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Let P1>=5, P2, P3 be consecutive primes, with P2-P1=2. a(n)=(P1+P2)/12 when P3-P2 sets a record.
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%I #5 Nov 08 2019 18:18:08

%S 1,5,23,33,87,278,495,1293,2027,2690,4245,6773,13283,24938,28893,

%T 44270,67475,139708,224922,315893,971000,1723960,3319792,6228255,

%U 7013717,13194622,25321985,31864375,32163975,65155398,86090027,381175405,452803425

%N Let P1>=5, P2, P3 be consecutive primes, with P2-P1=2. a(n)=(P1+P2)/12 when P3-P2 sets a record.

%C 6*a(n)-1, 6*a(n)+1 are twin primes such that the prime gap immediately following 6*a(n)+1 sets a record. The corresponding gap lengths are provided in A329161.

%e See A329161.

%o (PARI) p1=5; p2=7; r=0; forprime(p3=11, 1e9, if(p2-p1==2, d=p3-p2; if(d>r, r=d; print1((p1+p2)/12, ", "))); p1=p2; p2=p3)

%Y Cf. A002822, A084105, A329158, A329159, A329161, A329164, A329165.

%K nonn

%O 1,2

%A _Hugo Pfoertner_, Nov 08 2019