I think that it does, but I would have trouble showing that there is sand in the Sahara, so I'm not going to attempt such a proof.
However, I have been having some fun with the problem by endeavoring to compute the polynomials that give the result of the original expression as a function of k for a given n. For example,
scott1328 has shown that for n=2 the polynomial is 1, for n=3, the polynomial is k+1, and for n=4 the polynomial is 2k+1.
Again, I have not proven anything, so I can only say that it appears as if the expression always results in an integer, and that the polynomials that I computed give the same result as the original expression for n, k.
Anyway, my method was:
1) Calculate the values of the original expression for each n, k.
2) From the values, derive the polynomial coefficients to use with powers of k for each n.
3) Try to come up with a formula for the coefficients for all n, k.
4) Try to find a look-up table to make it easier - for hand computations, at least. These are the polynomial coefficients mentioned in 2).
So here are the results. The formula I arrived at is (don't worry, it gets easier):
- rs_2020_02_09_expr_latex.JPG (17.31 KiB) Viewed 498 times
For example, for n=4:
- rs_2020_02_09_expr_n_eq_4_latex.JPG (12.55 KiB) Viewed 498 times
as previously given.
But let's just go to a look-up table instead of relying on the formula. A look-up table may be found at
OEIS A102426 by scrolling down to the EXAMPLE section. The first few polynomials are given in terms of x, but in order to calculate the expression result, we just replace x with k.
- a102426_ex.JPG (25.45 KiB) Viewed 498 times