

A222566


a(1)=2; for n>0, a(n+1) is the least prime p>a(n) such that 2*(a(n)+1)p is prime.


5



2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, 43, 47, 53, 61, 71, 73, 89, 97, 107, 109, 113, 127, 149, 151, 167, 173, 181, 191, 193, 197, 199, 227, 229, 233, 239, 241, 251, 263, 271, 281, 283, 311, 313, 317, 353, 359, 367, 383, 389, 397, 443, 449, 457, 467, 479, 499, 509, 521, 523, 557, 569, 571, 587, 599, 601, 617, 619, 641, 643, 647, 653, 661, 677, 683, 691, 701, 727, 773, 787, 857, 859, 863, 907, 929, 941, 947, 967, 983, 991, 1013, 1019, 1021, 1031, 1033, 1049, 1051, 1091, 1093
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OFFSET

1,1


COMMENTS

For a prime p define g(p) as the least prime q>p such that 2*(p+1)q is prime. Construct a simple (undirected) graph G as follows: The vertex set of G is the set of all primes, and for the vertices p and q>p there is an edge connecting p and q if and only if g(p)=q. Clearly G contains no cycle.
Conjecture: The graph G constructed above is connected and hence it is a tree!


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.


EXAMPLE

a(2)=3 since 2(2+1)=3+3, and a(3)=5 since 2(3+1)=5+3.


MATHEMATICA

k=1
n=1
Do[If[m==1, Print[n, " ", 2]]; If[m==k, n=n+1; Do[If[PrimeQ[2(Prime[m]+1)Prime[j]]==True, k=j; Print[n, " ", Prime[j]]; Goto[aa]], {j, m+1, PrimePi[2Prime[m]]}]];
Label[aa]; Continue, {m, 1, 1000}]


CROSSREFS

Cf. A000040, A222532, A163846, A163847.
Sequence in context: A296915 A171032 A171045 * A299171 A002267 A178762
Adjacent sequences: A222563 A222564 A222565 * A222567 A222568 A222569


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Feb 25 2013


STATUS

approved



