aarjaneiro/planetmath huggingface dataset:
proof of all positive integers are polite numbers except powers of two

proof of all positive integers are polite numbers except powers of two

Theorem

.

All

positive integers

are

polite numbers

(that is, can be expressed as the sum of consecutive

nonnegative integers

in at least one way), with the exception of the

powers of two

.

Proof. Let k be a positive integer.

Let

k

=

2

h

for

h

>

0

. Let’s suppose that

k

is a polite number. We can write

k

=

∑

i

=

a

b

i

with

b

>

a

and

a

>

-

1

.

Then,

k

=

2

h

=

∑

i

=

a

b

i

=

∑

j

=

0

b

-

a

(

j

+

a

)

=

∑

j

=

0

b

-

a

j

+

∑

j

=

0

b

-

a

a

=

(

b

-

a

)

⁢

(

b

-

a

+

1

)

2

+

(

b

-

a

+

1

)

⁢

a

Multiplying both sides by two,

2

h

+

1

=

(

b

-

a

)

⁢

(

b

-

a

+

1

)

+

2

⁢

a

⁢

(

b

-

a

+

1

)

=

(

b

-

a

+

1

)

⁢

(

b

-

a

-

2

⁢

a

)

If

b

-

a

is even, then

b

-

a

+

1

is odd and then

2

h

+

1

would have an odd factor,

contradiction

.
If

b

-

a

is odd, then

b

-

a

-

2

⁢

a

is odd and then

2

h

+

1

would have an odd factor, contradiction.
Then,

k

is not a polite number.

If

k

=

1

, then

k

is a polite number trivially.

Now, suppose that

k

>

1

is not a power of two. Then we can factor

k

as

a

⁢

h

with

h

=

2

⁢

b

+

1

>

1

odd,

a

>

0

. Then,

∑

i

=

a

-

b

a

+

b

i

=

∑

j

=

-

b

b

(

j

+

a

)

=

∑

j

=

-

b

b

j

+

∑

j

=

-

b

b

a

=

0

+

(

2

⁢

b

+

1

)

⁢

a

=

k

If

a

-

b

≥

0

then

k

is clearly a polite number.

If

a

-

b

<

0

then

k

=

∑

i

=

a

-

b

a

+

b

i

=

∑

i

=

a

-

b

0

i

+

∑

i

=

1

|

a

-

b

|

i

+

∑

|

a

-

b

|

+

1

a

+

b

i

=

∑

|

a

-

b

|

+

1

a

+

b

i

so

k

is a polite number.

Title

proof of all positive integers are polite numbers except powers of two

Canonical name

ProofOfAllPositiveIntegersArePoliteNumbersExceptPowersOfTwo

Date of creation

2013-03-22 18:42:47

Last modified on

2013-03-22 18:42:47

Owner

n847530 (22696)

Last modified by

n847530 (22696)

Numerical id

8

Author

n847530 (22696)

Entry type

Proof

Classification

msc 11A25

Generated on Fri Feb 9 15:14:26 2018 by

LaTeXML
proof of square root of square root binomial

proof of square root of square root binomial

We square the expression on the right-hand-side and expand
using the

binomial formula

:

(

a

+

a

2

-

b

2

±

a

-

a

2

-

b

2

)

2

=

(

a

+

a

2

-

b

2

)

2

+

(

a

-

a

2

-

b

2

)

2

±

2

⁢

a

+

a

2

-

b

2

⁢

a

-

a

2

-

b

2

Since the squaring operation undoes the

square roots

, we
obtain the following:

(

a

+

a

2

-

b

2

)

2

+

(

a

-

a

2

-

b

2

)

2

=

a

+

a

2

-

b

2

+

a

-

a

2

-

b

2

=

a

Since the product of square roots equals the square root
of the product, we have the following:

a

+

a

2

-

b

2

⁢

a

-

a

2

-

b

2

=

a

+

a

2

-

b

2

⋅

a

-

a

2

-

b

2

=

a

2

-

(

a

2

-

b

)

2

4

=

a

2

-

(

a

2

-

b

)

4

=

b

4

=

b

2

Combining what we have calculated above, we obtain

(

a

+

a

2

-

b

2

±

a

-

a

2

-

b

2

)

2

=

a

±

b

.

Because the square of the asserted value of the square root equals the radicand (

a

±

b

) of the square root, and the asserted value of the square root is clearly non-negative, we have justified the validity of the formulas

a

±

b

=

a

+

a

2

-

b

2

±

a

-

a

2

-

b

2

.

Title

proof of square root of square root binomial

Canonical name

ProofOfSquareRootOfSquareRootBinomial

Date of creation

2013-03-22 17:42:45

Last modified on

2013-03-22 17:42:45

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

5

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 11A25

Generated on Fri Feb 9 15:14:28 2018 by

LaTeXML
proof of the correspondence between even 2-superperfect numbers and Mersenne primes

proof of the correspondence between even 2-superperfect numbers and Mersenne primes

Statement. Among the

even numbers

, only

powers of two

2

x

(with

x

being a

nonnegative integer

) can be 2-superperfect numbers (

http://planetmath.org/SuperperfectNumber

), and then if and only if

2

x

+

1

-

1

is a

Mersenne prime

. (The default multiplier

m

=

2

is tacitly assumed from this point forward).

Proof. The only

divisors

of

n

=

2

x

are smaller powers of 2 and itself,

1

,

2

,

…

,

2

x

-

1

,

2

x

. Therefore, the first iteration of the

sum of divisors function

is

σ

⁢

(

n

)

=

∑

i

=

0

x

2

i

=

2

x

+

1

-

1

=

2

⁢

n

-

1

.

If

2

⁢

n

-

1

is prime, that means its only other divisor is 1, and thus for the second iteration

σ

⁢

(

2

⁢

n

-

1

)

=

2

⁢

n

, and is thus a 2-superperfect number. But if

2

⁢

n

-

1

is

composite

then it is clear that

σ

⁢

(

2

⁢

n

-

1

)

>

2

⁢

n

by at least 2. So, for example,

σ

⁢

(

8

)

=

15

and

σ

2

⁢

(

8

)

=

24

, so 8 is not 2-superperfect. One more example:

σ

⁢

(

16

)

=

31

and since 31 is prime,

σ

2

⁢

(

16

)

=

32

.

Now it only remains to prove that no other even number

n

can be 2-superperfect. Any other even number can of course still be

divisible

by one or more powers of two, but it also must be divisible by some

odd prime

p

>

2

. Since the sum of divisors function is a

multiplicative function

, it follows that if

n

=

2

x

⁢

p

then

σ

⁢

(

n

)

=

σ

⁢

(

2

x

)

⁢

σ

⁢

(

p

)

. So, if, say,

p

=

3

, it is clear that

(

2

x

+

3

-

4

)

>

2

x

+

1

⁢

3

, and that on the second iteration this value that already exceeded twice the original value will be even greater. For example,

12

=

2

2

⁢

3

, and

σ

⁢

(

12

)

=

2

5

-

4

which is greater than

2

3

⁢

3

by 4. With any larger

p

the excess will be much greater.

Title

proof of the correspondence between even 2-superperfect numbers and Mersenne primes

Canonical name

ProofOfTheCorrespondenceBetweenEven2superperfectNumbersAndMersennePrimes

Date of creation

2013-03-22 17:03:48

Last modified on

2013-03-22 17:03:48

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

5

Author

PrimeFan (13766)

Entry type

Proof

Classification

msc 11A25

Generated on Fri Feb 9 15:14:30 2018 by

LaTeXML
proof that 3 is the only prime perfect totient number

proof that 3 is the only prime perfect totient number

Given a

prime number

p

, only

p

=

3

satisfies the equation

p

=

∑

i

=

1

c

+

1

ϕ

i

⁢

(

n

)

,

where

ϕ

i

⁢

(

x

)

is the

iterated totient function

and

c

is the

integer

such that

ϕ

c

⁢

(

n

)

=

2

. That is, 3 is the only

perfect totient number

that is prime.

The first four primes are most easily examined empirically. Since

ϕ

⁢

(

2

)

=

1

, 2 is deficient

totient

number.

ϕ

⁢

(

3

)

=

2

, so, per the previous remark, it is a perfect totient number. For 5, the

iterates

are 4, 2 and 1, adding up to 7, hence 5 is an abundant totient number. The same goes for 7, with its iterates being 7, 6, 2, 1.

It is for

p

>

7

that we can avail ourselves of the

inequality

ϕ

⁢

(

n

)

>

n

(true for all

n

>

6

). It is obvious that

ϕ

⁢

(

p

)

=

p

-

1

, and by the foregoing,

ϕ

2

⁢

(

p

)

>

3.162278

(that is, it is sure to be more than the

square root

of 10), so it follows that

ϕ

⁢

(

p

)

+

ϕ

2

⁢

(

p

)

>

p

+

2.162278

and thus it is not necessary to examine any further iterates to see that all such primes are abundant totient numbers.

Title

proof that 3 is the only prime perfect totient number

Canonical name

ProofThat3IsTheOnlyPrimePerfectTotientNumber

Date of creation

2013-03-22 16:34:29

Last modified on

2013-03-22 16:34:29

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

5

Author

PrimeFan (13766)

Entry type

Proof

Classification

msc 11A25

Generated on Fri Feb 9 15:14:32 2018 by

LaTeXML
proof that all powers of 3 are perfect totient numbers

proof that all powers of 3 are perfect totient numbers

Given an

integer

x

>

0

, it is always the case that

3

x

=

∑

i

=

1

c

+

1

ϕ

i

⁢

(

3

x

)

,

where

ϕ

i

⁢

(

x

)

is the

iterated totient function

and

c

is the integer such that

ϕ

c

⁢

(

n

)

=

2

. That is, all integer powers of three are

perfect totient numbers

.

The proof of this is easy and even considered trivial. Here it goes anyway:

Accepting as proven that

ϕ

⁢

(

p

x

)

=

(

p

-

1

)

⁢

p

x

-

1

, we can plug in

p

=

3

and see that

ϕ

⁢

(

3

x

)

=

2

⁢

(

3

x

-

1

)

, which falls short of

3

x

by

3

x

-

1

. Given the proof that Euler

ϕ

is a

multiplicative function

(

http://planetmath.org/ProofThatEulerPhiIsAMultiplicativeFunction

) and the fact that

ϕ

⁢

(

2

)

=

1

, it is obvious that

ϕ

⁢

(

2

⁢

(

3

x

-

1

)

)

=

ϕ

⁢

(

3

x

-

1

)

. Therefore, each

iterate

will be twice a power of three, with the

exponent

gradually

decreasing

as the iterator nears

c

. To put it algebraically,

ϕ

i

⁢

(

3

x

)

=

2

⁢

(

3

c

-

i

)

for

i

≤

c

. Adding up in

ascending order

starting at the

(

c

+

1

)

th iterate, we obtain

1

+

2

+

6

+

18

+

⋯

+

2

⁢

(

3

x

-

2

)

+

2

⁢

(

3

x

-

1

)

=

3

x

.

References

1

D. E. Ianucci, D. Moujie & G. L. Cohen, “On Perfect Totient Numbers”

Journal of

Integer Sequences

,

6

, 2003: 03.4.5

Title

proof that all powers of 3 are perfect totient numbers

Canonical name

ProofThatAllPowersOf3ArePerfectTotientNumbers

Date of creation

2013-03-22 16:34:32

Last modified on

2013-03-22 16:34:32

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

11

Author

PrimeFan (13766)

Entry type

Proof

Classification

msc 11A25

Generated on Fri Feb 9 15:14:34 2018 by

LaTeXML
proof that Euler φ function is multiplicative

proof that Euler

φ

function is multiplicative

Suppose that

t

=

m

⁢

n

where

m

,

n

are

coprime

. The

Chinese remainder theorem

(

http://planetmath.org/ChineseRemainderTheorem

) states that

gcd

⁡

(

a

,

t

)

=

1

if and only if

gcd

⁡

(

a

,

m

)

=

1

and

gcd

⁡

(

a

,

n

)

=

1

.

In other words, there is a

bijective

correspondence between these two sets:

•

{

a

:

a

≡

1

(

mod

t

)

}

•

{

a

:

a

≡

1

(

mod

m

)

and

a

≡

1

(

mod

n

)

}

Now the number of

positive integers

not greater than

t

and coprime with

t

is precisely

φ

⁢

(

t

)

, but it is also the number of pairs

(

u

,

v

)

, where

u

not greater than

m

and coprime with

m

, and

v

not greater than

n

and coprime with

n

. Thus,

φ

⁢

(

m

⁢

n

)

=

φ

⁢

(

m

)

⁢

φ

⁢

(

n

)

.

Title

proof that Euler

φ

function

is

multiplicative

Canonical name

ProofThatEulervarphiFunctionIsMultiplicative

Date of creation

2013-03-22 15:03:40

Last modified on

2013-03-22 15:03:40

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

12

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 11A25

Related topic

EulerPhiFunction

Related topic

MultiplicativeFunction

Related topic

EulerPhifunction

Generated on Fri Feb 9 15:14:35 2018 by

LaTeXML
proof that the sum of the iterated totient function is always odd

proof that the sum of the iterated totient function is always odd

Given a

positive integer

n

, it is always the case that

2

|̸

∑

i

=

1

c

+

1

ϕ

i

(

n

)

,

where

ϕ

i

⁢

(

x

)

is the

iterated totient function

and

c

is the

integer

such that

ϕ

c

⁢

(

n

)

=

2

.

Accepting as proven that

n

>

ϕ

⁢

(

n

)

and

2

|

ϕ

(

n

)

for

n

>

2

, it is clear that

summing

up the

iterates

of the

totient function

up to

c

is summing up a series of

even numbers

in

descending order

and that this sum is therefore itself even. Then, when we add the

c

+

1

iterate, the sum turns odd.

As a bonus, this proves that no even number can be a

perfect totient number

.

Title

proof that the sum of the iterated totient function is always odd

Canonical name

ProofThatTheSumOfTheIteratedTotientFunctionIsAlwaysOdd

Date of creation

2013-03-22 16:34:26

Last modified on

2013-03-22 16:34:26

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

6

Author

PrimeFan (13766)

Entry type

Proof

Classification

msc 11A25

Generated on Fri Feb 9 15:14:37 2018 by

LaTeXML
proof that τ⁢(n) is the number of positive divisors of n

proof that

τ

⁢

(

n

)

is the number of positive divisors of

n

The following is a proof that

τ

counts the

positive

divisors

of its input (which must be a

positive integer

).

Proof.

Recall that

τ

behaves according to the following two rules:

1.

If

p

is a prime and

k

is a

nonnegative integer

, then

τ

⁢

(

p

k

)

=

k

+

1

.

2.

If

gcd

⁡

(

a

,

b

)

=

1

, then

τ

⁢

(

a

⁢

b

)

=

τ

⁢

(

a

)

⁢

τ

⁢

(

b

)

.

Let

p

be a prime. Then

p

0

=

1

. Note that 1 is the only positive divisor of 1 and

τ

⁢

(

1

)

=

τ

⁢

(

p

0

)

=

0

+

1

=

1

.

Suppose that, for all positive integers

m

smaller than

z

∈

ℤ

with

z

>

1

, the number of positive divisors of

m

is

τ

⁢

(

m

)

. Since

z

>

1

, there exists a

prime divisor

p

of

z

. Let

k

be a positive integer such that

p

k

exactly divides

z

. Let

a

be a positive integer such that

z

=

p

k

⁢

a

. Then

gcd

⁡

(

a

,

p

)

=

1

. Thus,

gcd

⁡

(

a

,

p

k

)

=

1

. Since

1

≤

a

<

z

, by the

induction hypothesis

, there are

τ

⁢

(

a

)

positive divisors of

a

.

Let

d

be a positive divisor of

z

. Let

y

be a nonnegative integer such that

p

y

exactly divides

d

. Thus,

0

≤

y

≤

x

, and there are

k

+

1

choices for

y

. Let

c

be a positive integer such that

d

=

p

y

⁢

c

. Then

gcd

⁡

(

c

,

p

)

=

1

. Since

c

divides

d

and

d

divides

z

, we conclude that

c

divides

z

. Since

c

divides

p

k

⁢

a

and

gcd

⁡

(

c

,

p

)

=

1

, it must be the case that

c

divides

a

. Thus, there are

τ

⁢

(

a

)

choices for

c

. Since there are

k

+

1

choices for

y

and there are

τ

⁢

(

a

)

choices for

c

, there are

(

k

+

1

)

⁢

τ

⁢

(

a

)

choices for

d

. Hence, there are

(

k

+

1

)

⁢

τ

⁢

(

a

)

positive divisors of

z

. Since

τ

⁢

(

z

)

=

τ

⁢

(

p

k

⁢

a

)

=

τ

⁢

(

p

k

)

⁢

τ

⁢

(

a

)

=

(

k

+

1

)

⁢

τ

⁢

(

a

)

, it follows that, for every positive integer

n

, the number of positive divisors of

n

is

τ

⁢

(

n

)

.
∎

Title

proof that

τ

⁢

(

n

)

is the number of positive divisors of

n

Canonical

name

ProofThattaunIsTheNumberOfPositiveDivisorsOfN

Date of creation

2013-03-22 13:30:24

Last modified on

2013-03-22 13:30:24

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

11

Author

Wkbj79 (1863)

Entry type

Proof

Classification

msc 11A25

Generated on Fri Feb 9 15:14:40 2018 by

LaTeXML
Rare numbers

Rare numbers

Rare numbers

are the non

palindromic numbers

when added or subtracted to its reverse gives a

perfect square

. For example 65 is a rare number because 65 + 56 = 11¡sup¿2¡/sup¿ and 65 - 56 = 3¡sup¿2¡/sup¿. Other examples of rare numbers are 65, 621770, 281089082, 2022652202, 868591084757, 872546974178 … (

Sequence

A035519 of OEIS). If we consider palindromic rare numbers, there are infinitely many rare numbers. For example, the numbers in the series 242, 20402, 2004002, 200040002, 20000400002 … are palindromic rare numbers. There are 84 rare numbers less than 10¡sup¿20¡/sup¿.

Rare numbers exhibit certain properties. Rare numbers start always with an even digit and end in the digits 0, 2, 3, 7 and 8.

Digital root

of a rare number is always 2, 5, 8 and 9. Odd rare numbers are much fewer than even rare numbers. Also rare numbers with

odd number

of digits is fewer than rare numbers with

even number

of digits. After investigating rare numbers upto 10¡sup¿20¡/sup¿, Shyam Sunder Gupta conjectured that there are no

prime numbers

. Also it is not known whether there are infinitely many rare numbers.

Title

Rare numbers

Canonical name

RareNumbers

Date of creation

2013-03-22 19:20:33

Last modified on

2013-03-22 19:20:33

Owner

Kausthub (26471)

Last modified by

Kausthub (26471)

Numerical id

4

Author

Kausthub (26471)

Entry type

Definition

Classification

msc 11A25

Defines

Numbers when added or subtracted to its reverse gives a perfect square

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refactorable number

refactorable number

A

refactorable number

or

tau number

is an

integer

n

that is

divisible

by the count of its

divisors

, or to put it algebraically,

n

is such that

τ

(

n

)

|

n

, with

τ

⁢

(

n

)

being the

divisor function

. The first few refactorable numbers are listed in the OEIS A033950 are 1, 2, 8, 9, 12, 18, 24, 36, 40, 56, 60, 72, 80, 84, 88, 96. The equation

gcd

⁡

(

n

,

x

)

=

τ

⁢

(

n

)

has solutions only if

n

is a refactorable number.

Title

refactorable number

Canonical name

RefactorableNumber

Date of creation

2013-03-22 17:40:41

Last modified on

2013-03-22 17:40:41

Owner

CompositeFan (12809)

Last modified by

CompositeFan (12809)

Numerical id

4

Author

CompositeFan (12809)

Entry type

Definition

Classification

msc 11A25

Generated on Fri Feb 9 15:14:43 2018 by

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relationship between totatives and divisors

relationship between totatives and divisors

Theorem 1

.

Let

n

be a

positive integer

and define the sets

I

n

,

D

n

, and

T

n

as follows:

•

I

n

=

{

m

∈

ℤ

:

1

≤

m

≤

n

}

•

D

n

=

{

d

∈

I

n

:

d

>

1

and

d

|

n

}

•

T

n

=

{

t

∈

I

n

:

t

is a

totative

of

n

}

Then

D

n

∪

T

n

=

I

n

if and only if

n

=

1

,

n

=

4

, or

n

is prime.

Proof.

Necessity

:

If

n

=

1

, then

D

n

=

∅

and

T

n

=

{

1

}

. Thus,

D

n

∪

T

n

=

I

n

.

If

n

=

4

, then

D

n

=

{

2

,

4

}

and

T

n

=

{

1

,

3

}

. Thus,

D

n

∪

T

n

=

I

n

.

If

n

is prime, then

D

n

=

{

n

}

and

T

n

=

I

n

∖

{

n

}

. Thus,

D

n

∪

T

n

=

I

n

.

Sufficiency:

This will be proven by considering its

contrapositive

.

Suppose first that

n

is a power of

2

. Then

n

≥

8

. Thus,

6

∈

I

n

. On the other hand,

6

is neither a totative of

n

(since

gcd

⁡

(

6

,

n

)

=

2

) nor a

divisor

of

n

(since

n

is a power of

2

). Hence,

D

n

∪

T

n

≠

I

n

.

Now suppose that

n

is even and is not a power of

2

. Let

k

be a positive integer such that

2

k

exactly divides

n

. Since

n

is not a power of

2

, it must be the case that

n

=

2

k

⁢

r

for some

odd integer

r

≥

3

. Thus,

n

=

2

k

⁢

r

>

2

k

+

1

. Therefore,

2

k

+

1

∈

I

n

. On the other hand,

2

k

+

1

is neither a totative of

n

(since

n

is even) nor a divisor of

n

(since

2

k

exactly divides

n

). Hence,

D

n

∪

T

n

≠

I

n

.

Finally, suppose that

n

is odd. Let

p

be the smallest

prime divisor

of

n

. Since

n

is not prime, it must be the case that

n

=

p

⁢

s

for some odd integer

s

≥

3

. Thus,

n

=

p

⁢

s

>

2

⁢

p

. Therefore,

2

⁢

p

∈

I

n

. On the other hand,

2

⁢

p

is neither a totative of

n

(since

gcd

⁡

(

2

⁢

p

,

n

)

=

p

) nor a divisor of

n

(since

n

is odd). Hence,

D

n

∪

T

n

≠

I

n

.
∎

Title

relationship between totatives and divisors

Canonical name

RelationshipBetweenTotativesAndDivisors

Date of creation

2013-03-22 17:09:15

Last modified on

2013-03-22 17:09:15

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

15

Author

Wkbj79 (1863)

Entry type

Theorem

Classification

msc 11A25

Generated on Fri Feb 9 15:14:46 2018 by

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Robin’s theorem

Robin’s theorem

Let

σ

⁢

(

n

)

be the sum of the

positive

divisors

of an

integer

n

and

G

(

n

)

:=

σ

⁢

(

n

)

n

⁢

ln

⁡

(

ln

⁡

n

)

(

n

=

2

,

3

,

4

,

…

)

.

The

Riemann Hypothesis

is true if and only if

G

⁢

(

n

)

<

e

γ

for all

⁢

n

>

7

!

where

γ

is the Euler–Mascheroni constant.

References

1

G. Robin

: Grandes valeurs de la fonction somme des diviseurs et hypothèse de
Riemann.

-

J. Math. Pures Appl.

14

(1984) 187–213.

2

J. C. Lagarias

: An

elementary

problem

equivalent

to the Riemann hypothesis.

-

Amer. Math. Monthly

109

(2002), 534–543. Available at http://arxiv.org/abs/math/0008177

Title

Robin’s

theorem

Canonical name

RobinsTheorem

Date of creation

2013-03-22 19:33:47

Last modified on

2013-03-22 19:33:47

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 11A25

Classification

msc 11M26

Related topic

GronwallsTheorem

Related topic

PropertiesOfXiFunction

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sparsely totient number

sparsely totient number

A

sparsely totient number

is the largest

integer

with a given

totient

. That is, given an sparsely totient number

n

, the inequality

ϕ

⁢

(

m

)

>

ϕ

⁢

(

n

)

(with

ϕ

⁢

(

x

)

being Euler’s

totient function

) is true for any

m

>

n

. For example,

ϕ

⁢

(

12

)

=

4

. We can then verify that

ϕ

⁢

(

13

)

=

12

,

ϕ

⁢

(

14

)

=

6

and

ϕ

⁢

(

15

)

=

8

. Accepting as true that the inequality

ϕ

⁢

(

n

)

>

n

holds for all

n

>

6

, we don’t need to check any larger numbers to confirm that 12 is the largest integer with 4 as its totient.

The first few sparsely totient numbers are 2, 6, 12, 18, 30, 42, 60, 66, 90, 120, 126, 150, 210, 240, 270, 330, 420, 462, 510, 630, 660, 690, 840, 870, etc., listed in A036913 of Sloane’s OEIS.

All sparsely totient numbers are even. In 1986, Masser and Shiu proved that the

i

th

primorial

multiplied by the

i

th prime (for

i

>

1

, thus: 18, 150, 1470, 25410, 390390, etc.) is a sparsely totient number.

References

1

Roger C. Baker & Glyn Harman, “Sparsely totient numbers,”

Annales de la faculte des sciences de Toulouse Ser. 6

5

no. 2 (1996): 183 - 190

2

D. W. Masser & P. Shiu, “On sparsely totient numbers,”

Pacific J. Math.

121

, no. 2 (1986): 407 - 426.

Title

sparsely totient number

Canonical

name

SparselyTotientNumber

Date of creation

2013-03-22 16:48:59

Last modified on

2013-03-22 16:48:59

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

4

Author

PrimeFan (13766)

Entry type

Definition

Classification

msc 11A25

Generated on Fri Feb 9 15:14:49 2018 by

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square root of 2

square root of 2

The

square root of 2

is an

irrational number

, the first to have been proved irrational. Its

decimal expansion

begins 1.41421356237309504880168872420969807856… (

sequence

http://www.research.att.com/ njas/sequences/A002193

A002194 in Sloane’s OEIS) Its

simple continued fraction

is

1

+

1

2

+

1

2

+

1

2

+

1

2

+

…

,

periodically repeating the 2. Some call this number

Pythagoras’ constant

.

There are several different ways to express

2

as an infinite product. One way is

2

=

∏

i

=

0

∞

(

4

⁢

i

+

2

)

2

(

4

⁢

i

+

1

)

⁢

(

4

⁢

i

+

3

)

,

another is

2

=

∑

i

=

0

∞

(

-

1

)

i

+

1

⁢

(

2

⁢

i

-

3

)

!!

(

2

⁢

i

)

!!

.

References

1

Flannery, David.

The square root of 2 : a dialogue concerning a number and a sequence

. New York: Copernicus, 2006.

Title

square root of 2

Canonical name

SquareRootOf2

Date of creation

2013-03-22 17:29:12

Last modified on

2013-03-22 17:29:12

Owner

MathNerd (17818)

Last modified by

MathNerd (17818)

Numerical id

10

Author

MathNerd (17818)

Entry type

Definition

Classification

msc 11A25

Synonym

Pythagoras’ constant

Related topic

Surd

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