Given any number it is easy to find a rule based on a polynomial of order 4 such that the first four numbers are as listed in the question and the next is the given number. There are also non-polynomial solutions.
The simplest solution, in the form of a polynomial of order 3 is
Un = (3469n3 - 20661n2 + 37712n - 20346)/6 for n = 1, 2, 3, ... and accordingly, the next number is 14219.
127.
16 21 22 29
58
47 4+7 = 11 7+11 = 18 11+18=29 18+29=47
-47, -41, -35, -29, -23, -17, -11, -5, 1, etc.
127.
16 21 22 29
42 is the next number in this sequence. This number sequence is adding the next prime number to the last number. So 1 + 2 = 3. Then 3 + 3 = 6, 6 + 5 = 11, 11 + 7 = 18, 18 + 11 = 29. The next prime number after 11 is 13, so 29 + 13 = 42. The next numbers would be 59 (42+17), 78 (59+19), and 101 (78+23)
29, assuming it is an algebraically reclusive sequence.
One possibility: -3
58
21
47 4+7 = 11 7+11 = 18 11+18=29 18+29=47
It looks like it is +6, -5, +12, -5. So I would assume the next number is +18, or 29.
-47, -41, -35, -29, -23, -17, -11, -5, 1, etc.
assuming it keeps the same pattern of +10 -5 alternating then the next number would be 26
These are just the first prime numbers. The next few are 13, 17, 19, 23, 29, 31...