A329164 - OEIS

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 #31 May 25 2020 07:56:30

%S 1,23,322,495,3407,8113,28893,139708,716182,2497092,5130198,5761777,

%T 7315173,13194622,145995245,201544467,417649822,566513637,833667068,

%U 2266818768,4710228962,5186737183,5192311957,7454170028,9853412390,11817808908

%N 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 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.

%H Tomáš Brada and Natalia Makarova, <a href="/A329164/b329164.txt">Table of n, a(n) for n = 1..58</a>

%e Values of P1, P2, P3, P4 corresponding to record gaps:

%e P3-P1 P1 P2 P3 P4 a(n)

%e 6 5 7 11 13 (5+7)/12 = 1

%e 12 137 139 149 151 (137+139)/12 = 23

%e 18 1931 1933 1949 1951 (1931+1933)/12 = 322

%e 30 2969 2971 2999 3001 (2969+2971)/12 = 495

%o (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 Cf. A053778, A329158, A329159, A329160, A329161, A329165.

%K nonn

%O 1,2

%A _Hugo Pfoertner_, Nov 07 2019