Any number that you choose can be the next number. It is easy to find a rule based on a polynomial of order 6 such that the first six numbers are as listed in the question followed by the chosen next number. There are also non-polynomial solutions. Short of reading the mind of the person who posed the question, there is no way of determining which of the infinitely many solutions is the "correct" one.
In this particular example, the simplest solution, based on a polynomial or order 5, is
t(n) = (6n^5 - 125n^4 + 920n^3 - 2995n^2 + 4474n - 2280)/120 for n = 1, 2, 3, ...
and, accordingly, the next term is -12.
The next term is -2.
It is: 0 1 2 3 ... etc
4
12110 0r 1210
This sequence is all the single digits, written in reverse alphabetical order. The first value is 0, and zero comes last alphabetically. The next value is 2, and two comes second last alphabetically, and so on. From this, it is clear that the next few numbers will be 9,4,5 and 8 (eight being the first number alphabetically) and then the sequence will end.
The next term is -2.
The number 0 and the next is 2.
1, 2, 0, 3, -1, 4, -2, 5, -3, 6
4
312 211 110 3-1=2 2-1=1 1-1=0
The next whole number after 1 is 2. However, the decimal after 1 can be an infinite chain of 0's, with a 1 at the end.
This series is of the function f(x) = x2+1, starting with x=0.The next number in the series is 26. The number after that is 37.
The next number is 4, followed by -2
It is: 0 1 2 3 ... etc
It could be 34, if you use the position to value rule Un = (-9n4 + 106n3 - 429n2 + 692n - 348)/12 for n = 1, 2, 3, ...
0.1 These are all powers of 10: 10^3 = 1000 10^2 = 100 10^1 = 10 10^0 = 1 10^-1 = 0.1
4.