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Let g(n) be the n-th maximal prime gap; a(n) = 1 if g(n) has record merit, 0 if it does not.
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%I #93 Jul 08 2024 02:22:56

%S 1,1,1,0,0,1,0,0,1,1,0,0,1,1,1,1,1,0,0,1,1,0,1,1,0,0,0,1,0,1,0,0,1,1,

%T 1,1,0,0,1,0,0,0,0,0,0,1,0,1,0,0,1,1,1,1,0,1,1,0,0,0,1,0,0,1,0,0,0,0,

%U 0,0,1,1,1,1,1,0,0,0,0,0,0,1

%N Let g(n) be the n-th maximal prime gap; a(n) = 1 if g(n) has record merit, 0 if it does not.

%C a(n) = 1 if A002386(n) is in A111870, a(n) = 0 if A002386(n) is not in A111870.

%C The merit M of a prime gap of measure g following the prime p_1 is defined as M=g/ln(p_1). It is the ratio of the measure of the gap to the "average" measure of gaps near that point. As an example, the merit of the sixth maximal gap, of size 14, after prime 113 is 2.96.

%C a(81) = 0 because there are previous maximal gaps with higher merits. - _Rodolfo Ruiz-Huidobro_, Jan 23 2024

%C a(82) = 1 as the merit of the gap is 1572/log(18571673432051830099)=1572/44.37=35.43 (which is a record merit). - _Rodolfo Ruiz-Huidobro_, May 10 2024

%H Prime Gap List Community, <a href="https://primegap-list-project.github.io/lists/prime-gaps-high-watermarks/">Record prime gaps</a>, 2021.

%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>

%e The 5th record prime gap from 89 to 97 does not have record merit, so a(5) = 0.

%e The 10th record prime gap from 1327 to 1361 has record merit, so a(10) = 1.

%t Block[{nn = 10^6, s, t, u, v}, s = Prime@ Range[nn]; t = Differences@ s; u = Map[(#2 - #1)/Log[#1] & @@ # &, Partition[Prime@ Range[nn], 2, 1]]; v = Map[Prime@ FirstPosition[u, #][[1]] &, Union@ FoldList[Max, u]]; Boole[! FreeQ[v, s[[FirstPosition[t, #][[1]] ]] ] ] & /@ Union@ FoldList[Max, t]] (* _Michael De Vlieger_, Jul 19 2018 *)

%Y Cf. A000101, A002386, A111870, A111871, A005250.

%K nonn,more

%O 1,1

%A _Rodolfo Ruiz-Huidobro_, Jul 16 2018

%E a(81) from _Rodolfo Ruiz-Huidobro_, Jan 23 2024

%E a(82) from _Rodolfo Ruiz-Huidobro_, May 10 2024