A358200 - OEIS

%I #33 Apr 07 2023 17:08:02

%S 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 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 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 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 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 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 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 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 In the first 2^17 primes the median of all consecutive prime-gaps ratios is 16 and there are 621 different ratios.

%C Conjectures:

%C 1) a(n)=n or a(n) < x*n^y with x ~ 6 and y ~ 0.4 (verified for first 2^16 primes).

%C 2) The sequence is unbounded. (Equivalent to conjecture in A001223 on Sep 29 2018.)

%C 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 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    n  prime(n)  r(n)  psr(n)                        a(n)

%e    1    2       -     {}                             -

%e    2    3       -     {}                             -

%e    3    5       2     {}                             -

%e    4    7       1     {2}                            4

%e    5   11       2     {1, 2}                         2

%e    6   13       1/2   {2, 1}                         6

%e    7   17       2     {2, 1/2, 1}                    1

%e    8   19       1/2   {2, 1/2, 1}                    2

%e    9   23       2     {2, 1/2, 1}                    1

%e   10   29       3/2   {2, 1/2, 1}                    10

%e   11   31       1/3   {2, 1/2, 1, 3/2}               11

%e   12   37       3     {2, 1/2, 1/3, 1, 3/2}          12

%e   13   41       2/3   {2, 1/2, 1/3, 1, 3/2, 3}       13

%e   14   43       1/2   {2, 1/2, 1/3, 2/3, 1, 3/2, 3}  2

%e 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 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 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 p[n_]:= Prime[n];

%t (* consecutive prime-gaps ratio *)

%t r[n_]:= (p[n] - p[n - 1])/(p[n - 1] - p[n - 2]);

%t (* sorted ratios according to increasing frequency and decreasing value *)

%t fracs[n_]:= Transpose[SortBy[Tally[r[Range[3, n]]], {-#[[2]] &, #[[1]] &}]][[1]];

%t SetAttributes[fracs, Listable];

%t (* Position of the new ratio r[j] in previous list, or j if not present *)

%t a[j_] := Append[Position[fracs[j - 1], r[j]], {j}] // First // First;

%t SetAttributes[a, Listable];

%t (* First 100 terms starting from n=4 *)

%t a[Range[4,103]]

%Y Cf. A001223 (Prime gaps), A275785, A276812, A272863, A274225.

%K nonn

%O 4,1

%A _Andres Cicuttin_, Feb 22 2023