This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A358200 #37 Mar 24 2025 05:53:32
%S A358200 4,2,6,1,2,1,10,11,12,13,2,1,6,6,5,7,7,2,7,5,7,25,2,1,2,1,2,31,32,4,6,
%T A358200 35,36,6,7,6,4,7,7,12,9,2,2,47,6,5,1,2,5,4,9,55,5,4,4,7,7,1,8,63,10,1,
%U A358200 2,14,68,69,9,2,5,14,74,4,6,5,11,1,2,81,9,9,9,8,6,4,10,1,1,2,7,6,1,2,1,3,2,99,100,6,19,16
%N A358200 Frequency ranking position of the ratio r(n) between consecutive prime gaps, among all previous ratios {r(i) : 2 < i < n, r(i) = (prime(i) - prime(i-1))/(prime(i-1) - prime(i-2))}. If the ratio r(n) is not among previous ratios, then a(n)=n.
%C A358200 Given all primes up to prime(n-1) with n > 3, if a(n) != n then a(n) is the number of attempts to find the prime(n) using to the following algorithm:
%C A358200 1) Calculate all previous consecutive prime-gaps ratios psr(n) = {r(i) : 2 < i < n, r(i) = (prime(i) - prime(i-1))/(prime(i-1) - prime(i-2))}.
%C A358200 2) Tally the elements {r(i)} and sort them according to their frequencies in descending order first, and then by their values in ascending order.
%C A358200 3) Check the primality of candidates q in the same order as the ratio (q - prime(n-1))/(prime(n-1) - prime(n-2)) appears in the ordered set obtained above.
%C A358200 In the first 2^17 primes the median of all consecutive prime-gaps ratios is 16 and there are 621 different ratios.
%C A358200 Conjectures:
%C A358200 1) a(n)=n or a(n) < x*n^y with x ~ 6 and y ~ 0.4 (verified for first 2^16 primes).
%C A358200 2) The sequence is unbounded. (Equivalent to conjecture in A001223 on Sep 29 2018.)
%C A358200 3) The relative frequencies (probabilities) of the consecutive prime-gap ratios approach constants as the length of the list of first primes approaches infinity. (Equivalent to conjecture in A001223 on Sep 01 2019.)
%e A358200 In the table below for the first terms, the columns are: index n, primes(n), consecutive prime-gaps ratio r(n), previous sorted ratios psr(n), and a(n).
%e A358200 n prime(n) r(n) psr(n) a(n)
%e A358200 1 2 - {} -
%e A358200 2 3 - {} -
%e A358200 3 5 2 {} -
%e A358200 4 7 1 {2} 4
%e A358200 5 11 2 {1, 2} 2
%e A358200 6 13 1/2 {2, 1} 6
%e A358200 7 17 2 {2, 1/2, 1} 1
%e A358200 8 19 1/2 {2, 1/2, 1} 2
%e A358200 9 23 2 {2, 1/2, 1} 1
%e A358200 10 29 3/2 {2, 1/2, 1} 10
%e A358200 11 31 1/3 {2, 1/2, 1, 3/2} 11
%e A358200 12 37 3 {2, 1/2, 1/3, 1, 3/2} 12
%e A358200 13 41 2/3 {2, 1/2, 1/3, 1, 3/2, 3} 13
%e A358200 14 43 1/2 {2, 1/2, 1/3, 2/3, 1, 3/2, 3} 2
%e A358200 a(4), a(6), a(10), a(11), a(12) and a(13) are respectively 4, 6, 10, 11, 12 and 13 because the corresponding ratios 1, 1/2, 3/2, 1/3, 3 and 2/3 are ratios that appear for the first time.
%e A358200 a(5) = 2 because the corresponding ratio r(5)=2 is at the second position in the ordered set of previous ratios psr(5)={1, 2}.
%e A358200 a(9) = 1 because the corresponding ratio r(9)=2 is at the first position in the ordered set of previous ratios psr(7)={2, 1/2, 1}.
%t A358200 p[n_]:= Prime[n];
%t A358200 (* consecutive prime-gaps ratio *)
%t A358200 r[n_]:= (p[n] - p[n - 1])/(p[n - 1] - p[n - 2]);
%t A358200 (* sorted ratios according to increasing frequency and decreasing value *)
%t A358200 fracs[n_]:= Transpose[SortBy[Tally[r[Range[3, n]]], {-#[[2]] &, #[[1]] &}]][[1]];
%t A358200 SetAttributes[fracs, Listable];
%t A358200 (* Position of the new ratio r[j] in previous list, or j if not present *)
%t A358200 a[j_] := Append[Position[fracs[j - 1], r[j]], {j}] // First // First;
%t A358200 SetAttributes[a, Listable];
%t A358200 (* First 100 terms starting from n=4 *)
%t A358200 a[Range[4,103]]
%Y A358200 Cf. A001223 (Prime gaps), A275785, A276812, A272863, A274225.
%K A358200 nonn
%O A358200 4,1
%A A358200 _Andres Cicuttin_, Feb 22 2023