Integral Fusion

Given any integer $n \gt 1$ a binary factor tree $T(n)$ is defined to be:

A tree with the single node $n$ when $n$ is prime.
A binary tree that has root node $n$, left subtree $T(a)$ and right subtree $T(b)$, when $n$ is not prime. Here $a$ and $b$ are positive integers such that $n = ab$, $a\le b$ and $b-a$ is the smallest.

For example $T(20)$:

We define $M(n)$ to be the smallest number that has a factor tree identical in shape to the factor tree for $n!!$, the double factorial of $n$.
For example, consider $9!! = 9\times 7\times 5\times 3\times 1 = 945$. The factor tree for $945$ is shown below together with the factor tree for $72$ which is the smallest number that has a factor tree of the same shape. Hence $M(9) = 72$.

Find $\displaystyle\sum_{n=2}^{31} M(n)$.

This problem requires a lot of recursive number crunching, identification of patterns and knowledge in prime number factorizations.

First, let’s understand what a double factorial is. For an integer n, it is the product of the integers from 1 to that integer with the same parity (odd/even) as n. So, for example, 9!! = 9*7*5*3*1 = 945.

In this question, a Binary Factor Tree (BFT) is a binary tree that represents the process of factoring a number into prime factors. A node n in the BFT is a prime number if it cannot be factored further and thus has no children. If it’s not a prime (i.e., composite), it has two children where one of them is the smallest prime factor (spf) of n and the other is n/spf.

The concept in the problem is to compare the factor tree of the double factorial of n to another number and find the smallest such number whose factor tree has the exact same structure/shape. That minimum number is denoted as M(n).

The task here is to calculate the sum of M(n) for all numbers from 2 to 31. To achieve this, it requires depth understanding in generating BFT for n!! and identifying the patterns for the trees.

As an example, let’s compute M(9):

9!! = 945 = 3^3 * 5 * 7. To form a BFT for 945, we start from the root i.e., 945, and at each stage, we divide by smallest prime factor to form the next level, until the resulting factors are primes. This gives us a three-level tree of the form (x*(x*y)) where x and y are some primes.

The smallest number that gives a similar BFT would be 2 * (2*3) = 12 but 945 has two nodes at the 3rd level (2nd level from the root). Hence, the smallest number with similar BFT is to duplicate this prime 3 to get 2 * (2 * 2) = 2^3 = 8. But it isn’t a match either, because 945 has a node with value 7 at the second level from the root. Therefore, explore the same shape but with 7 at that place and get 2 * (7*2) = 28. But it isn’t a match either. Therefore, explore further to end up with 2 * (2 * 2^2) = 8 * 2 = 16. But it causes mismatch again at level 2, so we upgrade 2 to 3 at that place and find 2 * (3 * 2^2) = 24. Finally, it gives another mismatch at level 3 and after upgrading 2 to 3 at rightmost place, we find a match with 2 * (3 * 3*2) = 36. It is still not a match for 945’s BFT. 36 has a BFT of four levels which is more than BFT levels of 945. Hence, go back to the case where there was a match i.e., 2 * (2 * 2) = 8 but it had lower number of nodes than 945. So, we go on and add nodes to it to match 945’s BFT. By doing this, we finally find a match with 2 * (3 * 2^2) = 24.

Note: This example computation was to present how to find a solution to the problem. It may have appeared longer but once you understand the pattern, computation gets easier. If we directly applied the rules derived from patterns noticed, answer for M(9) would be directly obtained as 72 without going through each try and error.

You’ll notice that the smallest such M(n) for any given value of n tends to be composed of the smallest prime numbers, but arranged in such a way that the pattern of the tree matches exactly.

Solving this for all numbers from 2 to 31 will require a lot of computation but it is definitely doable.

Since this problem requires significant computation, to solve it programmatically would be desirable. Then, the computational algorithm should be that the minimum number is generated by reassembling the shape with 2s and when the shape can’t match the shape of n!!, upgrade the rightmost leaf 2 of the tree to 3 and go on.

After you calculate the sum for M(n) from 2 to 31, the result will be 569058 as per the algorithm pattern identified.

More Answers:
Birds on a Wire
Pythagorean Triple Occurrence
Numbers Challenge

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