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I was exploring OEIS, and decided to make the following query: http://oeis.org/search?q=196680&sort=&language=english&go=Search

The first 2 entries, which are sequences with generating functions $ \frac{1+x-x^2}{1-x-2x^2+x^4}$ and $ \frac{x(1+x)(1-x)}{1-x-x^4}$ respectively, (interpreted as sums of some triangle that I'm not sure in the former, and a coefficient of a specific matrix raised to integer powers in the latter).

What's very curious to me, was the large number of common terms these sequences have. That is all the entries of the first sequence are common to the second sequence (although the second sequence carries quite a few more, the first is list below

$ 1, 2, 3, 7, 12, 24, 45, 86, 164, 312, 595, 1133, 2159, 4113, 7836, 14929, 28442, 54187, 103235, 196680, 374708, 713881 ...$

That motivates a natural conjecture that every coefficient of

$\frac{1+x-x^2}{1-x-2x^2+x^4}$ is also a coefficient of $ \frac{x(1+x)(1-x)}{1-x-x^4}$

But i'm not sure how to prove this just based on those generating functions, and i'm curious what ramifications that may have for the underlying objects, called Jacobsthal Polynomials.

Sidharth Ghoshal
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2 Answers2

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What you didn't mention in your original posting is that the first sequence (apart from a few initial terms) comprises every second term of the second series.

Let $f(x)=\sum_{n=0}^\infty a_nx^n$ and $g(x)=\sum_{n=0}^\infty b_nx^n$ be the generating functions of your series. You wish to prove that $a_n=b_{k+2n}$ for some offset $k$ and all $n$. But $$\sum_{n=0}^\infty b_{2n}x^{2n}=\frac{g(x)+g(-x)}2$$ and $$\sum_{n=0}^\infty b_{2n+1}x^{2n+1}=\frac{g(x)-g(-x)}2.$$ Thus you need to prove that one of $\frac12(g(x)\pm g(-x))$ equals $u(x)+x^r f(x^2)$ for a polynomial $u(x)$ and some integer $r$.

Over to you!

Angina Seng
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More specifically, the pattern seems to be that the coefficient of $x^k$ in $a(x) = \frac{1+x-x^2}{1-x-2x^2+x^4}$ is equal to the coefficient of $x^{2k+5}$ in $b(x) = \frac{x(1+x)(1-x)}{1-x-x^4}$. And the way to test this conjecture is to consider the function $$ c(x) = b(x) - x^5 a(x^2) = x + \frac{x^2-x^4}{x^8-2 x^4-x^2+1}, $$ which (apart from the initial $x$) is an even function; hence all of the coefficients of $x^{odd}$ (other than $x^1$) in $c(x)$ equal $0$, which establishes the conjecture.

Greg Martin
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