Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto - "CONNECTION BERNOULLI NUMBERS Bn AND...

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CONNECTION BERNOULLI NUMBERS B N AND RIEMANN

ς(s) ZETA FUNCTION WITH ITS ZEROS

Ing. Pier Franz Roggero, Dott. Michele Nardelli, P.A. Francesco Di Noto

Abstract:

In this paper we focus attention on a relationship between the denominators

of Bernoulli numbers Bn and prime numbers.

We can define the Bernoulli's function as the analytic continuation of the

Bernoulli's formula in the field of complex numbers.

So we find an interesting correlation on the Riemann ς (s) zeta function and

the Bernoulli numbers in its zeros.



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1. BERNOULLI NUMBERs ....................................................................................................... 3 1.1 TABLE OF FACTORIZATION OF DENOMINATOR BERMOULLI NUMBERS AND

ζ(1−n) IN CORRELATION WITH PRIME NUMBERS ..................................................... 7 1.1.1 CALCULATION OF den(BN) ................................................................................ 11 1.1.2 NUMERATOR OF BN num(BN) ............................................................................ 14 1.1.3 FINDING A PROBABLY PRIME NUMBER p LARGE AS DESIRED TO 60% ....... 18 1.2 CONNECTION ζ(s) ZETA RIEMANN FUNCTION WITH PRIME NUMBERS .......... 21 1.3.1 SPECIFIC VALUES ............................................................................................. 24

2. BERNOULLI'S FORMULA EXTENDED IN THE FIELD OF REAL NUMBERS xεR.............. 27 3. BERNOULLI's FORMULA EXTENDED IN THE FIELD OF COMPLEX NUMBERS zεC, z=σ ±it

.............................................................................................................................................. 34 4. ZEROS ζ(½ ± ix)=0 ON THE CRITICAL LINE ..................................................................... 37

4.1 GRAPHICS OF THE REAL AND IMAGINARY ς(p) ................................................ 47 5. A NEW REPRESENTATION OF THE BERNOULLI FUNCTION CONNECTED WITH THE

RIEMANN ZETA FUNCTION................................................................................................. 50 5.1 A REPRESENTATION OF THE ABSOLUTE VALUES OF BERNOULLI NUMBERS

CONNECTED WITH THE RIEMANN ZETA FUNCTION FOR n≥1 ............................... 53 6. REFERENCES .................................................................................................................... 55



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1. BERNOULLI NUMBERS

The Bernoulli numbers Bn are a sequence of rational numbers with deep connections to

the number theory.

The values of the first few Bernoulli numbers are

B0=1, B1=±1 ⁄ 2, B2=1 ⁄ 6, B3=0, B4=−1 ⁄ 30, B5=0, B6=

1 ⁄ 42, B7=0, B8=−1 ⁄ 30.

If the convention B1=−1 ⁄ 2 is used, this sequence is also known as the first Bernoulli

numbers, with the convention B1=+1 ⁄ 2 is known as the second Bernoulli numbers.

Except for this one difference, the first and second Bernoulli numbers agree. Since Bn=0

for all odd n>1, and many formulas only involve even-index Bernoulli numbers, some

authors write Bn instead of B2n.

For m, n ≥ 0 we define

( ) ∑=

+++==n

k

mmmm

m nk nS 1

...21 .

This expression can always be rewritten as a polynomial in n of degree m + 1. The

coefficients of these polynomials are related to the Bernoulli numbers by Bernoulli's

formula:

( ) ∑=

−+

+

+=

m

k

k m

k m n Bk

m

mnS

0

11

1

1,

where the convention B1=+1/2 is used. (

+

k

m 1denotes the binomial coefficient, m+1

choose k .)

For example, taking m to be 1 gives the triangular numbers 0, 1, 3, 6,....

( ) ( )nnn Bn Bn +=+=+++21

1

2

02

12

2

1...21 .



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Taking m to be 2 gives the square pyramidal numbers 0, 1, 5, 14,....

Some authors use the convention B1=−1/2 and state Bernoulli's formula in this way:

( ) ( )∑=

−+

+−

+=

m

k

k m

k

k

m n Bk

m

mnS

0

11

11

1.

The Riemann zeta function ζ (s) is a function of a complex variable s=σ +it .

The following infinite series converges for all complex numbers s with real part greaterthan 1, and defines ζ (s) in this case:

( ) ∑∞

=

−+++==

1

...3

1

2

1

1

1

nsss

snsζ ( ) 1>= sR σ .

There is a relationship between the denominators of Bernoulli numbers Bn and prime

numbers



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We can define:

Denominator( Bn) = den( Bn) → n+1

den( B n ) = last factor of the factorization( B n ) if the last number is n +1 we have thatit is a prime number p

Example:

n=12

den(B12)= 2730

Denominator( B12)= last factor of the factorization( B12) = 2730 = 2 * 3 * 5 * 7 * 13 → 13 prime number.

When n odd integers we have that all the values of Bn are zero and so we find all

the n+1 even integer.

When we have a den(Bn ) that repeats again we will have all the odd integers that

aren’t prime numbers.

Examples:



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den(B8)= 30 already appeared with den(B4) →9 (not a prime number)

den(B14)= 6 already appeared with den(B2) → 15 (not a prime number)

den(B24)= 2730 already appeared with den(B12) → 25 (not prime)






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1.1 TABLE OF FACTORIZATION OF DENOMINATOR BERNOULLI NUMBERS

AND ζ (1−n) IN CORRELATION WITH PRIME NUMBERS

TAB. 1

den(Bn) Factorization n+1 den (ς(1-n))

1 2 2 2 2

2 6 2 * 3 3 124 30 2 * 3 * 5 5 120

6 42 2 * 3 * 7 7 252

8 30 2 * 3 * 5 9 240

10 66 2 * 3 * 11 11 660

12 2730 2 * 3 * 5 * 7 * 13 13 32760

14 6 2 * 3 15 84

16 510 2 * 3 * 5 * 17 17 8160

18 798 2 * 3 * 7 * 19 19 14364

20 330 2 * 3 * 5 * 11 21 6600

22 138 2 * 3 * 23 23 3036

24 2730 2 * 3 * 5 * 7 * 13 25 65520

26 6 2 * 3 27 156

28 870 2 * 3 * 5 * 29 29 24360

30 14322 2 * 3 * 7 * 11 * 31 31 429660

32 510 2 * 3 * 5 * 17 33 16320

34 6 2 * 3 35 204

36 1919190 2 * 3 * 5 * 7 * 13 * 19 * 37 37 69090840

38 6 2 * 3 39 228

40 13530 2 * 3 * 5 * 11 * 41 41 541200

42 1806 2 * 3 * 7 * 43 43 75852

44 690 2 * 3 * 5 * 23 45 30360

46 282 2 * 3 * 47 47 12972

48 46410 2 * 3 * 5 * 7 * 13 * 17 49 2227680

50 66 2 * 3 * 11 51 3300

52 1590 2 * 3 * 5 * 53 53 82680

54 798 2 * 3 * 7 * 19 55 43092



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56 870 2 * 3 * 5 * 29 57 4872058 354 2 * 3 * 59 59 20532

6056786730

2 * 3 * 5 * 7 * 11 * 13 * 31

* 61 613407203800

62 6 2 * 3 63 372

64 510 2 * 3 * 5 * 17 65 32640

66 64722 2 * 3 * 7 * 23 * 67 67 4271652

68 30 2 * 3 * 5 69 2040

70 4686 2 * 3 * 11 * 71 71 328020

72140100870

2 * 3 * 5 * 7 * 13 * 19 * 37* 73 73

10087262640

74 6 2 * 3 75 444

76 30 2 * 3 * 5 77 2280

78 3318 2 * 3 * 7 * 79 79 258804

80 230010 2 * 3 * 5 * 11 * 17 * 41 81 18400800

82 498 2 * 3 * 83 83 40836

84 3404310 2 * 3 * 5 * 7 * 13 * 29 * 43 85 285962040

86 6 2 * 3 87 516

88 61410 2 * 3 * 5 * 23 * 89 89 540408090 272118 2 * 3 * 7 * 11 * 19 * 31 91 24490620

92 1410 2 * 3 * 5 * 47 93 129720

94 6 2 * 3 95 564

96 4501770 2 * 3 * 5 * 7 * 13 * 17 * 97 97 432169920

98 6 2 * 3 99 588

100 33330 2 * 3 * 5 * 11 * 101 101 3333000

498 3499986 2 * 3 * 7 * 167 * 499 499 1742993028

We note that all the numbers of the column den (ς(1-n)) in bleu are divisible for 24, i.e.the number that is related to the physical vibrations of the bosonic strings by the

following Ramanujan function:



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( )

++

+

⋅

=

−

∞ −

∫

4

2710

4

21110log

'

142

'

cosh

'cos

log4

24

2

'

'4

0

'

2

2

wt itwe

dxe x

txw

anti

w

wt

w x

φ

π π

π

π

.

So we have a relation between the den(Bn) and the prime numbers that are hidden in the

den(Bn).

From the recursive formula

Bm = 1-∑−

=

1

0

m

k k

m Bk /(m-k+1)

For example:

B1 = 1-B0 /2=1-1/2=1/2

B2 = 1-B0 /3-(2/2)B1=1-1/3-1/2=1/6

B3 = 0

B4 = 1-B0 /5-(4/4)B1-(6/3)B2= 1- 1/5-1/2-1/3 = -1/30

B5 = 0

B6 = 1-B0 /7-(6/6)B1-(15/5)B2-(15/3)B4=1-1/7-1/2-1/2+1/6=1/42

B7 = 0

B8 = 1-B0 /9-(8/8)B1-(28/7)B2-(70/5)B4-(28/3)B6 = 1-1/9-1/2-4/6+14/30-2/9 = -3/90 = -

1/30

B4 = B8 = -1/30



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We can find any Bm starting with all the previous Bm-1.

It follows that any prime number can be found by knowing all the prime numbers thatprecede it.

We can predict the next prime number knowing all the prime numbers that precede it.



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1.1.1 CALCULATION OF den(B N )

1° method:

The denominator den( Bn) is thus the product of all divisors di+1 of the index of Bn

minus those who aren't prime number, except 2

In fact we have:

Example:

n=6

divisors of 6 di: 1, 2, 3, 6

prime numbers di+1: 2, 3, 4, 7

4 is discarded, because is not a prime number, and so:

den( B6 ) = 2*3*7

2° method:

Another way to calculate the den( Bn) considering all the product of prime numbers up ton+1 and remove the ones that are not the divisors di-1 of the index.



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den( B6 ) = 2*3*5*7 (all the prime up to n+1)

divisors di-1: 1, 2, 4, 6

4 is discarded because it isn't a divisor of 6 and so we discard 5 in the product:

den( B6 ) = 2*3*7

We can write:

den( Bn) = p∏ with p-1 divisors of n

Observation:

In both two methods in den(Bn) are always the factors 2 and 3 because all den(Bn) aremultiple of 6.



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Example:

For calculate den( B498) with the first method:

n=498

divisors of 498 di: 1, 2, 3, 6, 83, 166, 249, 498

prime numbers di+1: 2, 3, 4, 7, 84, 167, 250, 499

4, 84 and 250 are discarded, because not prime numbers, and so:

den( B498) = 2*3*7*167*499 = 3.499.986



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1.1.2 NUMERATOR OF B N num(B N )

We can define:

Numerator( B2n) = num( B2n) → n

num( B 2n ) = first factor of the factorization( B2n ) if the first number is n we have that

it is a prime number p

Example:

n=11

num(B22)= 854.513

Numerator( B22)= first factor of the factorization( B22) = 854.513 = 11*131*593 → 11prime number.

When n odd integers we have that all the values of Bn are zero and so we find all

the n+1 even integer.

When we have a num( B 2n ) that gives a number n that isn’t a prime number or

multiple of only 2 and 3, the factorization of num( B 2n ) is starting with a new prime

number.

If n=kp or a multiple of a prime number, but not multiple of only 2 and 3, we have

that the factorization of num( B 2n ) is always starting with p



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Example:

n=14

num(B28)= 23.749.461.029

Numerator( B28)= first factor of the factorization( B28) = 23.749.461.029=

7*9349*362903 → 7 prime number

Index of B28 → 28=4*7, 7 prime number

----------------------------------------------------------------------------------------------------------

n=16

num(B32)= 7.709.321.041.217

Numerator( B32)= first factor of the factorization( B32) = 7.709.321.041.217=

37*683*305065927 → 16 is not valid

Index of B32 → 32=25, not valid because it isn’t a multiple of a prime number, it is

multiple of only 2.



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TAB. 2

2n num(B2n) Factorization prime frequency

10 5 5 5 20k+10

12 691 691 7 42k+14, 42k+28

14 7 7 11 22k

16 3617 3617 13 26k

18 43867 43867 17 34k

20 174611 283 617 19 38k

22 854513 11 131 593 23 46k

24 236364091 103 2294797 29 58k

26 8553103 13 657931 31 62k

28 23749461029 7 9349 362903 37 36k+32, 74k

30 8,61584E+12 5 1721 1001259881 41 82k

32 7,70932E+12 37 683 305065927 43 86k

34 2,57769E+12 17 151628697551 47 94k

36 2,63153E+19 2,63153E+19 53 106k

38 2,92999E+15 19 154210205991661 59 59k+44, 118k

40 2,61083E+20 137616929 1897170067619 61 122k

42 1,5201E+21 1,5201E+21 67 66k+58, 134k

44 2,78333E+22 11 59 8089 2947939 1798482437 71 142k

46 5,96451E+23 23 383799511 67568238839737 73 146k

48 5,6094E+27 653 56039 153289748932447906241 79 158k

50 4,95057E+26 5^2 417202699 47464429777438199 83 166k

52 8,01166E+2913 577 58741 4010291774534045619429

89 178k

54 2,915E+31 39409 97 194k

56 2,47939E+337 113161 163979

19088082706840550550313



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58 8,44836E+3429 67 186707 623524204937349583369104129

60 1,21523E+42 2003

62 1,23006E+37 31 157 266689

64 1,06784E+41 1,06784E+41

66 1,4726E+45

11 839

1595622518286201813903585901562

39282938769

68 7,87731E+4317 37 101 123143 1822329343

5525473366510930028227481

70 1,50538E+48 5 7 688531

72 5,82795E+54 5,82795E+54

74 3,41524E+49

37

9230383051140856220089209116614

22572613197507651

76 2,46551E+52 19 58231

78 4,14846E+56 1380 4,60378E+60 631 10589

We can observe that irregular primes have a different behavior from other regular

primes in their frequencies.

In fact irregular primes 37, 59, 67,… have two different frequencies while all other

regular primes have only one regular frequency:

for p prime number we have frequency = 2p

We have an exception with the number prime 7. It has two frequencies but both are

regular. In fact 7 occurs with a frequency of:

2n = 14, 28, 56, 70, 98, 112, …or 42k+14, 42k+28



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1.1.3 FINDING A PROBABLY PRIME NUMBER p LARGE AS DESIRED TO

60%

We can find a large prime number as we like by simply considering this formula withthe den( Bn)

p = den( Bn)+1 or p = den( Bn)-1

We thus have two prime numbers or only one that is a prime number, or none of the twois a prime number.

The latter case is, however, the least likely of the 3 possible cases.

In the following table are highlighted the prime numbers that derive:



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TAB. 3

n den(Bn) Factorization den(Bn)-1 den(Bn)+1

1 2 2 1 3

2 6 2 * 3 5 7

4 30 2 * 3 * 5 29 31

6 42 2 * 3 * 7 41 43

8 30 2 * 3 * 5 29 3110 66 2 * 3 * 11 65 67

12 2730 2 * 3 * 5 * 7 * 13 2729 2731

14 6 2 * 3 5 7

16 510 2 * 3 * 5 * 17 509 511

18 798 2 * 3 * 7 * 19 797 799

20 330 2 * 3 * 5 * 11 329 331

22 138 2 * 3 * 23 137 139

24 2730 2 * 3 * 5 * 7 * 13 2729 2731

26 6 2 * 3 5 728 870 2 * 3 * 5 * 29 869 871

30 14322 2 * 3 * 7 * 11 * 31 14321 14323

32 510 2 * 3 * 5 * 17 509 511

34 6 2 * 3 5 7

36 19191902 * 3 * 5 * 7 * 13 * 19 *

37 1919189 1919191

38 6 2 * 3 5 7

40 13530 2 * 3 * 5 * 11 * 41 13529 13531

42 1806 2 * 3 * 7 * 43 1805 180744 690 2 * 3 * 5 * 23 689 691

46 282 2 * 3 * 47 281 283

48 46410 2 * 3 * 5 * 7 * 13 * 17 46409 46411

50 66 2 * 3 * 11 65 67

52 1590 2 * 3 * 5 * 53 1589 1591

54 798 2 * 3 * 7 * 19 797 799

56 870 2 * 3 * 5 * 29 869 871

58 354 2 * 3 * 59 353 355



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6056786730

2 * 3 * 5 * 7 * 11 * 13 *31 * 61 56786729 56786731

62 6 2 * 3 5 7

64 510 2 * 3 * 5 * 17 509 511

66 64722 2 * 3 * 7 * 23 * 67 64721 64723

68 30 2 * 3 * 5 29 31

70 4686 2 * 3 * 11 * 71 4685 4687

721,4E+08

2 * 3 * 5 * 7 * 13 * 19 *

37 * 73 140100869 140100871

74 6 2 * 3 5 776 30 2 * 3 * 5 29 31

78 3318 2 * 3 * 7 * 79 3317 3319

80 230010 2 * 3 * 5 * 11 * 17 * 41 230009 230011

82 498 2 * 3 * 83 497 499

843404310

2 * 3 * 5 * 7 * 13 * 29 *43 3404309 3404311

86 6 2 * 3 5 7

88 61410 2 * 3 * 5 * 23 * 89 61409 61411

90 272118 2 * 3 * 7 * 11 * 19 * 31 272117 27211992 1410 2 * 3 * 5 * 47 1409 1411

94 6 2 * 3 5 7

964501770

2 * 3 * 5 * 7 * 13 * 17 *

97 4501769 4501771

98 6 2 * 3 5 7

100 33330 2 * 3 * 5 * 11 * 101 33329 33331



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1.2 CONNECTION ζ (s) ZETA RIEMANN FUNCTION WITH PRIME NUMBERS

Bernoulli numbers as values of the Riemann zeta function.

Associated sequence: 1, +1/2, 1/6, 0, …

Using this convention, the values of the Riemann zeta function satisfy

ζ(1−n)= -n

Bn

for all integers n≥0.

For n=0 is to be understood as B0=lim x→0 xζ(1− x)=1

So we have:

Denominator(nζ(1−n)) = den(nζ(1−n)) = den( Bn) → n+1

We can eliminate the minus sign “-” because we consider only the positive integers asdenominator.

Denominator (nζ(1−n))= last factor of the factorization (nζ(1−n)) if the last number is n

+1 we have that it is a prime number p;

or also:

den(ζ(1−n))= last factor of the factorization (ζ(1−n)) if the last number is n +1 we havethat it is a prime number p



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Obviously, for negative even integers, we have that:

ζ(−2n)= 0 for n ≥ 1

ζ vanishes at the negative even integers because Bm = 0 for all odd m other than 1

So there is a link with

ζ(−2n−1) for n ≥ 0

or with the denominator ζ(negative odd integers) and the prime numbers:

ζ(0) = -2

1→ det(2) → 2

ζ(−1) = -12

1→ det(12) = 2

2 * 3 → 3

ζ(−3) = +120

1→ det(120) = 2

3 * 3 * 5 → 5

ζ(−5) = -252

1→ det(252) = 2

2 * 3

2 * 7 → 7

Being present n we have a factorization with exponents while with regard to Bn wereonly distinct prime factors with exponent equal to 1.

We note that for

ζ(−3) = +120

1→ det(120) = 2

3 * 3 * 5 → 5

we have that 120 = 24 * 5 , where 24 is related to the physical vibrations of the bosonic

strings by the following Ramanujan function:



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( )

++

+

⋅

=

−

∞−

∫

4

2710

4

21110log

'

142

'

cosh

'cos

log4

24

2

'

'4

0

'

2

2

wt itwe

dxe x

txw

anti

w

wt

w x

φ

π

π

π

π

,

and that 2, 3 and 5 are all Fibonacci’s numbers.



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1.3.1 SPECIFIC VALUES

Case s=-1

ζ(-1) = 1+2+3+4+…..= ...6

7

4

5

2

32 =

s prime p

−−

∏1

1= -

12

1

gives a way to assign a finite result to the divergent series 1 + 2 + 3 + 4 + …..·

----------------------------------------------------------------------------------------------------------

Case s=0

ζ(0) = 1+1+1+1+…. =0

1

0

1

0

1

0

1…. = -

2

1

gives a way to assign a finite result to the divergent series 1+1+1+1+….



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Case s=1, Pole of ζ(s)

( ) ∞=+++++= ...5

1

4

1

3

1

2

111ζ

thus,

∞=⋅⋅⋅⋅⋅

⋅⋅⋅⋅⋅

=+++++ ...106421

...117532

...5

1

4

1

3

1

2

1

1

for x→1+ ζ(1

+)=+∞

for x→1- ζ(1

-)=-∞

B0=lim x→0 xζ(1− x)=1

----------------------------------------------------------------------------------------------------------

Case s=2

( ) 645.16

...3

1

2

112

2

22 ≈=+++=

π ζ ;

ζ(2) = 48

49

24

25

8

9

3

4

…. = 6

2π

We note that 1.645 is very near to the aurea frequency 1.64809064 that is given by

0,61803398 · 4 · 2/3, where 0,61803398 is the aurea section, i.e.2

15 −

----------------------------------------------------------------------------------------------------------



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Case s=+∞

ζ(+∞) = 1+



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2. BERNOULLI'S FORMULA EXTENDED IN THE FIELD OF REAL

NUMBERS x ∈R

We can define the Bernoulli's formula as the analytic continuation of the formula in thefield of real numbers ∈ x R, as it had done with the Riemann zeta function.

The Cartesian plane of the zeta function ζ(x) for real numbers ∈R between -18.5≤ n ≤10



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We know that

ζ(1−n)= -n

Bn

for all integers n≥0.

We extend this formula with ∈ x R considering also the negative values of Bernoulli’snumbers and all the real positive and negative numbers, and so we have:

x B = ( ) x B = x− ζ ( ) x−1

We must consider the zeta function ζ(1-x), shifted to the left side of one, so that the y-axis becomes the x=0 pole.

Then we must draw the graph symmetric to the y-axis as if it were reflected.

Finally, we must multiply by the function bisector line of the second and fourthquadrant y=-x

The table with -10≤x≤25 is the following:



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TAB. 5

x B(x)

-10 10,004941

-9 9,008951

-8 8,016000

-7 7,028000-6 6,050094

-5 5,071262

-4 4,147600

-3 3,246969

-2 2,404113

-1 1,644934

0 1

1 0,500000

2 0,1666673 0

4 -0,033333

5 0

6 0,023810

7 0

8 -0,033333

9 0

10 0,075758

11 012 -0,253114

13 0

14 1,166667

15 0

16 -7,092156863

17 0

18 54,971177945

19 0



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20 -529,124242421 0

22 6192,123188

23 0

24 -86580,25311

25 0

And the graph is:

In x=0 is the pole.

We note that for x→-∞ ( ) x x B x −=−∞→lim



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and for x→+∞ we have oscillations increasingly large and always larger with all oddvalues of B(x) equal to 0.

For negative integers values:

B(-1/2)= 1/2 (3/2)= (1/2)*2,612...= 1,306

B(-1)= ζ(2)= 6

2π

= 1,6449...

B(-3/2)= 3/2 (5/2)= (3/2)*1,341...= 2,01...

B(-2)= 2ζ(3)= 2* 1,202056…= 2,404113…

B(-5/2) = 5/2 (7/2)= (5/2)*1,127= 2,8175...

B(-3)= 3ζ(4)= 3*

90

4π =3,246969

This value is very near to the following aurea frequency 3,24611797 that is equal about

to (1,61803398…)-5

· 12 · 3 = 3,246118, where2

15...61803398,1

+= is the aurea ratio.

…

B(-x)≈ -x per x large x>10

For positive integers values:

B(0)= 0*ζ(1)= 1= lim x→0 xζ(1− x)

B(1/2)= -1/2 (1/2)=-(1/2)(-1.4603...)= 0,73..

B(1)= -ζ(0)= ½



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B(2)= -2ζ(-1)= 1/6

B(3)= -3ζ(-2)= 0

B(4)= -4 (-3)= -1/30

So all the odd integers of B(x)=0 and therefore coincide with the even negative

integers trivial zeros of ζ(1-x)=0

for x=2k+1

k = 1, 2, 3, ...

B(2k+1)=ζ(-2k)=0

or

x=3, 5, 7, ....

B(x)=ζ(1-x)=0

The (1-x) even negative integers of ζ(1-x), with x=3, 5, 7, ...., match the negative

axis y=0 of the Cartesian plane and coincide with the zeros of the odd integer of

B(x).

In the range 0≤x≤12 is the following



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3. BERNOULLI's FORMULA EXTENDED IN THE FIELD OF COMPLEX

NUMBERS z∈C, z=σ ±it

We can define the Bernoulli's formula as the analytic continuation of the formula in thefield of complex numbers z∈C, z=σ ±it , as it had done with the Riemann zeta function.

We extend this formula with z∈C, z=σ ±it , considering also complex values ofBernoulli’s numbers and so we have:

B z = B(z)=-zζ(1− z)

( ) ∑=

+++==n

k

mmmm

m nk nS 1

...21 .

for m, n ≥ 0

We know that the Riemann zeta function ζ (s) is a function of a complex variable s=σ +it .

The following infinite series converges for all the complex numbers s with real partgreater than 1, or Re(s)>1, and defines ζ (s) in this case:

( ) ∑∞

=

−+++==

1

...31

21

11

nsss

snsζ ( ) 1>= sR σ .

We have

ζ(1−n)= -n

Bn



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for all integers n>0.

If we compare the 2 function Sm(n) for n→+∞ and ζ(s) the single factors of the twoseries is one the reciprocal of the other, or equivalently

ζ(-n) =∑∞

=1k

nk

for all integers n≤0.

for all integers n≤0.

Besides we have for s>1:

for s<1 we can write in the same way:

ζ(-s) = ...71

1

51

1

31

1

21

1ssss

−−−−=

s prime p−

∏1

1

Going from the product representation of the Riemann zeta-function and the aboveseries expansions of the Bernoulli numbers, the following shows:

( )

( )

( )

( )∏∈

−

−

−

−

=

−=

P p

nnn

nnnn

n

p

n

...5

11

3

11

2

11

1

2

!2211

2

!22

222

2

1

22π π

β



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for n≥0

Here, the product extends over all primes (Euler product of the Riemann zeta function)and

βn = |B(2n)| the absolute value

We can calculate, for example, B(2)

n=1

B(2) = β1 =( )

2

2 11

1

)2(

!22

p

prime

−

∏π

=64

4 2

2

π

π =

6

1



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4. ZEROS ζ(½ ± ix)=0 ON THE CRITICAL LINE

The functional equation shows that the Riemann zeta function has zeros at −2, −4, ... .

These are called the trivial zeros. They are trivial in the sense that their existence is

relatively easy to prove, for example, from sin(πs /2) being 0 in the functional equation

(see below). The non-trivial zeros have captured far more attention because theirdistribution not only is far less understood but, more importantly, their study yieldsimpressive results concerning prime numbers and related objects in number theory.

The Riemann zeta function satisfies the functional equation:

( ) ( ) ( )sss

sss

−−Γ

=

− 112

sin2 1 ζ π

π ζ ,

where Γ(s) is the gamma function, which is an equality of meromorphic functions validon the whole complex plane.

It is known that any non-trivial zero lies in the open strip {s∈C: 0<Re(s)<1}, which is

called the critical strip. The Riemann hypothesis, asserts that any non-trivial zero s has

Re(s)=1/2. In the theory of the Riemann zeta function, the set {s∈C: Re(s)= 1/2} is

called the critical line.



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Blue is the real part and red is the imaginary part of the function ζ(½ ± ix) for 0≤x≤100

is shown so that we can clearly see the first non-trivial zeros and the values where real

and imaginary part are both equal to zero:



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Re(ζ(½±ix)) = IM(ζ(½±ix)) = 0

We have for all ∈s C\ { }1 the integral relation

( ) ( )

( ) ( )∫∞

−

++

−

−

=0

22

1

11

arctansin2

1

2dt

et

t s

s

s

t

s

ss

π

ζ ,

holds true, which may be used for a numerical evaluation of the zeta-function.

To find non-trivial zeros for ς(p) = 0 and substituting s with p (s=p) we must have

p = x + jy

ς(x + jy) =yj1-x

2 yj1-x

+

+

- yjx2 +

∫∞

+++

+

0

2 / 2 / 2 )1()1(

)arctan*)sin((dt

et

t yj xt jy x π

dividing by 2p ≠ 0 (always true) we have

1)-2(p1 = ∫

∞

++0

2 / 2 )1()1()arctansin( dt

et

t pt p π

Only with particular values of x=½ and y that are precisely the zeros of the function we

have:

for p= ½ + jy, with y that has a specific real value ∈ R



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∫∞

++0

2 / 2 )1()1(

)arctansin(dt

et

t pt p π

= ∫∞

−

−

0

2)1(2

1dt

p=

1)-2(p

1;

2yj1-

1

+= ∫

∞

+++

+

0

2 / y4 / 12 )1()1(

)arctan)y2

1sin((

dt et

t j

t j π =

= )

4

12

4

1y

4

12

(2

2

2

0

2

2

+

−

+

+

∫∞

y

j

y

ydy=

+ 4

1

4

1-

2

y

-j

+ 4

1

2

y

2

y

=24y1

1-

+ - j

24y1

2y

+

for the first zero of the Riemann zeta function:

y=14,134725...

p= ½ + j14,134725...

2yj1-

1

+ = .j28,26945..1-

1

+ = .800,1618..

1-

- ...1618,800

.y28,26945..

j = -0.0012497 - 0.0353296j

∫∞

+++

+

0

2 / 14,1347254 / 12 )1()1(

)arctan)14,1347252

1sin((

dt et

t j

t j π =

24y1

1-

+ - j

24y1

2y

+= -0.0012497 - 0.0353296j

The two quantities are equal and p = ½+j14.134725... is a non-trivial zero, and is also

the first zero of the Riemann zeta function.

In the table the first 30 zeros of Riemann zeta function ς(½ + jy) with y up to 100



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TAB. 6

n°Value of the IM

of zero, y

1 14,134725142

2 21,022039639

3 25,010857580

4 30,424876126

5 32,935061588

6 37,586178159

7 40,918719012

8 43,327073281

9 48,005150881

10 49,773832478

11 52,970321478

12 56,446247697

13 59,347044003

14 60,831778525

15 65,112544048

16 67,079810529

17 69,546401711

18 72,067157674

19 75,704690699

20 77,144840069

21 79,33737502022 82,910380854

23 84,735492981

24 87,425274613

25 88,809111208

26 92,491899271

27 94,651344041

28 95,870634228



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29 98,83119421830 101,317851006

Returning to the Bernoulli numbers we have:

B z = B(z)=-zζ(1− z)

with z=½ ± ix

B(½ + ix)=-(½ + ix)ζ(½ - ix)

B(½ - ix)=-(½ - ix)ζ(½ + ix)

Let's consider the zeros of ζ(½ + ix) with the particular values of x=b so that ζ(½+ib)=0.

So all the values of z=½ + ib of ζ(½+ib)=0 and therefore coincide with the values of

z=½ + ib of B(½-ib)=0 or with its complex conjugate.

The complex zeros of ζ(½+ib)=0 coincide with the complex conjugate zeros of B(½-ib)=0 and vice versa.

Example for the first three zeros

b=14,134725142 – 21,022039639 - 25,010857580.....

ζ(½+ib)=B(½-ib)=0

ζ(½-ib)=B(½+ib)=0



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But the function ζ(½ + ix) has others values where the IM(ζ(½ + ix))=0 and so we findthe correlation with the Gram points g IM(ζ(½ + ig))=0

The first Gram point is:

B(½ - i3,436)=-(½ - i3,436)ζ(½ + i3,436)

or

B(½ + i3,436)=-(½ + i3,436)ζ(½ - i3,436)

ζ(½ + i0) = -2B(½) ≈-1,4603545...

ζ(½ + i3,436)≈0,5641

ζ(½ + i9,667)≈1,53181

ζ(½ + i14,13)=0 the first zero of ζ(z)=0

ζ(½ + i17,8455) ≈ 2,34018

ζ(½ + i21,02)=0 the second zero of ζ(z)=0

ζ(½ + i23,17)≈1,45744

ζ(½ + i25,01)=0 the third zero of ζ(z)=0

ζ(½ + i27,67)≈2,84509

These are the Gram points.



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TAB. 7 with Gram Points:

x jy

0,000000 -1,46035

3,436218 -0,56415

9,666908 -1,53181

14,134725 017,845599 2,34018

21,022040 0

23,170283 -1,45744

25,010858 0

27,670182 2,84509

30,424876 0

31,717980 -0,92526

32,935062 0

35,467184 2,9381237,586178 0

38,999210 -1,78672

40,918719 0

42,363550 1,098756

43,327073 0

45,593029 -3,6629

48,005151 0

48,710777 0,688292

49,773832 051,733843 -2,01121

52,970321 0

54,675237 2,91239

56,446248 0

57,545165 -1,75816

59,347044 0

60,351812 0,53858

60,831779 0



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63,101868 -4,164465,112544 0

65,800888 1,053877

67,079811 0

68,453545 -1,54

69,546402 0

71,000000 1,9

72,067158 0

73,600000 -3,6

75,704691 076,200000 0,7

77,144840 0

78,700000 -1,3

79,337375 0

81,100000 3,97

82,910381 0

83,500000 -1,2

84,735493 0

86,000000 1,0987,425275 0

88,300000 0,8

88,809111 0

90,800000 3,97

92,491899 0

93,100000 1,39

94,651344 0

95,300000 0,7

95,870634 097,700000 2,9

98,831194 0

100,000000 2,8

101,317851 0



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4.1 GRAPHICS OF THE REAL AND IMAGINARY ς (p)

Re ς (1/2 + jy) and Im ς (1/2 + jy)



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Re ς (1/3 + jy) and Im ς (1/3 + jy)



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Re ς (3/4 + jy) and Im ς (3/4 + jy)



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5. A NEW REPRESENTATION OF THE BERNOULLI FUNCTION

CONNECTED WITH THE RIEMANN ZETA FUNCTION

Applying for all ∈s C\ { }1 the integral relation

( ) ( )

( ) ( )∫∞

−

++

−−

=0

22

1

11

arctansin2

1

2dt

et

t s

ss

t s

ss

π

ζ ,

and

B(z)=-zζ(1− z)

substituting s with z (z=p) we have

1-z → z

ς(1-z) = z

z 2*

1

− -

z2

2∫∞

−

++

−

0 2

1

2 )1()1(

)arctan*)1sin((dt

et

t z

t

z

π

B(z)= z21 +

z z

22 ∫

∞

−

++

−

0 2

1

2 )1()1(

)arctan*)1sin(( dt

et

t z

t

z

π

or

B(z)= z2

1[1 + 2z ∫

∞

−

++

−

0 2

1

2 )1()1(

)arctan*)1sin((dt

et

t z

t

z

π

]



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B(z)=-zζ(1− z)

z)-(1

B(z)

ζ =-z

ζ(½+ib)=B(½-ib)=0

ζ(½-ib)=B(½+ib)=0

To find non-trivial zeros for B(z)= 0

with

z = x + jy

2z ≠ 0 (always) we have

[1 + 2z ∫∞

−

++

−

0 2

1

2 )1()1(

)arctan*)1sin((dt

et

t z

t

z

π

] = 0; z2

1− = ∫

∞

−

++

−

0 2

1

2 )1()1(

)arctan*)1sin((dt

et

t z

t

z

π

That’s equivalence with (pag. 40)

1)-2(p1 = ∫

∞

++0

2 / 2 )1()1()arctansin( dt

et t p t p π

with 1-p →z



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Now, we have:

∫∞

++0

2 / 2 )1()1(

)arctansin(dt

et

t pt p π

= ∫∞

−

−

0

2)1(2

1dt

p=

1)-2(p

1;

For p = 1/2+yj:

∫∞

+++

+

0

2 / y4 / 12 )1()1()arctan)y2

1

sin(( dt et

t jt j π

=2yj1-

1+

=24y1

1-+

- j24y1

2y+

Multiplying this equation for 1/6, we obtain the following interesting expression:

6

1∫∞

+++

+

0

2 / y4 / 12 )1()1(

)arctan)y2

1sin((

dt et

t j

t j π =

24y26

1-

+ - j

24y26

2y

+;

This expression can be related with the number 24 that is connected to the “modes” thatcorrespond to the physical vibrations of the bosonic strings by the following Ramanujan

function:

( )

++

+

⋅

=

−

∞−

∫

4

2710

4

21110log

'

142

'

cosh

'cos

log4

24

2

'

'4

0

'

2

2

wt itwe

dxe x

txw

anti

w

wt

w x

φ

π

π

π

π

.

Thence, we have the following mathematical connection:



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6

1∫∞

+++

+

0

2 / y4 / 12 )1()1(

)arctan)y21sin((

dt et

t j

t j π =

24y26

1-

+ - j ⇒

+24y26

2y

( )

++

+

⋅

⇒

−

∞−

∫

4

2710

4

21110log

'

142

'

cosh

'cos

log42

'

'4

0

'

2

2

wt itwe

dxe x

txw

anti

w

wt

w x

φ

π

π

π

π

.

5.1 A REPRESENTATION OF THE ABSOLUTE VALUES OF BERNOULLI

NUMBERS CONNECTED WITH THE RIEMANN ZETA FUNCTION FOR n≥ 1

The *

n B =|B(2n)| absolute values of Bernoulli numbers may be calculated from the

integral:

∫∞

−∗

−=

0 2

12

14

t

n

ne

dt t n B

π ,

for n≥1*

n B = |B(2n)| the absolute value

from

B(z)=-zζ(1− z)

with 2n → z, we have

|B(2n)|= 2n|ζ(1−2n)|



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|ζ(1−2n)| = 2 ∫∞ −

−0

2

12

1dt

e

t t

n

π for n≥1

Multiplying this equation for 4, we obtain the following interesting relationship:

4 |ζ(1−2n)| = 8 ∫∞ −

−0

2

12

1dt

e

t t

n

π

We remember that 8 is a Fibonacci’s number and is connected with the “modes” that

correspond to the physical vibrations of a superstring by the following Ramanujan

function:

( )

++

+

⋅

=

−

∞−

∫

4

2710

4

21110log

'

142

'

cosh

'cos

log4

318

2

'

'4

0

'

2

2

wt itwe

dxe x

txw

anti

w

wt

w x

φ

π

π

π

π

,

We have the following new equation:

4 |ζ(1−2n)| = ( )

++

+

⋅

−

∞−

∫

4

2710

4

21110log

'

142

'

cosh

'cos

log4

31

2

'

'4

0

'

2

2

wt

itwe

dxe x

txw

anti

w

wt

w x

φ

π

π

π

π

∫∞ −

−0

2

12

1dt

et

t

n

π



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6. REFERENCES

1) Wikipedia

2) Tesi di laurea su i numeri Bernoulli e loro applicazioni

3) PROGETTO BEFZS POSSIBILI CONNESSIONI MATEMATICHE TRA:

a) i numeri Bernoulli (B), b) i numeri di Eulero (E),

c) i numeri di Fibonacci (F),

d) la funzione zeta (Z) di Riemann, e

e) la teoria di stringa (S)

Pier Francesco Roggero, Michele Nardelli, Francesco Di Noto - "CONNECTION BERNOULLI NUMBERS Bn AND...

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