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A329164 Let P1, P2, P3, P4 be consecutive primes, with P2-P1=P4-P3=2. a(n)=(P1+P2)/12 when P3-P2 sets a new record.

%I A329164 #34 Aug 01 2025 11:05:36
%S A329164 1,23,322,495,3407,8113,28893,139708,716182,2497092,5130198,5761777,
%T A329164 7315173,13194622,145995245,201544467,417649822,566513637,833667068,
%U A329164 2266818768,4710228962,5186737183,5192311957,7454170028,9853412390,11817808908
%N A329164 Let P1, P2, P3, P4 be consecutive primes, with P2-P1=P4-P3=2. a(n)=(P1+P2)/12 when P3-P2 sets a new record.
%C A329164 Position of record gaps with no primes bounded by consecutive pairs of twin primes. The length of the corresponding record gaps (P3-P1)/6 is given by A329165.
%C A329164 In the neighborhood of a(15), the growth of this sequence seems to change notably. See the plot2 graph in the links. Does this signify anything important? - _Peter Munn_, Aug 01 2025
%H A329164 Tomáš Brada and Natalia Makarova, <a href="/A329164/b329164.txt">Table of n, a(n) for n = 1..58</a>
%H A329164 Peter Munn, <a href="https://oeis.org/plot2a?name1=A329164&amp;name2=A000045&amp;tform1=log+base+10&amp;tform2=log+base+10&amp;shift=5&amp;radiop1=ratio&amp;drawlines=true">Plot2 graph of sequence against Fibonacci growth</a>
%e A329164 Values of P1, P2, P3, P4 corresponding to record gaps:
%e A329164   P3-P1 P1   P2   P3   P4                   a(n)
%e A329164    6     5    7   11   13        (5+7)/12 =   1
%e A329164   12   137  139  149  151    (137+139)/12 =  23
%e A329164   18  1931 1933 1949 1951  (1931+1933)/12 = 322
%e A329164   30  2969 2971 2999 3001  (2969+2971)/12 = 495
%o A329164 (PARI) p1=3;p2=5;p3=7;r=0;forprime(p4=11,1e9,if(p2-p1==2&&p4-p3==2,d=p3-p2;if(d>r,r=d;print1((p1+p2)/12,", ")));p1=p2;p2=p3;p3=p4)
%Y A329164 Cf. A053778, A329158, A329159, A329160, A329161, A329165.
%K A329164 nonn
%O A329164 1,2
%A A329164 _Hugo Pfoertner_, Nov 07 2019