Let f(a) be a function defined over the natural numbers.

Suppose there exists one or more solutions "n" such that f(a) = n for many different a. Let's call the solution set A

For such n, suppose we are interested in finding consecutive numbers x through x + k (k > 0) such that f(x+i) = n for every 0 <= i <= k

Let L = k + 1, the range length

Let the solutions x be partitioned into number of groups based on the range length. The set of groups is given by: product (p(i) ^ e(j) for j = 0 to ceiling (log(l)/log(p(i))) where each p is prime

We partition the solutions into these groups so that we can later efficiently iterate over them (for the sake finding valid solutions in a consecutive range.

Primes p > l are not in consideration in this exercise. This is a crude "first-pass" measure just to see what common solutions over a consecutive range would have to obey.

Background: This general question came about when I unsuccessfully tried to tackle the problem posed here: https://www.primepuzzles.net/puzzles/puzz_670.htm

Task: Compute S(L), the exact number of sets for each L >=1. Solutions have been computed through L = 19 using a C++ program. Currently, it stores the sets in an "unordered map". The code takes advantage of

symmetry to avoid obvious redundancies but this quasi-brute force approach in the code requires progressively larger amounts of memory.. Ideally, a formula could be determined.

Short of that, a program / heuristic that avoided storing large numbers of sets would be terrific.

https://github.com/grandpascorpion/miscMath

We only need to partition A into A(2) and A(1). In this context, A(2) means its members are divisible by any positive power of 2 (even) and A(1) means its members are all 2-unsmooth or odd.

We only have two sensible combinations for solutions to f(x) = f(x+1) = n

Member of A(1)

Even followed by odd

Member of A(2)

Odd followed by even

So, there are two classes of solutions for two consecutive numbers.

Let's write this in shorthand as tuples (2,1) and (1,2)

As we extend this to longer ranges, it's straightforward to compute the number of solution classes but I am interested in the number of sets.

For length two, there is only one unique combination. Let's establish a convention that they are sorted from low to high by subscript.

So, the lone "set" is (Member of A(1), Member of A(2)) or (1,2) for brevity

Note: That among 3 consecutive integers, at least one must be divisble by 3. We must partition A into two subsets: non-multiples and multiples of 3. Let's call these A'(3,1) and A'(3,3).

If the first number of the 3-reange, "x" is 2 mod 4 then x +2 must be a multiple of 4. So, with respect to powers of 2, we need to partition A into 3 sets:

2-unsmooth, mod 4 = 2 and mod 4 = 0 (which is why the ceiling function was used earlier).

We will call these subsets A'(2,1), A'(2,2), A'(2,4)

We use A' instead of A because these are intermediate sets. We want to further partition A with respect to divisibility by both 2 and 3.

If take the intersections of between the A'(2,_) and A'(3,_) sets, we have 3 * 2 = 6. We will label A(ij) = intersection of A'(2,i) and A'(3,j)

This gives us A(1), A(2), A(3), A(4), A(6), A(12).

Let's look at the possible valid combinations for powers of 2 and 3 and let's say mA'(x,y) means member of the set A'(x,y)

The notation above is a little cumbersome. Let's replace the entries with 2nd subscript and use that convention going forward and also indicate if they are symmetric and has a previously occurring mirror/inverse.

* The status is defined as either: (S)ymmetrical, Y - previous mirror processed, N - previous mirror not processed

If we take the cartesian product, we will get 4 * 3 = 12 solution classes

Cross-products

We are interested in the number of distinct sets. Now, we have to start considering symmetry and reflection.

This is a reordering of the tables above to get a better handle on how symmetry/reflection of the specific power tuples factor in our calculation

New set (1,2,12)

New set (1, 4, 6)

New set (2, 3, 4)

New set (1, 3, 4)

New set (1, 1, 12)

New set (1,2, 3)

New set (1, 1, 6)

7 unique sets in all

(2,1,4) is the mirror image of (4,1,2) so it cannot contrubute any new sets. (4,1,2) has already contributed them.

(1,4,1) and (1,2,1)'s behavior with respect to contributing new sets is the same (because they are symmetrical).

As (1,4,1) is symmetrical, when combined with (3,1,1) and (1,1,3) there will only be one new set contributed from those two cross-products.