On 14/02/2026 22:00, Rui Ferreira wrote:
> Hello SeqFan,
>
> I observed an interesting connection between the OEIS sequences
> A248234 and A183861. After running a combinatorial calculation based
> on a specific combinatorial formula based on differences of squares
> and sums of cubes of integer pairs (Python code at
>
https://github.com/rf-iasys/OEIS/blob/main/OEIS_A248234_A183861.py),
> the results I obtained match A248234 almost exactly being the
> differences between the two corresponding exactly to A183861.
>
> When you subtract the values of A183861 from A248234, you get the
> exact same numbers as the ones produced by my combinatorial calculation.
As with your previous calculation having this form, the business with x
and y=x(b-a) obfuscates what's really going on. Your compute_elements
function does the following:
for a<b with a+b <= n_end:
x = (b^2-a^2)(b^3+a^3)
y = x(b-a)
if y=x then add this value to your list
which is exactly equivalent to the following:
for all a with 2a+1 <= n_end:
b = a+1
x = (b^2-a^2)(b^3+a^3)
add this value to your list
(General note: If all your "combinatorial calculation" things are of the
form "for all a,b write x=... and y=x(a-b); then consider only the cases
where y=x" then they are all obfuscated versions of a simpler
calculation that just says "for all a write b=1 and then x=...".)
... which is exactly equivalent to the following:
for all a <= (n_end-1)/2:
add (2a+1)(a^3+(a+1)^3) to your list
or:
for all a <= (n_end-1)/2:
add (2a+1)^2 (a^2+a+1) to your list
So your observation is that it appears that
A248234(n) = floor(1/((n+1)^-5 + (n+2)^-5 + (n+3)^-5 + ...)
equals
(2n+1)^2 (n^2+n+1) + (n-2) + ceiling(((n-1)^2-1)/3)
for n >= 3. (For n=2 we need A183861(1) which is special-cased; for n=1,
A183861(0) isn't defined at all.)
This is in fact _already in A248234_, though not in the same form; it's
described as a "Conjecture" made by Kotesovec, but Sahu's paper linked
in the OEIS entry proves it (in, again, a slightly different form; there
are generally lots of different ways to write down things of the form
"nearest integer to polynomial in n with coefficients rational with
small denominators"; I have to say I really don't like Kotesovec's
version, which makes things look distinctly more complicated than they
really are).
There will be similar rounded-polynomial expressions for the remainder
terms in other zeta(n); the case n=3 is in A248230 (and, again, in
Sahu's paper).
Your observation amounts to writing the relevant integerified polynomial
as a combination of one particular quartic polynomial that has the right
n^4 and n^3 terms, and a quadratic correction that happens to be in
OEIS. I don't think there's any significance to the particular quartic
polynomial you've chosen, beyond the fact that its highest two
coefficients are the ones you need.
--
g