|
|
A260753
|
|
Least positive integer k such that both k and k*n belong to the set {m>0: prime(prime(m))-prime(m)+1 = prime(p) for some prime p}.
|
|
2
|
|
|
2, 2, 2279, 5806, 4, 1135, 816, 6556, 725, 2, 1333, 10839, 27, 829, 2279, 2838, 3881, 6540, 2564, 2, 7830, 6540, 27, 4905, 6121, 8220, 316, 1061, 2, 14691, 2, 1168, 738, 4707, 785, 12467, 5492, 1447, 542, 538, 12840, 829, 4732, 5637, 785, 1246, 1198, 433, 58, 573, 26280, 17387, 316, 430, 1198, 4315, 4315, 1479, 4315, 1497
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Conjecture: For any s and t in the set {1,-1}, every positive rational number r can be written as m/n with m and n in the set {k>0: prime(prime(k))+s*prime(k)+t = prime(p) for some prime p}.
In the case s = -1 and t = 1, the conjecture implies that A261136 has infinitely many terms.
|
|
REFERENCES
|
Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.
|
|
LINKS
|
|
|
EXAMPLE
|
a(3) = 2279 since prime(prime(2279))-prime(2279)+1 = prime(20147)-20147+1 = 226553-20146 = 206407 = prime(18503) with 18503 prime, and prime(prime(2279*3))-prime(2279*3)+1 = prime(68777)-68777+1 = 865757-68776 = 796981 = prime(63737) with 63737 prime.
|
|
MATHEMATICA
|
f[n_]:=Prime[Prime[n]]-Prime[n]+1
PQ[p_]:=PrimeQ[p]&&PrimeQ[PrimePi[p]]
Do[k=0; Label[bb]; k=k+1; If[PQ[f[k]]&&PQ[f[k*n]], Goto[aa], Goto[bb]]; Label[aa]; Print[n, " ", k]; Continue, {n, 1, 60}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|