Arithmetic Integral

n d x n x ? \Large \displaystyle\int n\text{ }dx \neq nx?

In calculus, when you take the derivative of a constant you get zero as an answer. In number theory, there is something called the arithmetic derivative which allows you to differentiate a number and get a nonzero answer. The arithmetic derivative works as follows.

Where n n' denotes the arithmetic derivative of n n :

p = 1 p' = 1 for all primes p p

( a b ) = a b + a b (ab)'=a'b+ab'

0 = 1 = 0 0'=1'=0

For example, 6 = ( 2 × 3 ) = ( 2 ) ( 3 ) + ( 2 ) ( 3 ) = ( 1 ) ( 3 ) + ( 2 ) ( 1 ) = 5 6'=(2\times3)'=(2')(3)+(2)(3')=(1)(3)+(2)(1)=5

While you clearly get a single answer when taking the arithmetic derivative of a number, multiple values of n n can lead to the same n n' . Let us define the arithmetic integral, denoted n \int n , as the function that when given a value of n n returns a set of all the positive integers m m such that m = n m'=n . What is the value of n n , where 1 < n < 100 1<n<100 , that gives us the set with the largest dimensions? In other words, what is the value of n n that has the most solutions m m to the equation m = n m'=n ?

This is a member of a set of problems on the Arithmetic Derivative .


The answer is 90.

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2 solutions

Patrick Corn
Jul 9, 2015

I cheated! https://oeis.org/A099302/list

Kazem Sepehrinia
Jul 10, 2015

If n n has a representation in the form p 2 p 3 . . . p k + p 1 p 3 . . . p k + p 1 p 2 . . . p k 1 p_2 p_3 ... p_k+ p_1 p_3 ... p_k + p_1 p_2 ... p_{k-1} for distinct primes p 1 , p 2 , . . . , p k p_1, p_2, ..., p_k and k 2 k \ge 2 , then m = p 1 p 2 . . . p k m=p_1 p_2 ... p_k is a solution to the equation m = n m'=n . I think these kind of solutions are easy to seek and even easier for k = 2 k=2 where n n can be represented in the form n = p + q n=p+q and m = p q m=pq is a solution to m = n m'=n .

For example m = 6 m=6 is a solution to the equation m = 5 m'=5 , because 6 = 2 × 3 + 3 × 2 = 3 + 2 = 5 6'=2' \times 3+ 3'\times 2=3+2=5 Or m = 30 m=30 is an answer for the equation m = 31 m'=31 because 3 0 = 2 × 3 + 2 × 5 + 3 × 5 = 31 30'=2 \times 3+2 \times 5+3 \times 5=31 For n < 100 n<100 the equation m = n m'=n has maximum possible values of answers when n = 90 n=90 , because 90 90 can be represented as sum of two primes in 9 9 ways: 90 = 7 + 83 = 11 + 79 = 17 + 73 = 19 + 71 = 23 + 67 = 29 + 61 = 31 + 59 = 37 + 53 = 43 + 47 \displaystyle \begin{array}{c}\\ 90 &&=7+83=11+79=17+73=19+71 \\ &&=23+67=29+61=31+59=37+53 \\ &&=43+47 \end{array} I think it will be too difficult to find all of solutions to the equation m = n m'=n .

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