The discussion revolves around finding the nth term of the sequence -1, 5, -17, 65. Participants explore various formulations for the nth term, considering whether it fits into known sequence types and discussing the implications of different representations.
Participants do not reach a consensus on a single nth term formulation, and multiple competing views on how to express the nth term remain. The discussion reflects uncertainty and exploration of various possibilities.
Participants acknowledge that the problem may have different interpretations based on imposed restrictions on the form of the nth term, and they discuss the implications of these restrictions on the uniqueness of the nth term representation.
scottshannon said:I have been trying to find the nth term of the following sequence:
-1, 5, -17, 65...
I am thinking that the nth term has (-1) ^n . It doesn't appear to be a geometric or arithmetic sequence. I am stymied.
ZaidAlyafey said:It seems if $a_n = (-1)^{\color{red}{n}} (4^{\color{red}{n-1}}+1) \,\,\,\, n\geq 1$
skeeter said:correction
If the first number has index 1, then these numbers are described by the following formula.scottshannon said:I have been trying to find the nth term of the following sequence:
-1, 5, -17, 65...
It is permissible in principle. So without any restrictions this problem is not interesting. One could write a formulascottshannon said:There are a couple of things that had never occurred to me:
1) That it is permissible to write the nth term this way.
One can do something silly such as changing an expression $E$ to $E+1+\dots+1-1-\dots-1$ where all 1's cancel. Strictly speaking, this expression specifies the same function, but it is a different from $E$. But one can also use write expressions such that proving their equality to $(-1)^{n} (4^{n-1}+1)$ would require highly nontrivial mathematics.scottshannon said:Also that there might be infinitely many nth term representations for a series where only the first 4 terms are given.
In some sense, but it is still infinite. An infinite set $A$ may be a subset of another set $B$ and yet have as many elements as $B$.scottshannon said:Is it valid then to say that as one sees more terms in the series that the number of possibilities for the nth term decreases?
For each sequence (function) there is an infinite number of formulas (expressions) that specify it. Also, there are infinitely many sequences with a given initial finite segment such as -1, 5, -17, 65.scottshannon said:If it is true that there are an infinite number of ways of writing the nth term representation, then will every one of them give the same 5th, 6th...terms?
Evgeny.Makarov said:It is permissible in principle. So without any restrictions this problem is not interesting. One could write a formula
\[
a_n=
\begin{cases}
-1&n=1\\
5&n=2\\
-17&n=3\\
65&n=4\\
a_n=1&n>4.
\end{cases}
\]
It specifies a valid function from natural numbers to natural numbers, and its values for $n=1,\dots,4$ are as required.
A specific problem may impose restrictions on how the function (i.e., the sequence) is specified. For example, it is possible to require that the expression specifying $a_n$ uses only addition, subtraction, multiplication and division. Then one solution is $a_n=22 n^3-146 n^2+290 n-167$ (it does not even use division). You can verify that $a_n$ are as required for $n=1,\dots,4$.
It is also possible to ask for a shortest expression. This problem may be interesting, but I haven't seen something like it.
One can do something silly such as changing an expression $E$ to $E+1+\dots+1-1-\dots-1$ where all 1's cancel. Strictly speaking, this expression specifies the same function, but it is a different from $E$. But one can also use write expressions such that proving their equality to $(-1)^{n} (4^{n-1}+1)$ would require highly nontrivial mathematics.
In some sense, but it is still infinite. An infinite set $A$ may be a subset of another set $B$ and yet have as many elements as $B$.
For each sequence (function) there is an infinite number of formulas (expressions) that specify it. Also, there are infinitely many sequences with a given initial finite segment such as -1, 5, -17, 65.