Given any 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 and the next is the given number. There are also non-polynomial solutions.
The simplest rule, based on a polynomial of order 5, is
Un = (-4n5 + 75n4 - 510n3 + 1665n2 - 2306n + 1440)/120 for n = 1, 2, 3, ...
Another possibility is a Fibonacci-like sequence with
U1 = 3, U2 = 4 and Un+2 = Un + Un+1 for n = 1, 2, 3, ...
Adding consecutive odd numbers
65/18 as a mixed number is 3 and 11/18
The number is 1.
47 4+7 = 11 7+11 = 18 11+18=29 18+29=47
The number progression is n+prime numbers progressively increasing starting from 2. The missing number is 29.
Adding consecutive odd numbers
18 over 11 as a mixed number = 17/11
36, possibly. The sequnce being *2, *2, -6, *2, *2, -6 etc
11/18 cannot be simplified - 11 is a prime number !
7 + 11 = 18 11 + 18 = 29 18 + 29 = 47 Add the two previous terms together.
If you mean 4 11 18 25 then the next number is 32 because the numbers are increasing by increments of 7
-18 is less than -11 or -18 < -11
With each new term, you add a number n. This number is always odd, and increases by 2 with each term. So, starting with the first term, 2, you add n = 1 to get 3; add 3 to n = 3 to get 6; add 6 to 5 to get 11; and 7 to 11 to get 18. Here is the pattern written out: 2 + 1 = 3 3 + 3 = 6 6 + 5 = 11 11 + 7 = 18 ...
65/18 as a mixed number is 3 and 11/18
The number is 1.
47 4+7 = 11 7+11 = 18 11+18=29 18+29=47
No.