Euler's Totient Function

From Wikipedia:

Euler's totient function counts the positive integers up to a given integer n that are relatively prime to n. It is written using the Greek letter phi as φ(n) or ϕ(n), and may also be called Euler's phi function.

Today I turned 25545 days old and one of the points of interest about this number is that it forms a pair with 25546, both having the property that their totient values are the same (12480). The first member of the pairs of numbers with this property up to and including 25545 is given by OEIS A001274: numbers \(n\) such that \( \phi(n) = \phi(n+1) \):

1, 3, 15, 104, 164, 194, 255, 495, 584, 975, 2204, 2625, 2834, 3255, 3705, 5186, 5187, 10604, 11715, 13365, 18315, 22935, 25545

Note that for all prime numbers \(p\), \( \phi(p)=p-1 \). These totient values (totatives) represent the maximum possible for numbers up to \(p\) and their points lie on the maximal line as shown in Figure 1:

Figure 1: generated using SageMathCell using
plot(lambda x:euler_phi(int(x)), (x,1,100))

From Figure 1 it can be seen that the totient values (totatives) for 1 and 2, 3 and 4, 15 and 16 are the same. Note that 5186 and 5187 themselves form a pair, meaning that the totient values of 5186, 5187 and 5188 are the same (2592). This is the only triplet that occurs for numbers up to \( 10^{13} \).

Euler's totient function is a multiplicative function, meaning that if two numbers \(m\) and \(n\) are relatively prime, then \( \phi(m \times n) = \phi(m) \times \phi(n) \). This is important in developing Euler's product formula along with the fact that: $$\phi(p^k)=p^{k-1} \bigg (1-\frac{1}{p} \bigg )$$The product formula is: $$\phi(n)=n \prod_{p|n} \bigg (1-\frac{1}{p} \bigg )$$Applying this to today's number 25545=3 * 5 * 13 * 131, we note that:$$ \phi(25545) = 25545\bigg (1-\frac{1}{3} \bigg ) \bigg (1-\frac{1}{5} \bigg ) \bigg (1-\frac{1}{13} \bigg ) \bigg (1-\frac{1}{131} \bigg )=12480$$