%I #9 Oct 16 2015 06:03:56
%S 7,5,251,353,137,2393,109,1931,1753,883,3733,7351,12007,2969,8887,
%T 27697,1321,22811,38377,62987,183823,15679,124001,180563,45887,48677,
%U 100847,178693,152993,557087,34057,367949,294551,134507,173357,1802407,531359,1134311,933067
%N Smallest prime starting a sequence of 4 consecutive odd primes such that the center of the symmetrical gaps is 2n.
%C The sequence is generalizable with primes starting a sequence of 2k consecutive odd primes.
%C Conjecture: a(n) exists for all n>0.
%e a(2)=5 because the 4 consecutive primes 5, 7, 11, 13 have gaps 2, 4, 2, which is symmetric about its center 4 = 2*2.
%p with(numtheory):nn:=500000:l:=2:T:=array(1..2*l-1)):
%p for n from 1 to 35 do:ii:=0:
%p for k from 1 to nn while(ii=0) do:
%p lst:={}:lst1:={}:
%p for m from 1 to 2*l do:
%p lst:=lst union {ithprime(k+m-1)}
%p od:
%p for p from 1 to 2*l do:
%p lst1:=lst1 union {lst[p]+lst[2*l-p+1]}
%p od:
%p n0:=nops(lst1):
%p if n0=1
%p then
%p for a from 1 to 2*l-1 do:
%p T[a]:=lst[a+1]-lst[a]:
%p od:
%p if T[2]=2*n then ii:=1:printf(`%d, `,lst[1]):
%p else fi :fi:
%p od :
%p od:
%o (PARI) a(n) = {pa = 3; pb = 5; pc = 7; forprime(p=8, , if (((pc-pb) == 2*n) && ((pb-pa) == (p-pc)), return(pa)); pa = pb; pb = pc; pc = p;);} \\ _Michel Marcus_, Oct 16 2015
%Y Cf. A001223, A055380, A055381, A055382, A175309.
%K nonn
%O 1,1
%A _Michel Lagneau_, Oct 11 2015