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.

1, 23, 322, 495, 3407, 8113, 28893, 139708, 716182, 2497092, 5130198, 5761777, 7315173, 13194622, 145995245, 201544467, 417649822, 566513637, 833667068, 2266818768, 4710228962, 5186737183, 5192311957, 7454170028, 9853412390, 11817808908

Offset: 1

Author

Hugo Pfoertner, Nov 07 2019

Keywords

• nonn

Comments

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.

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

Examples

			Values of P1, P2, P3, P4 corresponding to record gaps:
  P3-P1 P1   P2   P3   P4                   a(n)
   6     5    7   11   13        (5+7)/12 =   1
  12   137  139  149  151    (137+139)/12 =  23
  18  1931 1933 1949 1951  (1931+1933)/12 = 322
  30  2969 2971 2999 3001  (2969+2971)/12 = 495
		

Links

• Tomáš Brada and Natalia Makarova, Table of n, a(n) for n = 1..58
• Peter Munn, Plot2 graph of sequence against Fibonacci growth

Crossrefs

Cf. A053778, A329158, A329159, A329160, A329161, A329165.

Programs

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)