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

%I #41 Sep 08 2022 08:44:52

%S 1,3,7,11,15,19,27,31,35,39,47,55,59,67,71,75,87,91,99,111,115,119,

%T 127,131,151,155,159,167,171,175,179,187,195,207,211,231,235,239,255,

%U 259,267,279,287,295,299,307,311,319,327,335,339,347,371,375,379,391

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

%C Related to hyperperfect numbers of a certain form.

%C 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

%H J. S. McCranie, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL3/mccranie.html">A study of hyperperfect numbers</a>, J. Int. Seqs. Vol. 3 (2000) #P00.1.3.

%F a(n) = A175505(A003627(n)). - _Jaroslav Krizek_, Aug 01 2010

%e 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

%t Select[Range[500], PrimeQ[(3# + 1)/2] &] (* _Harvey P. Dale_, Jan 15 2011 *)

%o (Magma) [ n: n in [1..400 by 2] | IsPrime((3*n+1) div 2) ];

%o (PARI) is(n)=isprime((3*n+1)/2) \\ _Charles R Greathouse IV_, Feb 20 2017

%Y Cf. A038536, A045309.

%K nonn

%O 1,2

%A _Jud McCranie_

%E Corrected by _Vincenzo Librandi_, Mar 24 2010