arithmetic sequence

2. Multiplicative Functions
• An arithmetical function, or 'number-theoretic function' is a complex-
valued function defined for all positive integers. It can be viewed as a
sequence of complex numbers.
• Examples: n!,ϕ(n),π(n) which denotes the number of primes less than
or equal to n.
• An arithmetical function
is multiplicative if f(mn)=f(m)f(n) whenever gcd(m,n)=1, and totally
multiplicative or completely multiplicative if this holds for any m,n.
Thus f(1)=1 unless f is the zero function, and a multiplicative function
is completely determined by its behaviour on the prime powers.

3. Examples: We have seen that the Euler totient
function ϕ is multiplicative but not totally multiplicative (this
is one reason it is convenient to have ϕ(1)=1). The
function n2 is totally multiplicative. The product of (totally)
multiplicative functions is (totally) multiplicative.

4. • Theorem: Let f(n) be a multiplicative function, and define
• F(n)=∑d|nf(d).
• Then F(n) is also a multiplicative function.

5. • We start by defining the Mobius function which investigates integers
in terms of their prime decomposition. We then determine the
Mobius inversion formula which determines the values of the a
function f at a given integer in terms of its summatory function.

6. MOEBIUS FUNCTION
• IT IS DENOTED AS “μ” and defined as follows:
μ (n) = 1 if n=1
0 if n has a square factor > 1
−1𝐾
𝑖𝑓 𝑛 > 1
n = 𝑃1
𝑑1
𝑃1
𝑑2
… … . 𝑃𝑘
𝑑𝑘
and 𝑑1 = 𝑑2 = ….. 𝑑𝑘 = 1

7. MOEBIUS INVERSION FORMULA
f(n) = 𝒅/𝒏 μ (𝒅)𝑭
𝒏
𝒅
n 1 2 3 4 5 6
μ (n) 1 -1 -1 0 -1 1

8. EULER PHI FUNCTION
• It is a function of a positive integer n is the number of integer
(1,2,3,,,n) which are relatively prime to n.

9. EULER PHI FUNCTION FORMULA
• PRIME NUMBERS: Ø(P)= P – 1
• COMPOSITE NUMBER : Ø(𝑃𝑎)= 𝑃𝑎 - 𝑃𝑎−1

10. CARMICHAEL CONJECTURE
It takes as input an integer, n. The function is written as a Lambda: 𝝀(n).
It’s output, m, is the lowest value such that every integer coprime to n, between 1 & n, when raised to the power m, and mod n, gives 1.
n = 8
𝝀(8) = ?> Find all coprimes of n, between 1 & n
1, 3, 5, 7> We need to find the smallest value of 'm' which satisfies...
1ᵐ ≡ 1 (mod 8)
3ᵐ ≡ 1 (mod 8)
5ᵐ ≡ 1 (mod 8)
7ᵐ ≡ 1 (mod 8)> for n = 8, m = 2
1² % 8 = 1
3² % 8 = 1
5² % 8 = 1
7² % 8 = 1𝝀(8) = 2

12. PERFECT NUMBERS
• In number theory, a perfect number is a positive integer that is equal
to the sum of its positive factors, excluding the number itself. The
most popular and the smallest perfect number is 6, which is equal to
the sum of 1, 2, and 3. Other examples of perfect numbers are 28,
496, and 8128.

14. • How to Find a Perfect Number?
• Euclid said that (2n - 1) multiplied by 2n - 1, can be a perfect number if
the term in the bracket, that is, (2n - 1) is a prime number. In other
words, [2n - 1 × (2n - 1) = perfect number], if (2n - 1) is a prime number.

15. n 𝟐𝒏−𝟏 (𝟐𝒏
−𝟏) 𝟐𝒏−𝟏 x (𝟐𝒏−𝟏)
1 1 1 -
2 2 3 (prime number) 6 (perfect number)
3 4 7 (prime number) 28 (perfect number)
4 8 15 -
5 16 31 (prime number) 496 (perfect number)
6 32 63 -
7 64 127 (prime number) 8128 (perfect number)
8 128 255 -
9 256 511 -
10 512 1023 -

16. The Sum-of-Divisors Function
The sum of divisors function, denoted by σ(n)σ(n), is the sum of all positive divisors of n.
σ(12)=1+2+3+4+6+12=28.σ(12)=1+2+3+4+6+12=28.
Note that we can express σ(n)σ(n) as σ(n)=∑d∣ndσ(n)=∑d∣nd.
We now prove that σ(n)σ(n) is a multiplicative function.
The sum of divisors function σ(n)σ(n) is multiplicative.
Let p be a prime and let n=𝑃1
𝑎1
𝑃2
𝑎2
... 𝑃𝑡
𝑎𝑡
be a positive integer. Then
σ(𝑝𝑎)= (𝑝𝑎+1
−1)/𝑝 − 1
(200)=σ(2352)= ((24−1)/(2 − 1) ) ((53−1)/(5 − 1)) =15.31=465.

17. The Number-of-Divisors Function
• The number of divisors function, denoted by τ(n) , is the sum of all
positive divisors of n .
Let p be a prime and let n=𝑃1
𝑎1
𝑃2
𝑎2
... 𝑃𝑡
𝑎𝑡
be a positive integer.
Then
τ (𝑝𝑎
)= a + 1
τ(200)=τ(2352)=(3+1)(2+1)=12