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
-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...
This appears to be two interwoven sequences. The first sequence, increases by 9 then 19......7 + 9 = 16........8 + 19 = 27 The second sequence decreases by 8 then 18....16 - 8 = 8........27 - 9 = 18 The next number would thus be 9 + 29 = 38.