ))=−λλ ( F++21 ( s )) ( F n ( s )) we get
λ(sFs2222122nn ( ))=−=λλλ ( F+++ ( s )) ( F n ( s )) ( F n ( s )) for n > 0 and
λ(sFs21nnn+++ ( ))=−=−λλ ( Fs 23 ( )) ( Fs 22 ( )) λ ( Fs 22 n + ( )).
* The sequence (λ ()Fsn ()) begins with
0,1,1, 1, 3,3,17, 17, 155,155,2073, 2073, 38227.
Therefore
* λ (()Fs221nn+ F+ ())[0] s== n (4.39) and
30
* n λ (()())(1)FsFs21nn−++=−+ 2( GG 2 nn 22) . (4.40)
Thus we are led to consider the basis consisting of the polynomials FsFs22kk−−()+ 21 ().
Here we get
Theorem 4.8
Let
k ank1(,)=≤ [ k n ]∑ anj (,). (4.41) j=0
Then
n Fs21nkk++()=+∑ ankFsFs 1 (, )() 2 () 21 (). (4.42) k=0
The first entries of (ank1(,)) are
100000 1 1 0 0 0 0 3 21000 . 1711 3100 155 100 26 4 1 0 2073 1337 346 50 5 1
Proof
n−1 We know that Fs221nk()=−∑ ankFs ( 1,)+ (). Therefore k=0 n−1
Fs221nn()+=− F+++ () s∑ ankFs ( 1,) 2121 k () + F n (). s This implies that in (4.42) the matrix (aij1(, )) k=0 is the inverse of IB+ , where I denotes the identity matrix and bnk(,)= an (− 1,) k for kn< .
We have to show that ank1(,)is given by (4.41).
⎛⎞j It suffices to show that ()[](,).I +≤B ⎜⎟j iaiI∑ A = ⎝⎠A=0
31
This means that for each k n−1 ∑an(−+++++== 1,) j() a (,0) j a (,1) j"" a (,) jk an (,0) ank (,) [ n k ]. jk=
n−1 Let kn< . By definition aa(1,) n−= k∑ a (1,)(,). n − j a j k Therefore j=0 nk−−11 aan(−− 1,) i∑∑ an ( − 1,) ja (,) ji = an ( − 1,) ja (,). ji jk== j0
Using (4.37) and (4.10) the left-hand side is equivalent with
kk−−11 aan(−− 1, 0)∑∑ an ( − 1, jaj ) ( , 0) + aan ( −− 1,1) an ( − 1, jaj ) ( ,1) +" jj==01 +−−−−−−−+−+aankankakk( 1, 1) ( 1, 1) ( 1, 1) aankanan ( 1, ) ( , 0) + ( ,1) ++" ank ( , ) ki−1 =−+aa(1,0)(,0)(1,1)(,1) n a n +−+++−+ aa n a n" aa (1,)(,) n k a n k −∑∑ a (1,)(,) n − i a i j i==00j k−1 =−+−++−−−an(1,0)(1,1) an" an (1,1)(1,)0. k∑ an −= j j=0
For kn= the first sum vanishes and an(,0)+"+= ann (,) 1.
Corollary 4.9
Let
ank211(,)=≤ [ k nank ]( (,) − an ( + 1,). k) (4.43)
Then
n FsFs21nn++()+= 22 ()∑ ankFsFs 2 (, )() 2 kk () + 21 + (). (4.44) k=0
Proof
By Theorem 4.7 we have
nn−1 Fs21nkkkknn++()=+=+++∑∑ ankFsFs 1 (, )()() 2 () 21 () ankFsFsFsFs 1 (, ) 2 () 21 ++ () 2 () 21 (). kk==00
Therefore
32
n Fs22nkk++()=−∑ ankFsFs 1 ( + 1,)() 2 () + 21 () k=0
n and thus FsFs21nn++()+= 22 ()∑ ankFsFs 2 (, )() 2 kk () + 21 + (). k=0
The first terms of (ank2 (,)) are
20000 4 3 0 0 0 20 13 4 0 0 . 172 111 29 5 0 2228 1437 372 54 6
n Since ank(,)=+++++∑ Tn ( 1, j 1)( j 1)( t j 1, k 1) we see that ank2 (,)= Fnj (,) if in Lemma j=0 3.11 we choose wn()=+ ( n 1)2 and F()AA= + 1.
n wwˆˆ Therefore ank2 (,)=+∑ S (,) nAA F() 1 s (,). A k A=0
This gives
Theorem 4.10
Let wn()=+ ( n 1)2 and thus wnˆ ()=+ ( n 2).2 Then
nn−−11 ()aij(,)nn−−11==+ Swwˆˆ (,) ij() [ i ji ](2) s (,) ij . (4.45) 2 ij,0==()ij,0== ij ,0( ) ij ,0
For example
1000020000 10000 41000 03000 4 1 0 0 0 16 13 1 0 0 . 00400 . 36 13 1 0 0 . 64 133 29 1 0 00050 576 244 29 1 0 256 1261 597 54 1 00006 14400 6676 969 54 1
Next we consider the linear functional μ defined by μ(Fs22k + ( ))= [ k= 0]. From (4.2) we see that 33
μμ(())Fs ezzn++ 1 (()) Fs ze221 1 z+ 21nn+ zznB21nn+ 2 2 (2 1) . ∑∑===+zz ∑2n (4.46) (2nene+− 1)! 1 (2 )! 2 − 1n (2 n + 1)!
This implies
μ (F21nn+ ()snB) =+ (2 1) 2 (4.47) and
n ⎛⎞21n + B22nj− Fs21nj++()= ∑⎜⎟ Fs 22 (). (4.48) j=0 ⎝⎠21j + j +1
Comparing (4.6) with (4.48) we see that
−1 −1 ⎛⎞⎛⎞21j + B22jk− Aajk==()(,)⎜⎟ . (4.49) ij,0≥ ⎜⎟21k + k +1 ⎝⎠⎝⎠ ij,0≥
The first terms of the sequence (2nB+ 1) are 1, , , , , , , , . ( 2n )n≥0
⎛⎞21j + B22jk− Let w(,)j kk=≤⎜⎟ [j ]. Then ⎝⎠21k + k +1
100000 0000 000 5 wjk(,) = . ()jk,0= 00 1 0
By (4.15) we have
⎛⎞⎛⎞i + 1 −1 ⎛⎞⎛⎞i +1 ⎜⎟[]ji≤ ⎜⎟Aji=≤⎜⎟[]⎜⎟ . ⎝⎠⎝⎠22ij−+ 1 ⎝⎠22ij− ij,0≥ ⎝⎠ij,0≥ Considering the first column we get
34
nn⎛⎞nn++11 ⎛⎞ ∑∑⎜⎟(2jB+= 1)222j ⎜⎟ (2 njB −+ 2 1) nj− jj==00⎝⎠221nj−+ ⎝⎠ 21 j +
n ⎛⎞n +1 =+−+==∑⎜⎟(21)[0].nn j Bnn+−2 j n j=0 ⎝⎠nj− 2
Since B21i+ = 0 for i > 0 this is equivalent with n+1 ⎛⎞n +1 ∑⎜⎟(1)0ni++ Bni+ = for n > 1. But it also holds for n = 0 and n = 1. i=0 ⎝⎠i
Therefore we get
Kaneko's identity ([7],[10],[13])
n+1 ⎛⎞n +1 ∑⎜⎟(1)0.ni+ += Bni+ (4.50) i=0 ⎝⎠i
This identity has first been proved by A.v. Ettingshausen [7] and has been rediscovered by L. Seidel [13],VIII, and by M. Kaneko [10].
(4.49) implies that
n−1 nn−−11⎛⎞1 n − 1 wjk(,)=++ Ti ( 1, j 1) [ i = j ] ti ( ++ 1, j 1) . (4.51) ()(jk,0== ) i ,0 j⎜⎟() i ,0 j = ⎝⎠j +1 ij,0=
nn−11− The inverse of ()aij(,)nn−−11==+ Swwˆˆ (,) ij() [ i ji ](2) s (,) ij is 2 ij,0==()ij,0= ij ,0( ) ij ,0=
n−1 n−1 wwˆˆnn−−11⎛⎞1 zi(,) j== S (,)[] i j i j s (,). i j ()ij,0= ()ij,0==⎜⎟() ij ,0 ⎝⎠i + 2 ij,0=
k By Lemma 3.11 this implies that znk(,)=−+∑() wn (,) j wn ( 1,) j for kn≤ and znk(,)= 0 j=0 else.
As an example consider
35
00000 0000 000 5 ()zi(, j ) = . ij,0= 00 0
13 7 For example we have zww(3,0)=−=+= (3,0) (4,0) , 61015 77 67 zwwww(3,1)=−+−=−−=− (3, 0) (4, 0) (3,1) (4,1) 1 ," . 15 12 60
So we have
n Fs221nn()+= F+++ () s∑ znkF (, )() 2122 k () s + F k (). s (4.52) k=0
5. Analogous results for Lucas polynomials
For the Lucas polynomials we get
Ls() ez − 1 Ls () 21nn+ zz21nn+ = 2 2. (5.1) ∑∑(2nen++ 1)!z 1 (2 )!
This follows from
eeeαβzz+=(1−− β ) z + e (1 α ) zzz = eee( −− α + β z). (5.2)
ez −1 We write the series expansion of in the form ez +1
eTzzk−1 21+ (1)k 21k+ , zk=−∑ 21+ (5.3) ek++12(21)!k≥0
where Tn , n ≥1, are the tangent numbers 1,2,16,272,7936," .
Therefore
36
YeXz −1 nnzz21nn+ = 2 (5.4) ∑∑(2nen++ 1)!z 1 (2 )!
is equivalent with
nnTT⎛⎞21nn++ ⎛⎞ 21 YX(1)knk21knk+−+ (1)− 221 X . nnkk=−∑∑21knk+−+⎜⎟− =− 221 ⎜⎟ kk==0022⎝⎠21kk+ ⎝⎠ 2
nk−−TG221nk−+⎛⎞21nn++ nk1 ⎛⎞ 21 222 nk −+ If we set bnk(,)=− (1)221nk−+⎜⎟ =− (1) ⎜⎟ , then (5.4) is the same as 221⎝⎠22kk ⎝⎠nk− +
n YbnjXnj= ∑ (,) . (5.5) j=0
We call the matrices
⎛⎞ij− T22ij−+ 1⎛⎞21i + Bbij==−()(, )⎜⎟ ( 1) (5.6) ij,0≥ 222ij−+ 1⎜⎟2 j ⎝⎠⎝⎠ij,0≥
tangent-matrices.
n If we compare (5.1) with (5.5) we see that L21nj+ ()sbnjLs= ∑ (, ) 2 (). This is again (2.3). j=0
⎢⎥n ⎣⎦⎢⎥2 n n k k k Let Ln ()slnss= ∑ (,) . Then L2n ()slnks= ∑ (2, ) and L21n+ ()slnks=+∑ (2 1,) . k =0 k=0 k=0
Therefore we get
−1 Blij=+(2 1, ) lij (2 , ) . (5.7) ()()ij,0≥≥( ij ,0)
For example
0000 000 B = 00 . 5 0 63 21
37
2 ⎛⎞21n + w w Let wn()= ⎜⎟ and let Unk(,)= S (,) nk and unk(,)= s (,). nk respectively. ⎝⎠2
The first values of the numbers 4(,)nk− Unkand 4(,)nk− unk are given by the following tables.
10 0 0 0 00 11 0 0 0 00 1101 0 0 00 ij− 6 4(,)Uij = 191351 0 00 ()ij,0= 1 820 966 84 1 0 0 1 7381 24970 5082 165 1 0 1 66430 631631 273988 18447 286 1 and
6 4(,)ij− ui j = ( )ij,0= 1000000 1 1 0 0 0 0 0 9 1010000 225 259 35 1 0 0 0 11025 12916 1974 84 1 0 0 893025 1057221 172810 8778 165 1 0 108056025 128816766 21967231 1234948 28743 286 1
Here we have
22kk 2n xx ∑Unkx(,) ==22 2k . nk≥ ⎛⎞1322 ⎛⎞ ⎛ 21k + ⎞ 2 ⎛⎞21j + (1−−⎜⎟xx )(1 ⎜⎟ )" (1 − ⎜ ⎟ x ) ∏ (1− ⎜⎟x ) ⎝⎠22 ⎝⎠ ⎝ 2 ⎠ jk=− −1 ⎝⎠2
From the partial fraction expansion
x21k+ 1121k+ ⎛⎞21k + =−(1)j 22 2(2k + 1)! ∑ ⎜⎟j 212kj+− ⎛⎞1322 ⎛⎞ ⎛ 21k + ⎞ 2 j=0 ⎝⎠1− x (1−−⎜⎟xx )(1 ⎜⎟ )" (1 − ⎜ ⎟ x ) ⎝⎠22 ⎝⎠ ⎝ 2 ⎠ 2 we see that
38
21k+ 21n+ 1212j ⎛⎞21k + ⎛⎞kj+− Unk(,)=−∑ (1)⎜⎟⎜⎟ . (5.8) (2k + 1)!j=0 ⎝⎠j ⎝⎠ 2
The exponential generating function is therefore (cf. [12])
zz21k+ ⎛⎞− zee21n+ 1 22− Uz()== Unk (,)⎜⎟ . (5.9) k ∑ ⎜⎟ n (2nk++ 1)! (2 1)!⎜⎟ 2 ⎝⎠
Proposition 5.1
z21n+ The formal power series Uzk ()= ∑ Unk (,) is the uniquely determined solution of the n (2n + 1)! ekz −+121 differential equation Uz′()= Uz ()with Ukk(,)= 1. ez +12kk
21k + 135 Since bkk(,)= the eigenvalues of B are ,,," . 2 222
Theorem 5.2
The tangent matrix B has the factorization
⎛⎞21j + BUij==(, ) [ ij ] uij (, ) , ()ij,0≥≥⎜⎟() ij ,0 (5.10) ⎝⎠2 ij,0≥ where ui(, j )and Ui(, j ) are the generalized Stirling numbers corresponding to 2 ⎛⎞21n + wn()= ⎜⎟ . ⎝⎠2
21k + To prove this observe that the eigenvector Uk() corresponding to the eigenvalue of 2 bi(, j ) satisfies ()ij,0≥
n 21k + ∑bn(,) jUjn () k= U () k j=0 2
39 which is equivalent with
eUkkz −+1()21() Uk nnzz221nn= + . (5.11) enz ++1(2)!∑∑ 2 (21)! n
Uk() Let fz()= n z21n+ . Then (5.11) says that k ∑ (2n + 1)! ekz −+121 f ′()zfz= (). By Proposition 5.1 we see that Uk()= Unk (,) as asserted. ez +12kk n
This implies (5.10).
Thus we get again (2.3), i.e.
n L21nk+ ()s= ∑ bnkL (, ) 2 () s (5.12) k=1 with
nk− T221nk−+⎛⎞21n + bnk(,)=− (1)221nk−+⎜⎟ . (5.13) 2 ⎝⎠2k
Proposition 5.3
The inverse of B is
−1 ⎛⎞⎛⎞2iB22ij− Bji=≤⎜⎟[]⎜⎟ . (5.14) ⎝⎠⎝⎠2 j 21j +
zezn+1 z2 B . This follows from (5.4) and the well-known series z = ∑ 2n 21en− n≥0 (2)!
Then from (5.1) we get again (2.4).
There is also an analogue of Theorem 3.6.
To this end we introduce an analogue of the Legendre-Stirling numbers.
(2nn−+ 1)(2 1) Let wn()= and Vnk ( , )= Sw ( nk , ) and vnk ( , )= sw ( nk , ). 4
40
Then we get
Theorem 5.4
n ∑lnjVjk(2 , ) ( , )= 2 Unk ( , ) (5.15) j=0 and
n ∑ln(2+=+ 1, jVjk ) ( , ) (2 k 1) Unk ( , ). (5.16) j=0
Proof
We use again induction. Let both identities be already proved for n −1. Then
nn−−11 n ∑∑lnjVjk(2,)(,)=− ln (2 1,)(,) jVjk +−− ∑ ln (2 2, j 1)(,) Vjk jj==00 j = 0 nn−−11 =−++−∑∑ln(2( 1) 1, jVjk ) ( , ) ln (2( 1), jVjk ) ( + 1, ) jj==00 nn−−11(2kk−+ 1)(2 1) =+(2k 1) Unk ( −+ 1, )∑∑ ln (2( − 1), jVjk ) ( , −+ 1) ln (2( − 1), jVjk ) ( , ) jj==004 (2kk−+ 1)(2 1) =+(2kUnkUnk 1) ( −+ 1, ) 2 ( −−+ 1, 1) 2 Unk ( − 1, ) 4 ⎛⎞(2k + 1)2 =−−+2(1,1)⎜⎟Un k Un (1,)2(,). −= k Unk ⎝⎠4
And
nnn−1 ∑∑∑l(2 n+= 1, jV ) ( jk , ) l (2 n , jV ) ( jk , ) +−− l (2 n 1, j 1) V ( jk , ) jjj===000 n−1 =+−+2Unk ( , )∑ l (2 n 1, jVj ) ( 1, k ) j=0 nn−−11(2kk−+ 1)(2 1) =+−−+2(,)Unk l (21,)(, n jVjk 1) l (2(1),)(,) n − jVjk ∑∑4 jj==00 (2k − 1)(2k + 1) =+−−−+2(,)(21)(Unk k Un 1,1) k (2kUnk+− 1) ( 1, ) 4 ⎛⎞21kk−+⎛⎞ (21)2 =+2(,)⎜⎟Unk⎜⎟ Un (1,1) −−+− k Un (1,) k ⎝⎠24⎝⎠ ⎛⎞21k − =+2⎜⎟Unk (,) Unk (,) =+ (2 k 1)(,). Unk ⎝⎠2 41
This implies again (5.10).
6. Some interesting identities
n ⎛⎞j−1 For any sequence an() the sequence cn()=− (1)jw wi () S (,)() n ja j is the first ( )n≥0 ∑∏⎜⎟ ji==00⎝⎠ ∞∞ column in Sijww( , )() aii ( )([= j ]∞ sij ( , ) . ()ij,0==ij,0= ( ) ij ,0
More precisely we have
Theorem 6.1
Let
X==( xij( , )) Sww ( ij , )( ai ( )([ i = j ]) s ( ij , ) . (6.1) ij,0≥≥()ij,0≥≥ ij ,0( ) ij ,0
Then
n j−1 jw ⎛⎞ x(,0)nSnjajwi=−∑∏ (1) (,)()⎜⎟ () (6.2) ji==00⎝⎠ and
nn−1 ∑∏snjxjwn(,)(,0)(1)()=− an wj (). (6.3) jj==00
The sequence cn() can be simply computed with the Akiyama-Tanigawa algorithm (cf. [1], ( )n≥0 [9], [11], [15] ):
Theorem 6.2 ( Akiyama-Tanigawa algorithm) ( [1],[9],[11],[15])
n j−1 jw⎛⎞ Suppose that wn()≠ 0 for all n. Let cn()=−∑∏ (1)⎜⎟ wi () S (,)() n ja j . ji==00⎝⎠
Define a matrix Mmij= (, ) by ( )ij,0≥
42 mjaj(0, )= ( ) for j ∈` and
mi(, j )=−−−+ w ( j )( mi ( 1,) j mi ( 1, j 1)) (6.4)
n j−1 jw⎛⎞ Then mn(,0)==− cn ()∑∏ (1)⎜⎟ wi () S (,)(). n ja j ji==00⎝⎠
Proof
To prove this we show more generally that
(1)− k k mnk(,)= k−1 ∑ skicn (,)(+ i ). (6.5) ∏ w()A i=0 A=0
This holds for n = 0 because (1)−−kk (1) i ⎛⎞j−1 wwjwA kk−−11∑∑∑∏s (,)()kici=− s (,) ki (1)⎜⎟ w () S (,)() i ja j ∏∏ww()AAiij≥≥==0000 () ⎝⎠A AA==00 k j−1 (1)− jww⎛⎞ =−k−1 ∑∏(1)⎜⎟wajskiSijak ()A () ∑ (,) (,) = (). ∏ w()A ji≥=00⎝⎠A ≥ 0 A=0
Now suppose that (6.5) holds for n − 1. Then wk()(1,)(1,1)() mn−−−+ k mn k ⎛⎞ ⎜⎟kk+1 (1)−−ww (1) =−+−+−+wk()⎜⎟kk−1 ∑∑ skicn (,)( 1 i ) sk ( 1,)( icn 1 i ) ⎜⎟ii≥≥00 ⎜⎟∏∏ww()AA () ⎝⎠AA==00 k (1)− ww =++−+k−1 ∑()wks() (,) ki s ( k 1,) i cn ( 1 i ) ∏ w()A i≥0 A=0 k k (1)− w (1)− w =−−+k−1 ∑ski(, 1)( cn 1 i)(,)()(,).=+=k−1 ∑skicni mnk ∏ w()A i≥1 ∏ w()A i≥0 A=0 A=0
We used the fact that wksk()ww (,0)++= sk ( 1,0)0.
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1 For wn()=+ n 1and an()= this reduces to the original Akiyama-Tanigawa algorithm for n +1 (bn()) as shown in [11].
Now we give a list of some formulas.
1 For wn()=+ n 1, an()= we get from (2.16) and (3.10) n +1
n j! ∑Sn(1,1)(1)++− jj = bn (). (6.6) j=0 j +1
Another proof of (6.6) can be found in [11], but I suspect that this result must be much older.
Formula (6.3) gives
n n! ∑sn(1,1)()(1)++ j b j =−n . (6.7) j=0 (1)n +
2 n For wn()=+ ( n 1), an()=+ n 1 we know from (4.7) and (4.16) that cn()=− (1) G22n+ .
This gives
n nk−−11 2 (1)−=−GTnkkk2n ∑ (1) (,)() ( − 1)! (6.8) k=1 and
n nk− ∑(1)−=−tnkG (,)2k n !( n 1)!. (6.9) k=1
The Akiyama-Tanigawa algorithm applied to (6.8) gives another method for computing the Genocchi numbers. Choose wn()=+ ( n 1)2 and an()= n+ 1 in Theorem 7.2.
Then the left upper part of the corresponding matrix M = (mi(, j )) is given by
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12 3 4 5 6 1 4 9 16 25 36 3 20 63 144 275 468 . 17 172 729 2096 4825 9612 155 2228 12303 43664 119675 276660 2073 40300 282249 1216176 3924625 10444428
n In the first column we get cn()=− (1) G22n+ .
From (4.30) we deduce the following identity (cf. [14], Exercise 5.8):
n nk−−112 (1)−=−GTnkk22n+ ∑ (1) (,)()!.() (6.10) k=1
The left upper part of the corresponding Akiyama-Tanigawa matrix is
14 9 162536 3 20 63 144 275 468 17 172 729 2096 4825 9612 . 155 2228 12303 43664 119675 276660 2073 40300 282249 1216176 3924625 10444428 38227 967796 8405343 43335184 162995075 495672372 (6.3) gives the companion formula
n nk− 2 ∑(1)−=tnkG (,)22k+ () n ! . (6.11) k=0
2 For example for n = 3 we have 45GGG468++=++== 4151736(3!).
(4.21) and Lemma 4.3 give
n nk− 2 ∑(1)−+++=LS ( n 1, k 1)(() k 1)! H 23n+ (6.12) k=0
From (4.45) and (4.44) we deduce for wn()=+ ( n 2)2 and an()= n+ 2
n nk− w ∑(−++=+ 1)Snkk ( , )( 1)!( k 2)! G22nn+ G 24+ (6.13) k=0 and
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n nk− w ∑(−+=++ 1)snkG ( , )()22kk++ G 24 ( n 1)!( n 2)!. (6.14) k=0
1 For wn()=+ n 1 and an()= we get from (4.49) n +1
n 2 j (!)j (2nB+=− 1)2n ∑ ( 1) Tnj ( ++ 1, 1). (6.15) j=0 j +1
Here the left upper part of the Akiyama-Tanigawa matrix begins with
1 .
From (5.6) and (5.10) we deduce
n nk−− nk 2 ∑(1)4−+−=Unk (,)(2 k 1)(2() k 1)!! T21n+ . (6.16) k=0
Finally Proposition 5.3 gives
n k 1 2 ∑(1)(,)−−=Unkk () (2 k 1)!! B2n . (6.17) k=0 (2k + 1)4
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References
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