LINEAR RELATIONS FOR BERNOULLI NUMBERS AND ITS APPLICATION TO CONGRUENCES INVOLVING HARMONIC SUMS

LINEAR RELATIONS FOR

BERNOULLI NUMBERS AND ITS APPLICATION TO CONGRUENCES

INVOLVING HARMONIC SUMS ^∗

Kazufumi Kimoto

Abstract

We show certain linear relations among Bernoulli numbers by using umbral calculus. As an application, we prove some congruence relations involving bino- mial coefficients and harmonic sums which appear in a certain supercongruence problem.

2020 Mathematics Subject Classification. Primary 11A07, 11B68; Secondary 05A40.

Key words and phrases. Binomial coefficients, harmonic sums, congruence rela- tions, umbral calculus, Bernoulli numbers.

1 Introduction

In this short note, we give a simple way to produce linear relations among Bernoulli numbers by using umbral calculus, and use it to prove some congruence relations involving binomial coefficients and harmonic sums, which appear in a certain super- congruence problem [3].

In §2, we first introduce a linear map ψ:R[x]→ Rwhich sends each monomial x^k to the Bernoulli numberBk, and describe the very basic properties of it. For any polynomial f(x) ∈ kerψ, the equation ψ(f(x)) = 0 gives a certain linear relation among Bernoulli numbers. Thus it is natural to seek a sufficient condition for a polynomial f(x) to be in the kernel of this umbral map ψ. We give such a simple sufficient condition. Our calculation in §2 is essentially the same with the one given by Momiyama [2]. Actually, if we discuss over thep-adic integer ringZ^p, then the umbral mapψis realized as the Volkenborn integral. As we will see, however, we do not need to bring the Volkenborn integral to obtain linear relations among Bernoulli numbers in a similar manner; We only needs the standard properties on Bernoulli numbers.

In §3, by using the facts given in §2 and the von Staudt-Clausen theorem, we give several congruence relations modulop², wherepis an odd prime, among certain

∗Received November 30, 2022.

Ryukyu Math. J., 35(2022), 1-11

sums involving binomial coefficients and harmonic sums. Such congruence relations are used to reduce a certain supercongruence i.e. a congruence relation modulo a power ofp to a lower power case.

Lin r r tion or B rnou i nu r

We denote by Bk and Bk(x) the Bernoulli numbers and Bernoulli polynomials re- spectively

!∞ k=0

Bk

t^k k! = t

e^t−1,

!∞ k=0

Bk(x)t^k

k! = te^tx e^t−1. We recall the standard facts For anyk∈Z≥0, we have

Bk(x) =

!k

j=0

"k j

#

B_k−jx^j, 2.

Bk(b)−Bk(a) =k !

a≤j<b

j^k−1 (a, b∈Z, a < b), 2.2

(−1)^kBk(1) =Bk. 2.3

1 A or t u r

efine aR-linear mapψ:R[x]→Rby

ψ:R[x]$f(x) =

!n

k=0

akx^k%−→ψ(f(x)) =

!n

k=0

akBk ∈R. 2.

The following fact is an immediately consequence of the basic properties 2. , 2.2 and 2.3 .

Proposition 2.1. For anyf(x)∈R[x], we have

ψ((x+a)^k) =Bk(a) (k∈Z≥0, a∈R), 2.

ψ(f(x+b)−f(x+a)) = !

a≤j<b

f^%(j) (a, b∈Z, a < b), 2.

ψ(f(−x−1)) =ψ(f(x)). 2.

Proof. First, by the linearity ofψ and 2. , we have

ψ((x+a)^k) =

!k

j=0

"

k j

#

a^jψ(x^k−j) =

!k

j=0

"

k j

#

a^jBk−j=Bk(a),

which is 2. . By the linearity ofψagain, it is enough to prove 2. and 2. when f(x) =x^k, k∈Z≥0. By 2.2 and 2.3 , we have

ψ(f(x+b)−f(x+a))− !

a≤j<b

f^#(j) =ψ((x+b)^k)−ψ((x+a)^k)− !

a≤j<b

kj^k−1

=Bk(b)−Bk(a)−k !

a≤j<b

j^k−1

= 0 and

ψ(f(−x−1))−ψ(f(x)) =ψ((−x−1)^k)−ψ(x^k)

= (−1)^kBk(1)−Bk= 0 as desired.

If f(x) ∈ kerψ, then we get some linear relation ψ(f(x)) = 0 among Bernoulli numbers. Thus it is convenient if we have a simple sufficient condition forf(x)to be killed byψ. One such would be as follows.

2.2. e L e a o ve n e er. e ha f(x) ∈ R[x] a e he fo ow n on on

A f(−x) =−f(x−L), A2 "L−1

i=1 f^#(−i) = 0.

hen ψ(f(x)) = 0.

Proof. By 2. and A , we have

ψ(f(x)) =ψ(f(−x−1)) =−ψ(f(−(−x−1)−L)) =−ψ(f(x−L+ 1)).

IfL= 1, then we haveψ(f(x)) = 0at this point. WhenL≥2, by addingψ(f(x))to the both side and using 2. , we get

2ψ(f(x)) =ψ(f(x)−f(x−L+ 1)) = !

−L+1≤j<0

f^#(j) =

L!−1

i=1

f^#(−i) = 0 by A2 .

We give a slightly weaker version of the lemma above. This is the main tool in our discussion below.

2. . e L e a o ve n e er. e ha F(x) ∈ R[x] a e he fo ow n on on

B F(−x) =F(x−L), B2 #L−1

i=1(x+i)³|F(x), hen ψ(F^#(x)) = 0.

Proof. It is clear thatF^#(x)satisfies A whenF(x)satisfies B . IfF(x)satisfies B2 , thenF^##(−i) = 0fori= 1, . . . , L−1, which implies thatF^#(x)satisfies A2 .

E

We give a few e amples obtained by emma 2.3.

p 2. . etsbe a non-negative integer. ut F(x) =x^s(x+ 1)^s.

It is immediate to see thatF(x)satisfies B and B2 with L= 1. ence we have ψ(F^!(x)) = 0by emma 2.3. Since

F^!(x) = d dx

!s

k=0

"s k

# x^s+k=

!s

k=0

(k+s)

"s k

# x^k+s−1, we get the formula

!s

k=0

(k+s)

"s k

#

Bk+s−1= 0. 2.

The formula 2. is due to von ttingshausen [ ]. For anyr≥0, the2r-th derivative F^(2r)(x)ofF(x)also satisfies B and B2 with L= 1. Since

F^(2r+1)(x) (2r+ 1)! =

!s

k=0

"s k

#"k+s 2r+ 1

#

x^k+s−2r−1,

we also get a slightly general formula

!s

k=0

"

s k

#"

k+s 2r+ 1

#

Bs−2r−1+k = 0, 2.

where we understand thatBi= 0wheni <0. As a special case, by lettings= 2r+ 1,

we have _s

!

k=0

"

s k

#"

k+s k

#

Bk = 0 2.

ifsiso . This equation does not hold when sis even.

e ar 2. . By puttings=n+ 1in 2. , we get B$2n=− 1

n+ 1

n−1!

i=0

"n+ 1 i

#B$n+i, 2.

whereB$k = (k+ 1)Bk [ ].

p 2. . etm, nbe non-negative integers. ut

F(x) = (−1)ⁿxⁿ⁺¹(x+ 1)^m+1+ (−1)^mx^m+1(x+ 1)ⁿ⁺¹.

It is immediate to see thatF(x)satisfies B and B2 with L= 1. ence we have ψ(F^!(x)) = 0by emma 2.3. Since

F(x) = (−1)ⁿ

m+1!

k=0

"m+ 1 k

#

x^n+k+1−(−1)^m

n+1!

k=0

"n+ 1 k

#

x^m+k+1,

we have

(−1)ⁿ

m+1!

k=0

"

m+ 1 k

#

(n+k+ 1)Bn+k+ (−1)^m

n+1!

k=0

"

n+ 1 k

#

(m+k+ 1)Bm+k = 0.

We notice that there occurs a cancellation between the last terms in these sums ((−1)ⁿ+ (−1)^m)Bm+n+1= 0whenm+n >0. Thus we get

(−1)ⁿ

!m

k=0

"

m+ 1 k

#

(n+k+ 1)Bn+k+ (−1)^m

!n

k=0

"

n+ 1 k

#

(m+k+ 1)Bm+k= 0.

This is the Momiyama s identity [2]. By the same argument as in ample 2. , we have

(−1)ⁿ

m+1!

k=0

"m+ 1 k

#"n+k+ 1 2r+ 1

#

B_n−2r+k

+ (−1)^m

n+1!

k=0

"n+ 1 k

#"m+k+ 1 2r+ 1

#

Bm−2r+k= 0

forr≥0.

e ar 2. . In general, for any positive integerLand any polynomialp(x)such that

$L−1

i=1(x−i)³|p(x),

F(x) =p(−x) +p(x+L)

satisfies B and B2 . For instance,p(x) =−xⁿ⁺¹(1−x)^m+1 and L= 1 give the last e ample.

Con ru nc in o in ino i co ci nt nd r onic u

We first recall the von Staudt-Clausen theorem

or .1. For any o ve n e er nan any o r ep,

B2n+ !

p:prime p−1|2n

1

p 3.

an n e er.

As a simple consequence of the theorem, for any odd primepand a positive integer k, we have

pBk≡

%−1 p−1|k

0 otherwise (modp). 3.2

This implies that if

f(x) =

!N

k=0

akx^k ∈Q[x]

and the denominator of the coefficientak is not divisible bypfor everyk, then

pψ(f(x)) =

!N

k=0

akpBk ≡ −

"

N p−1

#

!

i=1

a_(p−1)i (modp). 3.3 We give a lemma for later use.

.2. For any o r e p, p^k

k! ≡0 (modp²) ho fork≥3.

Proof. et us denote byνp(x)thep-adic valuation ofx∈Q\ {0}, that is, x=p^νp(x)a

b, a, b∈Z, p!a, p!b.

It is well known that

νp(k!) =!

i≥1

$k pⁱ

% .

ence we have νp

&p^k k!

'=k−!

i≥1

$k pⁱ

%

≥k−!

i≥1

k pⁱ ≥3&

1− p p−1

'≥ 3 2 >1

as desired.

1 R u t

In what follows, we fi an odd primep, and put m= ^p−₂¹ for short. We denote by Hn the harmonic sum, i.e. Hn =(n

k=11 k. or . . fp≥5, hen

!m

k=0

)m+k k

*4

(Hm+k−Hk)≡0 (modp²), 3.

!m

k=0

)m+k k

*6

(Hm+k−Hk)≡0 (modp²). 3.

Proof. For a positive integers, define Fs(x) :=

!x+m m

"s

=

#sm

i=0

e^(s)_i xⁱ.

otice that the denominator of the coefficiente^(s)_i ∈Qis not divisible bypfor every i. Since

F_s^!(x) Fs(x)=s

#m

i=1

1 x+i,

we have _m

#

k=0

!m+k k

"s

(Hm+k−Hk) =1 s

#m

k=0

F_s^!(k).

Thus it is enough to prove 1 s

#m

k=0

F_s^!(k)≡0 (mod p²)

fors= 4,6. otice that 1

sF_s^!(x) =

!x+m m

"s−1#m i=1

1 m!

$

1≤j≤m j$=i

(x+j),

so that the denominator of every coefficient ^ie(m)i_s of ¹_sF_s^!(x)∈Q[x]is not divisible by pregardless of whethersis divisible bypor not.

For a while, we only suppose thatsis even,s≥4 and p!s notice thats= 4,6 satisfy this condition . Since

1

sF_s^!(k) =

!k+m m

"^s−1#m i=1

1 m!

$

1≤j≤m j$=i

(k+j)≡0 (modp^s−1)

ifm+ 1≤k≤p−1, we have 1 s

#m

k=0

F_s^!(k)≡1 s

p−1

#

k=0

F_s^!(k) (mod p²).

By 2. and emma 3.2, we have

p−1

#

k=0

F_s^!(k) =ψ(Fs(x+p)−Fs(x))

=

#sm

k=1

p^k

k!ψ(F_s^(k)(x))

≡pψ(F_s^!(x)) +p²

2 ψ(F_s^!!(x)) (mod p²).

We see thatFs(x)satisfies B and B2 withL=m+ 1. Indeed, Fs(x) =

!m

i=1

(x+i)^s i^s is divisible by"m

i=1(x+i)³since we assume s≥4, and Fs(−x) =

#−x+m m

$s

=

#(x−m−1) +m m

$s

=Fs(x−m−1) by the relation%_−a

m

&

= (−1)^m%_a−m+1

m

&and the assumption thats is even. ence we have

ψ(F_s^"(x)) = 0

by emma 2.3. By using 3.3 , we get

pψ(F_s^""(x)) =pψ'^sm(⁻²

i=0

(i+ 2)(i+ 1)e^(s)_i+2xⁱ)

≡ −

*_sm−2

2m

+

(

i=1

(2im+ 2)(2im+ 1)e^(s)_2im+2 (modp)

≡ −

s 2−1

(

i=1

(i−1)(i−2)e^(s)_2im+2 (modp).

This is congruent to0 modulopifs≤6. Thus we have

p−1

(

k=0

F_s^"(k)≡pψ(F_s^"(x)) +p

2 ·pψ(F_s^""(x))≡0 (modp²)

fors= 4,6 as desired.

e ar 3. . In general, it isno true that (m

k=0

#m+k k

$s

(Hm+k−Hk)≡0 (modp²)

whensis even ands%= 4,6. When s= 2, we have (m

k=0

#m+k k

$2

(Hm+k−Hk)≡ p

2ψ(F₂^"(x)) (mod p²).

We see thatpψ(F₂^"(x))≡0 (modp), butpψ(F₂^"(x))%≡0 (modp²)in general. When s >6, we have

(m

k=0

#m+k k

$^s

(Hm+k−Hk)≡ −1 s

s 2−1

(

i=3

#i−1 2

$

e^(s)_2im+2 (mod p²), which is not congruent to0modulop² in general.

oro r . . For any o ve even n e ers, we have

!m

k=0

"m+k k

#s

(Hm+k−Hk)≡0 (mod p).

Proof. By the same discussion as in the proof above, we have

!m

k=0

"m+k k

#^s

(Hm+k−Hk) = 1 s

!m

k=0

F_s^!(k)≡1 s

p−1

!

k=0

F_s^!(k)≡pψ(F_s^!(x)/s) (mod p).

Whens≥4, we haveψ(F_s^!(x)/s) = 0. Whens= 2, we directly have

pψ(F₂^!(x)) =

!2m

k=1

ke⁽²⁾_k pBk−1≡0 (modp).

An ic tion

. .

"m+k k

#

≡(−1)^k

"m k

#$1 +p(Hm+k−Hm)%

(mod p²).

Proof. We have

"m+k k

#

=

"m k

#k−1&

j=0

m+j+ 1 m−j

= (−1)^k

"

m k

#^k&−1 j=0

1 +^p₂(j+¹₂)⁻¹ 1−^p2(j+¹₂)⁻¹

≡(−1)^k

"m k

#$1 +p

k!−1

j=0

1 j+¹₂

% (mod p²).

Since

Hm+k−Hm=

k−1

!

j=0

1 m+j+ 1 ≡

k!−1

j=0

1

j+¹₂ (mod p), we have the conclusion.

By the lemma, for anys≥1, we have

"

m+k k

#2s

≡

"

m k

#2s$

1 + 2sp(Hm+k−Hm)%

(modp²).

ence we have

!m

k=0

"m+k k

#2s

(Hm+k−Hk)

≡

!m

k=0

"m k

#2s

(Hm+k−Hk) + 2sp

!m

k=0

"m k

#2s

Hm+k(Hm+k−Hk)

−2spHm

!m

k=0

"

m k

#2s

(Hm+k−Hk) (modp²).

sing Corollary 3. , this implies that

!m

k=0

"

m k

#2s

(Hm+k−Hk)≡0 (mod p), 3.

and hence

!m

k=0

"m+k k

#2s

(Hm+k−Hk)

≡

!m

k=0

"m k

#2s

(Hm+k−Hk) + 2sp

!m

k=0

"m k

#2s

Hm+k(Hm+k−Hk) (modp²).

specially, whens= 2,3, Theorem 3.3 allows us to obtain the following e pressions Proposition . . e have

!m

k=0

"m k

#4

(Hm+k−Hk)≡ −4p

!m

k=0

"m k

#4

Hm+k(Hm+k−Hk) (modp²), 3.

!m

k=0

"

m k

#6

(Hm+k−Hk)≡ −6p

!m

k=0

"

m k

#6

Hm+k(Hm+k−Hk) (modp²). 3.

These e hibit thep-divisibility of the sums in an e plicit manner. These formulas could be used to reduce the analysis of the modp² behavior of the sums in the left- hand sides to that of the modpbehavior of the corresponding sums in the right-hand sides.

R t d con ctur con ru nc

In the final position, we give several con ectures on congruences involvingo powers of binomial coefficients and harmonic sums which we found by numerical e periments.

on t r . . fp≡1 (mod 4)an p >5, hen

!m

k=0

"

m+k k

#3

(Hm+k−Hk)≡0 (modp²), 3.

!m

k=0

"m+k k

#⁵

(Hm+k−Hk)≡0 (modp²). 3.

on t r . . fp≡3 (mod 4), hen

!m

k=0

"m+k k

#5

(H_m+k⁽²⁾ −H_k⁽²⁾−5(Hm+k−Hk)²)≡0 (modp²), 3.

!m

k=0

"m+k k

#7

(H_m+k⁽²⁾ −H_k⁽²⁾−7(Hm+k−Hk)²)≡0 (modp²), 3. 2

whereHn⁽²⁾ =$n i=1

1 i².

Ac no d nt

The author would like to thank obert Osburn for the discussions on supercongruence problems we had during his stay in OIST. The author was supported by ST C ST

rant umber M C and S S A I rant umber 32 ,

22 32 2.

R r nc

[ ] M. aneko, A recurrence formula for the Bernoulli numbers, roc. apan Acad.

Ser. A Math. Sci. 1 , 2 3.

[2] . Momiyama, A new recurrence formula for Bernoulli numbers. Fibonacci uart.

2 , 2 2 .

[3] . Osburn, private communication, 2 22.

[ ] A. von ttingshausen, or e n en er e h here a he a , vol. , Wien, 2 .

epartment of Mathematical Sciences Faculty of Science

niversity of the yukyus ishihara-cho, Okinawa 3- 2 3 A A