Witnesses:

Can you find any +5 witnesses smaller than 42? (I'm currently trying to prove a length 45 and a length 41 candidate.) Can you find a smaller +3 witness than 19, or a smaller +4 witness than 26?

(These witnesses prove that f(n) >= sollen. Note that a "good" witness will be one that can be exhaustively proved reasonably fast; some stacks take much longer than others to prove. I have run all the ones listed here exhaustively.) lbound is a lower bound of f(n) for n>14. The value shown is the length of an existing solution to the sequence shown. All of these witnesses have been proven by me.

nlboundstack
<1
101
212 1
331 3 2
442 4 1 3
552 5 3 1 4
+1
674 6 2 5 1 3
781 5 2 7 4 6 3
896 3 7 4 8 1 5 2
9101 7 5 3 8 6 9 4 2
10111 3 5 2 4 6 8 10 7 9
+2
11131 11 3 6 9 4 7 10 5 8 2
12141 10 2 12 4 6 8 5 9 7 11 3
13151 10 3 12 2 11 9 6 8 4 13 7 5
14161 14 12 4 7 10 5 8 11 6 9 3 13 2
15171 14 12 4 7 10 5 8 11 6 9 3 13 15 2
16181 14 12 4 7 10 5 8 11 6 9 3 15 13 16 2
171916 12 9 14 11 8 13 10 15 7 17 4 6 2 5 1 3
18201 10 13 8 3 6 4 7 5 2 12 9 11 14 16 18 15 17
+3
19221 3 7 5 2 6 4 8 12 14 10 13 11 9 15 18 16 19 17
20231 5 2 7 4 6 3 8 14 11 13 10 12 9 15 20 17 19 16 18
21241 3 7 5 2 6 4 8 10 14 12 9 13 11 15 17 21 19 16 20 18
222511 1 9 4 7 2 5 8 3 6 10 12 22 14 19 16 21 18 15 20 17 13
23261 4 8 3 6 2 7 5 9 14 11 13 16 12 10 15 17 22 20 23 19 21 18
24274 6 2 5 1 3 7 9 11 8 12 10 14 18 16 13 17 15 19 23 20 22 24 21
25282 6 4 1 5 3 7 9 13 11 8 12 10 14 16 20 18 15 19 17 21 23 25 22 24
+4
26301 5 3 7 4 2 6 8 12 14 10 13 11 9 15 21 19 17 20 16 18 22 25 23 26 24
27311 5 3 7 4 2 6 8 12 14 10 13 11 9 15 21 19 17 20 16 18 22 27 25 23 26 24
28321 5 2 7 4 6 3 8 14 11 13 10 12 9 15 19 21 18 16 20 17 22 26 23 28 25 27 24
29336 3 8 5 2 7 4 1 9 13 16 11 14 17 12 15 10 18 24 22 20 23 21 19 25 27 29 26 28
30341 5 8 3 6 9 4 7 2 10 14 17 12 15 18 13 16 11 19 25 23 21 24 22 20 26 28 30 27 29
313519 11 16 13 18 15 12 17 14 10 2 7 4 9 6 3 8 5 1 20 26 24 22 25 23 21 27 29 31 28 30
32361 7 5 3 6 4 2 8 16 10 12 14 11 13 15 9 17 23 21 19 22 20 18 24 32 26 28 30 27 29 31 25
33371 6 3 8 5 2 7 4 9 14 11 16 13 10 15 12 17 25 22 19 24 21 18 23 20 26 31 28 33 30 27 32 29
34381 9 6 3 8 5 2 7 4 10 15 12 17 14 11 16 13 18 23 20 25 22 19 24 21 26 34 31 28 33 30 27 32 29
35391 7 5 3 6 4 2 8 14 12 10 13 11 9 15 21 19 17 20 18 16 22 28 26 24 27 25 23 29 35 33 31 34 32 30
364015 9 11 13 10 12 14 8 2 4 6 3 5 7 1 16 22 20 18 21 19 17 23 29 27 25 28 26 24 30 36 34 32 35 33 31
374130 23 17 19 21 18 20 22 16 1 7 5 3 6 4 2 8 14 12 10 13 11 9 15 29 27 25 28 26 24 31 37 35 33 36 34 32
384231 24 26 28 25 27 29 15 9 11 13 10 12 14 8 2 4 6 3 5 7 1 16 22 20 18 21 19 17 23 30 32 38 36 34 37 35 33
394332 30 23 17 19 21 18 20 22 16 1 7 5 3 6 4 2 8 14 12 10 13 11 9 15 29 27 25 28 26 24 31 33 39 37 35 38 36 34
40441 3 5 2 4 6 8 10 7 9 11 13 15 12 14 16 18 20 17 19 21 23 25 22 24 26 28 30 27 29 31 33 35 32 34 36 38 40 37 39
41451 6 3 8 5 2 7 4 9 14 11 16 13 10 15 12 17 25 22 19 24 21 18 23 20 26 31 28 33 30 27 32 29 34 39 36 41 38 35 40 37
+5
42471 7 5 3 6 4 2 8 12 15 10 13 16 11 14 9 17 21 24 19 22 25 20 23 18 26 30 33 28 31 34 29 32 27 35 37 40 36 39 42 38 41
43??
44496 3 8 5 2 7 4 1 9 13 16 11 14 17 12 15 10 18 22 25 20 23 26 21 24 19 27 31 34 29 32 35 30 33 28 36 40 43 38 41 44 39 42 37
45501 5 8 3 6 9 4 7 2 10 14 17 12 15 18 13 16 11 19 23 26 21 24 27 22 25 20 28 32 35 30 33 36 31 34 29 37 41 44 39 42 45 40 43 38

What we know about f(n):

f(n) >= n+1 for n >= 6

We have proved the above values for f(n) up to f(14)=16. For 15..50, all we have is witnesses for lower bounds.

f(n) >= 15n/14 for n divisible by 14 (H&S)
f(n) >= 17n/16 for n divisible by 16 (G&P)
f(n+1) >= f(n)
f(n) >= f(n+1)-2

This is about all we know. I believe the H&S result also includes (though they do not state this in their paper) that

f(n) >= n+(n-7)/14 for n==7 mod 14

I also believe the H&S and G&P results combine, since the proof is essentially identical, allowing us to say that

f(n+m) >= 15n/14 + 17m/16 for n divisible by 14 and m divisible by 16

This helps us fill in some of the discrete gaps. Further, combining the above two, we also know that

f(n+m) >= n+(n-7)/14 + 17m/16 for n==7 mod 14 and m divisible by 16

For high values of n, where G&P no longer contributes, and H&S dominates, we can set the following bounds:

f(14*k+i) >= 14*k+i + k - v(i), 0<=i<14
iv(i)
00
11
22
33
43
52
61
70
81
92
103
113
122
131