What the nth term in the sequence of -1 1 -1 1 -1?

Answer 1

One possible solution is the nth term is (-1)ⁿ for n = 1, 2, .., 5.

Beyond n = 5, we have no information as to how the sequence continues, The first 10 terms may be:

• -1, 1, -1, 1, -1, 1, 2, 3, 4, 5, ... (with the counting numbers);
• -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, .... (repeating alternating between -1 and 1; t{n} = (-1)ⁿ);
• -1, 1, -1, 1, -1, -31, -129, -351, -769, -1471, ... (t{n} = -(2x⁴ - 24x³ + 100x² - 168x + 93) / 3);

Or some other following terms.

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The problem with a sequence is that it is the zeros of t{n} - a{n}.

The solution will be a polynomial of degree equal to the number of terms given.

However, there are infinitely many polynomials of a given degree that will go through the zeros given. There are some polynomials of a lesser degree which will go through the given zeros.

eg for a{n} = {1, 2, 3, 4, ...} the counting numbers, a polynomial of degree 1 will suffice: t{n} = n. However, it is possible to find a polynomial of degree 4 which will permit any value to follow;

eg for the given a{n}, t{n} = -(x⁴ - 10x³ +35x² - 62x + 24)/12 will also give a{n} = {1, 2, 3, 4} for n = 1, 2, 3, 4, but a{5} = 3; it has the same 4 first terms and then continues with a different 5th and subsequent terms.

Answer 2

According to Wittgenstein's Finite Rule Paradox every finite sequence of numbers can be a described in infinitely many ways and so can be continued any of these ways - some simple, some complicated but all equally valid.
The simplest polynomial rule here is T(n) = 5n2 - 15n - 1 for n = 1, 2, 3