A358200 - OEIS

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.

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, 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, 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

(list; graph; refs; listen; history; text; internal format)

OFFSET

4,1

COMMENTS

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:

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)}.

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.

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.

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

Conjectures:

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

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

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.)

LINKS

Table of n, a(n) for n=4..103.

EXAMPLE

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).

n prime(n) r(n) psr(n) a(n)

1 2 - {} -

2 3 - {} -

3 5 2 {} -

4 7 1 {2} 4

5 11 2 {1, 2} 2

6 13 1/2 {2, 1} 6

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

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

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

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

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

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

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

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

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.

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}.

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}.

MATHEMATICA

p[n_]:= Prime[n];

(* consecutive prime-gaps ratio *)

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

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

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

SetAttributes[fracs, Listable];

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

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

SetAttributes[a, Listable];