|
|
A034934
|
|
Numbers k such that (3*k + 1)/2 is prime.
|
|
2
|
|
|
1, 3, 7, 11, 15, 19, 27, 31, 35, 39, 47, 55, 59, 67, 71, 75, 87, 91, 99, 111, 115, 119, 127, 131, 151, 155, 159, 167, 171, 175, 179, 187, 195, 207, 211, 231, 235, 239, 255, 259, 267, 279, 287, 295, 299, 307, 311, 319, 327, 335, 339, 347, 371, 375, 379, 391
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Related to hyperperfect numbers of a certain form.
The formula by J. Krizek is explained as follows: If p=(3n+1)/2 is prime, then it is an integer, and p must be of the form p=3m-1, i.e., p=A003627(k). On the other hand, if p=A003627(k), then all k < p are coprime to p, so we have B(p) = (Sum_{k<p} k^2)/(Sum_{k<p} k) = (2p-1)/3. This is an integer, since p=3m-1, and for this number n = (2p-1)/3 = 2m-1, we have that (3n+1)/2 = p is prime. - M. F. Hasler, Nov 29 2010
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
|
|
MATHEMATICA
|
Select[Range[500], PrimeQ[(3# + 1)/2] &] (* Harvey P. Dale, Jan 15 2011 *)
|
|
PROG
|
(Magma) [ n: n in [1..400 by 2] | IsPrime((3*n+1) div 2) ];
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|