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Cyclic recursive summation (Posted on 2021-07-16) Difficulty: 5 of 5
Let N be the set of positive integers. Find all functions f:N->N that satisfy the equation

fabc-a(abc) + fabc-b(abc) + fabc-c(abc) = a + b + c

for all a, b, c ≥ 2.

(Here f1(n) = f(n) and fk(n) = f(fk-1(n)) for every integer k greater than 1)
(Also note: abc is the product a·b·c and not the concatenation 100a+10b+c)

No Solution Yet Submitted by Danish Ahmed Khan    
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Thoughts | Comment 1 of 9
The definition of this function puts a very heavy burden on the values of the sequence f^1(1), f^1(2), .. For instance, f^1 is tendentially increasing but cannot be monotonic, else the left hand side will grow too quickly. 

It is also interesting to look at values of a, b, c which give the same product.

It is annoying that f^1 never appears in any equation. We start with f^6. In fact, f^m is "missing" for many values of m. For instance, we have no equations involving f^p, where p is prime. But the list of allowable m-values appears to be irregular. I don't think we can renormalise the equation around a some base m value. 

______________
Further thoughts:

For any integer n > 1, there is a value m such that f(m) = n. We can also express m using the function, f:

Take a = b = c to be some natural number, n, then 

f^[n^3 - n](n^3) = n
so, if we take m = f^[n^3 - n - 1](n^3) we have f(m) = n. For instance, we have

a2 = f^5(8), a3 = f^23(27), a4 = f^59(64)   then
f(a2) = 2, f(a3) = 3, f(a4) = 4...



Edited on August 4, 2021, 12:40 pm
  Posted by FrankM on 2021-08-04 10:26:12

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