Need some help finding the equation (nth term) of a sequence

Question

The sequence goes like this

Sequence = 1 , 4 , 14 , 58 , 292

Each number is being multiplied by (n+1) and then 2 is added to them. Should i proceed to find the equation from

$ 2(n+1)^n $

this? . Thank you very much in advance!

Answer 1

Your recurrence equation is $a_{n+1} = (n+1) a_n + 2$ with $a_1 = 1$. Note that if $a_n = n! b_n$, we have $ b_{n+1} = b_n + \frac{2}{(n+1)!}$ with $b_1 = 1$. Thus $$b_n = 1 + \sum_{k=2}^n \frac{2}{k!}$$ and $$ a_n = n! + \sum_{k=2}^n \frac{2n!}{k!} $$ Unfortunately the sum can't be simplified in an elementary way, although it can be expressed using the incomplete Gamma function.

Answer 2

An exponential isn't quite right here, since you are multiplying by a different amount each step. At the first step, $n+1=2$, while at the second step, $n+1=3$, etc. It seems like factorials will be a much better fit for this sequence, since they have the same sort of "multiply by an increasing sequence of numbers behavior. Let's just play around a bit with the first few terms:

$$ 1 = 1 $$

$$ 4 = 2(1) + 2 $$

$$ 14 = 3(2(1) + 2) + 2 $$

$$ 58 = 4(3(2(1) + 2) + 2) + 2 $$

Let's take 58 as out example, and expand it out.

We get: $58 = 1(4\cdot3\cdot2) + 2(4\cdot3) + 2(4) + 2$

If we take a bigger example, like 292, we get: $292 = 1(5\cdot4\cdot3\cdot2) + 2(5\cdot4\cdot3) + 2(5\cdot4) + 2(5) + 2$. This expression is just the same as the previous one, but we've multiplied all the terms by 5, and then added another 2 onto the end. I'm noticing that all of the terms here begin with a 2, except for the first one. So let's just rewrite things so the first term also starts with a 2, to make it a bit neater.

$$ 292 = 2(5\cdot4\cdot3\cdot2) + 2(5\cdot4\cdot3) + 2(5\cdot4) + 2(5) + 2 - 5! $$

Here, I've both added and subtracted $5!$, and combined the positive one into the first term. Now we have: $ 292 = 2X - 5! $ where $X = (5\cdot4\cdot3\cdot2) + (5\cdot4\cdot3) + (5\cdot4) + (5) + 1$

Now, if we can just manage to find $X$, we'll be done. Let's rewrite $X$ as the following:

$$ X = \frac{5!}{1!} + \frac{5!}{2!} + \frac{5!}{3!} + \frac{5!}{4!} + \frac{5!}{5!} $$

I'm noticing the Taylor series for $e$ in that series of fractions. The series for $e$ is $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \frac{1}{6!} + \frac{1}{7!} + ...$.

Therefore: $5!(e-1) = X + \frac{5!}{6!} + \frac{5!}{7!} + ...$

Notice that that the terms $\frac{5!}{6!} + \frac{5!}{7!} + ...$ are equal to $\frac{1}{6} + \frac{1}{42} + ...$. In other words, their sum is less than $1$. So we could legitimately write the following expression for $X$:

$$ X = \left\lfloor{5!(e-1)}\right\rfloor $$

Now we did all this for the specific example of the $5^{th}$ term, 292, but we can also make a general formula out of this by replacing the 5s with $n$s.

For $X$, we have: $X = \left\lfloor{n!(e-1)}\right\rfloor$.

So my proposed formula is:

$$ t_n = 2\left\lfloor{n!(e-1)}\right\rfloor - n! $$

Obviously you still need to prove that this works, and taking the floor to find $X$ might fail for small $n$, meaning you'll need to add some special cases. But this gives you some idea of how one might go about finding a candidate formula in the first place.