

A103828


Sequence of odd numbers defined recursively by: a(1)=1 and a(n) is the first odd number greater than a(n1) such that a(n) + a(i) + 1 is prime for 1<=i<=n1.


9



1, 3, 9, 27, 69, 429, 1059, 56499, 166839, 5020059, 7681809, 274343589, 8316187179, 2866819175649, 7180244842749, 216549352241349, 22129340663539629, 2504509324460255499
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

Is the sequence infinite? Is each prime a(i)+a(j)+1, i<>j, always distinct?
The HardyLittlewood ktuple conjecture would imply that this sequence is infinite. Note that, for n>2, a(n)+2 and a(n)+4 are both primes, so a proof that this sequence is infinite would also show that there are infinitely many twin primes.  N. J. A. Sloane, Apr 21 2007
From the mod 30 property of A115760 we conclude that a(n) == 9 (mod 15) for n>4. This implies that either a(n) == 9 (mod 30) or == 24 (mod 30), but == 24 (mod 30) is impossible because then == 0 (mod 6). Therefore a(n) == 9 (mod 30) for n>4.  Don Reble, Aug 17 2021


LINKS



FORMULA



EXAMPLE

a(1)=1, a(2)=3, but 5+1+1=7, 5+3+1=9; 7+1+1=9, 7+3+1=11; 9+1+1=11, 9+3+1=13 so a(3)=9.


MAPLE

EP:=[]: for w to 1 do for n from 1 to 8*10^6 do s:=2*n1; Q:=map(z>z+s+1, EP); if andmap(isprime, Q) then EP:=[op(EP), s]; print(nops(EP), s); fi od od; EP;


MATHEMATICA

a[1] = 1; a[2] = 3; a[n_] := a[n] = Block[{k = a[n  1] + 6, t = Table[ a[i], {i, n  1}] + 1}, While[ First@ Union@ PrimeQ[k + t] == False, k += 6]; k]; Do[ Print[ a[n]], {n, 15}]  Robert G. Wilson v, Jun 03 2006


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



EXTENSIONS



STATUS

approved



