I have the following sequence:
$$\lbrace a_n\rbrace=\lbrace 10,48,126,254,438,690,1023,1451,1989...\rbrace$$
which I am trying to match a function to.
I understand I can use arithmetic progression to find a polynomial:
$$ \begin {array} {l l l l l l} 10&48&126&254&438&690&1023&1451&1989\\38&78&128&184&252&333&428&538\\40&50&56&68&81&95&110\\10&6&12&13&14&15\\-4&6&1&1&1 \end {array}$$
because a pattern does emerge for $n > 2 $. I can then derive a formula using typical methods, and get:
$$ a(n)= \frac{n^4 + 30n^3 + 215n^2 + 282n - 648}{24} $$
This function works for all $n > 2 $, but gives $a(1) = -5$ and $a(2) = 43$ (instead of $10$ and $48$ respectively).
Is there any advanced formula formulation technique that would allow me to make the formula work for all n, not just $n > 2 $? I figure the $-4$ and $6$ differences in the arithmetic progression method could be a clue, but I'm not sure what to do with it. I am unable to find such information online.
(I'm looking for a method that would work generally, rather than a quick fix for this specific sequence.)
You have a polynomial $p(n)$ so that $p(n) = a_n$ for all $n > 2$. If there is another polynomial $q(n)$ so that $q(n) = a_n$ for all $n$, then $p(n)$ and $q(n)$ agree on infinitely many points and so must be equal. But then $p(1) = q(1) = a_1$, a contradiction since the formula fails for $n=1$. So there can be no polynomial solution. I would just write it as a piecewise function - that is as simple as you will be able to get.
Often, the best solution to fit a formula to a sequence is to look it up in OEIS. Your sequence doesn't seem to be there, so if you think it is particularly interesting mathematically you may consider submitting it. You could also use the superseeker to try harder to find a fit.