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Some properties of Bernoullis numbers

Journal of the Indian Mathematical Society, III, 1911, 219 234

1. Let the well-known expansion ofx cot x (videEdwards Differential Calculus,149) bewritten in the form

x cot x= 1 B22!

(2x)2 B44!

(2x)4 B66!

(2x)6 , (1)

from which we infer that B0 may be supposed to be1. Now

cot x = cos xsin x =

1

x2

2!

+x4

4!x6

6!

+

x x

3

3! +

x5

5! x

7

7! +

= sin2x

1cos 2x =2x

1! (2x)

3

3! +

(2x)5

5!

(2x)2

2! (2x)

4

4! +

(2x)6

6!

= 1 + cos 2x

sin2x =

2(2x)2

2! +

(2x)4

4! (2x)

6

6! +

2x (2x)3

3! +

(2x)5

5! (2x)

7

7! +

Multiplying both sides in each of the above three relations by the deno minator of theright-hand side and equating the coefficients ofxn on both sides, we can write the resultsthus:

c1Bn1

2 c3 Bn3

23 + c5

Bn525

+(1)1

2(n1)

2n B0+

n

2n(1) 12 (n1) = 0, (2)

where n is any odd integer;

c2Bn2 c4Bn4+ c6Bn6 + (1)12 (n2)B0+ n2

(1) 12 (n2) = 0, (3)

where n is any even integer;

c1Bn1 c3Bn3+ c5Bn5 + (1)1

2(n1)B0+

n

2(1) 12 (n1) = 0, (4)

where n is any odd integer greater than unity.From any one of (2), (3), (4) we can calculate the Bs. But as n becomes greater andgreater the calculation will get tedious. So we shall try to find simpler methods.

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Some properties of Bernoullis numbers 5

for n in succession in (12). Thus

B0

=

1; B2

= 16

; B4

= 130

; B6

= 142

; B8

1

3B

2=

1

45;

B10 52B4= 1132 ; B1211B6 = 4455 ; B14 1434 B8+B2

5 = 1120 ;

B16 2863 B10+ 4B4= 1306 ; B18221B12+ 2045 B6= 3665 ;

B20 32307 B14+ 19387 B8B2

7 = 1231 ;

B22 35534 B16+ 71065 B10 112B4= 1552 ;

and so on. Hence we have finally the following values:

B2= 16 ; B4=

130 ; B6=

142 ; B8=

130 ; B10 =

566 ; B12=

6912730 ; B14=

76 ;

B16= 3617510; B18=

43867798 ; B20=

174611330 ; B22=

854513138 ; B24=

2363640912730 ;

B26= 8553103

6 ; B28= 23749461029

870 ; B30= 8615841276005

14322 ;

B32= 7709321041217

510 ; B34= 2577687858367

6 ; B36= 26315271553053477373

1919190 ;

B38= 29299939138415596 ; B40= 26108271849644912205113530 ; , B =.

6. It will be observed that, ifn is even but not equal to zero,(i) Bn is a fraction and the numerator ofBn/nin its lowest terms isa prime number, (14)(ii) the denominator ofBn contains each of the factors 2 and 3 onceand only once, (15)(iii) 2n(2n 1)Bn/nis an integer and consequently 2(2n 1)Bn is anodd integer. (16)

From (16) it can easily be shewn that the denominator of 2(2n1)Bn/nin its lowest termsis the greatest power of 2 which divides n; and consequently, ifn is not a multiple of 4,

then 4(2n 1)Bn/nis an odd integer. . . . (17)It follows from (14) that the numerator ofBn in its lowest terms is divisible by the greatestmeasure ofn prime to the denominator, and the quotient is a prime number. . . . (18)Examples: (a) 2 and 3 are the only prime factors of 12, 24 and 36 and they are found inthe denominators ofB12, B24 and B36 and their numerators are prime numbers.

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(b) 11 is not found in the denominator ofB22, and hence its numerator is divisible by 11;similarly, the numerators ofB26, B34, B38 are divisible by 13, 17, 19, respectively and the

quotients in all cases are prime numbers.(c) 5 is found in the denominator of B20 and not in that of B30, and consequently thenumerator ofB30 is divisible by 5 while that ofB20 is a prime number. Thus we may saythat if a prime number appearing in n is not found in the denominator it will appear inthe numerator, and vice versa.

7. Next, let us consider the denominators.All the denominators are divisible by 6; those of B4, B8, B12, by 5; those of B6, B12,B18, by 7; those ofB10, B20, B30, by 11; but those ofB8, B16, B24, are notdivisibleby 9; and those ofB14, B28, are notdivisible by 15. Hence we may infer that:the denominator of Bn is the continued product of prime numbers which are the nextnumbers (in the natural order) to the factors ofn(including unity and the number itself)

...(19)As an example take the denominator ofB24. Write all the factors of 24, viz. 1, 2, 3, 4, 6,8, 12, 24. The next numbers to these are 2, 3, 4, 5, 7, 9, 13, 25. Strike out the compositenumbers and we have the prime numbers 2, 3, 5, 7, 13. And the denominator ofB24 is theproduct of 2, 3, 5, 7, 13, i.e., 2730.It is unnecessary to write the oddfactors ofn except unity, as the next numbers to theseare even and hence composite.The following are some further examples:

Even factors ofn and unity Denominator of Bn

B2 1, 2, 23 = 6B6 1, 2, 6 237 = 42B12 1, 2, 4, 6, 12 235713 = 2730B20 1, 2, 4, 10, 20 23511 = 330B30 1, 2, 6, 10, 30 2371131 = 14322B42 1, 2, 6, 14, 42 23743 = 1806B56 1, 2, 4, 8, 14, 28, 56 23529 = 870B72 1, 2, 4, 6, 8, 12, 18, 24, 36, 72 235713193773 = 140100870B90 1, 2, 6, 10, 18, 30, 90 237111931 = 272118B110 1, 2, 10, 22, 110 231123 = 1518

8. Again taking the fractional part of any B and splitting it into partial fractions, we seethat:the fractional part ofBn = (1) 12n{the sum of the reciprocals of the prime factors of thedenominator ofBn} (1)12n. (20)

Thus the fractional part of B16 = 12 +

13 +

15 +

1171 = 47510 ;

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that of B22 = 1 12 13 123 = 17138 ;that of B28 =

12 +

13 +

15 +

129

1 = 59870 ; and so on.

9. It can be inferred from (20) that:IfG be the G.C.M. and L the L.C.M. of the denominators ofBm and Bn, then L/G is

the denominator ofBm (1) 12 (mn)Bn,and hence, if the denominators ofBmandBn areequal, thenBm(1)12 (mn)Bn is an integer. (21)Example: B24 B12 and B32 B16 are integers, while the denominator ofB10+ B20 is 5.It will be observed that:(1) Ifn is a multiple of 4, then the numerator ofBn 130 in its lowest terms is divisible by20; but ifn is not a multiple of 4 then that of Bn

n 112 in its lowest terms is divisible by 5;

(22)

(2) Ifnis any integer, then

2(24n+2 1) B4n+22n+ 1

, 2(28n+4 1) B8n+42n+ 1

, 2(28n+4 1) B16n+82n+ 1

are integers of the form 30p+ 1. (23)

10. If aB is known to lie between certain limits, then it is possible to find its exact valuefrom the above properties.Suppose we know that B22 lies between 6084 and 6244; its exact value can be found asfollows.The fractional part ofB22 =

17138 by (20), also B22 is divisible by 11 by (18). And by (22)

B22 116 must be divisible by 5. To satisfy these conditions B23 must be either 6137

17138 or

6192 17138 .But according to (18) the numerator ofB22 should be a primenumber after it is dividedby 11; and consequently B22 must be equal to 6192

17138 or

854513138 , since the numerator of

6137 17138 is divisible not only by 11 but also by 7 and 17.

11. It is known (Edwards Differential calculus, Ch. v, Ex.29) that

Bn = 2n!(2)n

1

1n+

1

2n+

1

3n+

,

or2

n!

(2)n =Bn

1 1

2n

1 1

3n

1 1

5n

(24)

where 2, 3, 5, . . . are prime numbers.Also

Bn2n

=

0

xn1

e2x 1 dx. (25)

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For

0

xn1

e2x 1 dx =0

xn1(e2x + e4x + )dx

= (n 1)!

(2)n

1

1n+

1

2n+

1

3n+

=

Bn2n

by (24). In a similar manner

0

xn

(ex ex)2 dx=Bn4

,

and0

xn2 log(1 e2x)dx= Bnn(n 1) . (26)

Take logarithms of both sides in (24) and write for log e n! the well known expansion ofloge(n + 1), as in Carrs Synopsis, viz.

(n + 12 )log n n + 12log 2 + B212n

B434n3 +

B656n5

(1)p B2p(2p 1)2pn2p1 , (27)

where 0<

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values of log10 e, loge10, , . . . , which are known to a great number of decimal places, weshould have to find the logarithms of certain numbers whose values are not found in the

tables to as many places of decimals as we require. Such difficulties are removed by themethod given in13.

12. Results (14) to (17), (20) and (21) can be obtained as follows. We have

1

2x2 +

1

(x + 1)2+

1

(x + 2)2+

1

(x+ 3)2+

1

(x+ 4)2+

1x 1

6(x3 x)+ 1

5(x5 x)+ 1

7(x7 x)+ 1

11(x11 x)+

= 1

x15 7

x17+

55

x19 529

x21 + (28)

where 5, 7, 11, 13, are prime numbers above 3. If we can prove that the left-hand side of(28) can be expanded in ascending powers of 1/x with integral coefficients, then (20) and(21) are at once deduced as follows.From (27) we have

d2 log (n + 1)

dn2 =

1

(n + 1)2+

1

(n + 2)2+

1

(n + 3)2+

= 1

n 1

2n2+

B2n3

B4n5

+B6n7

B8n9

+ ()pB2pn2p+1

, (29)

where

B2p

n2p+1 =

B2pn2p+1

B2p+2n2p+3

+

= 4

0

x2p

n2p+1(ex ex)2 dx40

x2p+2

n2p+3(ex ex)2 dx +

=

0

x2p

n2p+1 x

2p+2

n2p+3+

dx

sinh2 x

=

0

x2pn2p1(n2 + x2)

dxsinh2 x

=

0

x2p

(1 + x2)sinh2 nxdx.

Substituting the result of (29) in (28) we see that

B2x3

B4x5

+B6x7

16(x3 x)+

1

5(x5 x)+ 1

7(x7 x)+ 1

11(x11 x)+ ,

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where 5, 7, 11, . . . are prime numbers, can be expanded in ascending powers of 1 /x withintegral coefficients.

ThereforeB216 ,B4

16+

15 ,B6

16+

17 ,B8

16+

15 ,B10

16+

111 , ,which are the

coefficients of 1/x3, 1/x5, 1/x7, ,are integers.Writing 12 +

131 for16 we get the results of (20) and (21).

Again changing n to 12n in (29), and subtracting half of the result from (29), we have

1

(n + 1)2 1

(n + 2)2+

1

(n+ 3)2 = 1

2n2 (2

2 1)B2n3

+(24 1)B4

n5

(26 1)B6

n7 + + (1)p(22p 1)B2p

n2p+1, (30)

where 0<

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where 0<

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(ii) c3

1 12n4

Bn3 c9

1 1

2n10

Bn9+ c15

1 1

2n16

Bn15

= 23 n

2n{312 (n1) + (1)16 (n3) or (1) 16 (n+1) or (1)16 (n+5)}. (37)

17. The formul obtained in1, 3, 4 may be called the one interval, two interval andthree interval formula respectively. The p interval formul can be got by taking the pthroots of unity or ofi according as p is odd or even.For example, let us take the fifth roots of unity (1, , 2, 3, 4), and find the 5 intervalformul.

Let

(x) = sin x + sin x + sin x2 + sin x3 + sin x4

= 5

x5

5! x

15

15! +

x25

25!

.

Then it can easily be shewn that

16sin x sin x sin x2 sin x3 sin x4 = (2x) {2x(+ 4)} {2x(2 + 3)}= (2x) + {x(1 +

5)}+ {x(1

5)}.

Taking logarithms and differentiating both sides, we have

5

B0+ B10

x10

10!+ B20

x20

20!+ B30

x30

30!+

=x

x4

4!(1 + 5) x

14

14!(1 + 15) +

x5

5!(1 + 5) x

15

15!(1 + 15) +

, (38)

where

n =

1 +

5

2

n+

1 5

2

n, so that nm= n+m+ (1)nmn.

Similarly

(i) 5

B2x2

2! + B12x12

12! + B22x22

22! + =xx6

6!(1 + 7)

x16

16!(1 + 17) +

x55!

(1 + 5) x15

15!(1 + 15) +

, (39)

(ii) 5

B4

x4

4! + B14

x14

14!+ B24

x24

24!+

=x

x8

8!(71) x

18

18!(171) +

x5

5!(1 + 5) x

15

15!(1 + 15) +

, (40)

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(iii) 10

B6

x6

6! + B16

x16

16! + B26

x26

26!+

=x

x10

10!(103) x

20

20!(203) + x

30

30!(303)

x5

5!(1 + 5) x

15

15!(1 + 15) +

x25

25!(1 + 25)

, (41)

(iv) 5

B8

x8

8! + B18

x18

18!+ B28

x28

28!+

=x

x12

12!(111) x

22

22!(211) + x

32

32!(311)

x5

5!(1 + 5)x15

15! (1 + 15) +x25

25! (1 + 25) . (42)

Again from

16cos x cos x cos x2 cos x3 cos x4

= 1 + [2x] + [2x(+ 4)] + [2x(2 + 3)]

where

(x) = cos x + cos x + cos x2 + cos 3 + cos x4

= 5

1x10

10! +x20

20!x30

30! +

,

and similar relations, we can get many other identities.

18. The four interval formul can be got from the following identities: If

an =

1 +

12

n+

1 1

2

nand bn=

1 +

12

n

1 12

n,

so that am+n = aman amn/2n; and bm+n = ambn+ bmn/2n; then:

(i) 4

B0+ B8x8

8! + B16

x16

16! + B24

x24

24! +

=xx3

3!a2 x

11

11!a6+

x19

19!a10

x4

4!a2 x

12

12!a6+

x20

20!a10

, (43)

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(ii) 4

B2

x2

2! + B10

x10

10!+ B18

x18

18!+

=xx5

5!a3 x

13

13!a7+

x21

21!a11

x4

4!a2 x

12

12!a6+

x20

20!a10

, (44)

(iii) 4

2

B4

x4

4! + B12

x12

12! + B20

x20

20!+

=x

x7

7!b3x

15

15!b7+

x23

23!b11

x4

4!a2x12

12! a6+

x20

20! a10

, (45)

(iv) 4

2

B6

x6

6! + B14

x14

14! + B22

x22

22!+

=x

x9

9!b4x

17

17!b8+

x25

25!b12

x4

4!a2 x

12

12!a6+

x20

20!a10

. (46)

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