I need some help to make a generating function for

Question

. I need some help to make a generating function for those series (3,6,11,18,...) and (3a1,0,0,3^2 a2,..) .

Answer 1

Here is one function that gives the first sequence $3,6,11,18,\ldots$ from $1,2,3,4,\ldots$:

$$f(x)=x^2+2$$

Is that what you meant?

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There are many ways to get that formula from your sequence. Here is one way, close to what I did mentally.

Look at the differences between successive entries, then look at the differences in that sequence. Putting each sequence on a line, we get

$$\begin{matrix} 3 & & 6 & & 11 & & 18 \\ & 3 & & 5 & & 7 & \\ & & 2 & & 2 & & \\ \end{matrix}$$

We see that the entries in the third row are equal. That means that our desired formula is a quadratic polynomial. The repeated value in the third row is $2$. In a way similar to integrating the constant $2$ twice and getting $x^2+bx+c$, that repeated $2$ means our quadratic has a leading coefficient of $1$.

There are ways to quickly get the values of $b$ and $c$, but a quick look at your sequence immediately let me see that the correct formula is $x^2+2$.

This kind of analysis is called the Calculus of Finite Differences, and I got my introduction to that in one of Martin Gardner's columns in the Mathematical Games department of Scientific American magazine. That column was reprinted with comments in Gardner's book New Mathematical Diversions from Scientific American. The main theorem in that study is that an $n$'th degree polynomial gives constant values in the $n+1$'th row of finite differences, and that value is $n!$ times the leading coefficient of the polynomial.