If you fit the following cubic t(n) = (-n3 + 9n2 - 20n + 18)/3 where n = 1, 2, 3, ... then the next number is t(6) = 1
There does not appear to be a missing number. The sequence is n2 + 1. 12 + 1 = 2 : 22 + 1 = 5 : 52 + 1 = 26 : 262 + 1 = 677. The next number in the sequence is 6772 + 1 = 458330.
16 is the next squared number.
12110 0r 1210
It is a number system that uses the digits 0 and 1 only. So 0 is written 0 and 1 is 1, but two in base 10 is written 10 in binary. The first digit is 20 or 1, the next is 21 or 2, the next digits is 22 This keeps going and any number can be written in binary or base two.
If you fit the following cubic t(n) = (-n3 + 9n2 - 20n + 18)/3 where n = 1, 2, 3, ... then the next number is t(6) = 1
The answer is 8 because 1+1=2, 1+2=3, 2+3=5,and 3+5=8.
There does not appear to be a missing number. The sequence is n2 + 1. 12 + 1 = 2 : 22 + 1 = 5 : 52 + 1 = 26 : 262 + 1 = 677. The next number in the sequence is 6772 + 1 = 458330.
1
The next number is 4, followed by -2
16 is the next squared number.
The number 0 and the next is 2.
4.
1 + 1 = 2 2 * 2 = 4 4 - 1 = 3 3 + 3 = 6 6 + 2 = 8 8 - 1 = 7 so you just repeat that .. meaning that the next number is 15
4
1+7=8 8-7=1 1+3=4 4+3=7 7-7=0 0+7=7 7-7=0 x=0+3 the next number is 3
12110 0r 1210