Does this sequence of polynomials have a closed form?

Question

Consider the following sequence of polynomials in the variable $d$, which I encountered during a calculation:

$$ 1 + 3 d^2 \\ 1 + 10 d^2 + 5 d^4 \\ 1 + 21 d^2 + 35 d^4 + 7 d^6 \\ 1 + 36 d^2 + 126 d^4 + 84 d^6 + 9 d^8 \\ 1 + 55 d^2 + 330 d^4 + 462 d^6 + 165 d^8 + 11 d^{10} \\ 1 + 78 d^2 + 715 d^4 + 1716 d^6 + 1287 d^8 + 286 d^{10} + 13 d^{12} \\ 1 + 105 d^2 + 1365 d^4 + 5005 d^6 + 6435 d^8 + 3003 d^{10} + 455 d^{12} + 15 d^{14} \\ 1 + 136 d^2 + 2380 d^4 + 12376 d^6 + 24310 d^8 + 19448 d^{10} + 6188 d^{12} + 680 d^{14} + 17 d^{16} \\ ... $$

These polynomials came from a relatively simple calculation, so there might be a simple way to generate them directly. There are some patterns in the prime factorizations of the coefficients, but I can't exactly point out what.

Is there a closed form or something similar for these polynomials, perhaps involving special functions or binomial coefficients?

The $d^2$ coefficient is $n \cdot (2n + 1)$ and the $d^4$ coefficient is $\frac{1}{6}n(n - 1)(2n - 1) \cdot (2n + 1)$. The polynomial in front of the odd factor is the sum of squares, shifted. I was hopeful that the pattern would persist, but then the second $d^6$ coefficient (in the fourth row) would have to be $17 \cdot 9 = 153$, but alas, it's $84$. — Sammy Black, May 22 '14 at 08:07

score 2 · Accepted Answer · answered May 22 '14 at 08:02

2

It looks like $\dfrac{(1+d)^{2k+1}+(1-d)^{2k+1}}{2}$.

answered May 22 '14 at 08:02

André Nicolas

507,029

Ah, much simpler than I expected, thank you. How did you find this? – user111187 May 22 '14 at 08:06
I recognized the binomial coefficients, and that half were missing. The standard way to make half of them miss is through the $1+x$, $1-x$ trick. – André Nicolas May 22 '14 at 08:08
Very smart, thanks again. – user111187 May 22 '14 at 08:11
You are welcome. Happens a lot. For example in number-theory (and recurrences) one bumps into $(a+b\sqrt{d})^n +(a-b\sqrt{d})^n$. And to get rid of half the reciprocals of factorials, we use $\frac{e^x+e^{-x}}{2}$. – André Nicolas May 22 '14 at 08:15

Does this sequence of polynomials have a closed form?

1 Answers1