|
| |
|
|
A260120
|
|
Least integer k > 0 such that (prime(k*n)-1)^2 = prime(j*n)-1 for some j > 0.
|
|
8
|
|
|
|
1, 2, 14, 1, 12, 9, 30, 198, 69, 83, 66, 132, 44, 15, 4, 99, 71, 88, 339, 230, 10, 33, 167, 66, 42, 22, 126, 442, 318, 1185, 29, 289, 37, 174, 157, 44, 146, 301, 171, 403, 2, 5, 26, 699, 573, 144, 338, 33, 2032, 1212, 404, 11, 135, 267, 380, 221, 447, 159, 898, 1397
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Conjecture: a(n) exists for any n > 0. In general, if a,b,c and m are integers with a > 0, gcd(a,b,c-m) = 1 and c == (a+b+1)*(m+1) (mod 2) such that b^2-4a*(c-m) is not a square and gcd(a*m-b,b^2+b-a*c-1) is not divisible by 3, then for any positive integer n there are two elements x and y of the set {prime(k*n)+m: k = 1,2,3,...} with a*x^2+b*x+c = y.
This implies the conjecture in A259731.
|
|
|
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) = 14 since (prime(14*3)-1)^2 = 180^2 = prime(3477)-1 = prime(1159*3)-1.
a(63) = 5162 since (prime(5162*63)-1)^2 = 4642456^2 = 21552397711936 = prime(726521033763)-1 = prime(11532079901*63)-1.
|
|
|
MATHEMATICA
|
P[n_, p_]:=PrimeQ[p]&&Mod[PrimePi[p], n]==0
Do[k=0; Label[aa]; k=k+1; If[P[n, (Prime[k*n]-1)^2+1], Goto[bb]]; Goto[aa]; Label[bb]; Print[n, " ", k]; Continue, {n, 1, 60}]
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
