A034934 - OEIS

A034934

Numbers k such that (3*k + 1)/2 is prime.

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

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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

Table of n, a(n) for n=1..56.

J. S. McCranie, A study of hyperperfect numbers, J. Int. Seqs. Vol. 3 (2000) #P00.1.3.

FORMULA

a(n) = A175505(A003627(n)). - Jaroslav Krizek, Aug 01 2010

EXAMPLE

a(6) = 19 because for A003627(6) = 29, B(29) = A053818(29)/A023896(29) = 7714/406 = 19. Cf. A179871-A179891, A003627, A007645. - Jaroslav Krizek, Aug 01 2010

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) ];

(PARI) is(n)=isprime((3*n+1)/2) \\ Charles R Greathouse IV, Feb 20 2017