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A034934
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Numbers n such that (3*n + 1)/2 is prime.
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3
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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
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1,2
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COMMENTS
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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). OTOH, if p=A003627(k), then all k<p are coprime to p, so we have B(p) = sum(k^2, k<^p)/sum(k, k<p) = (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
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LINKS
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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.
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FORMULA
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a(n) = A175505(A003627(n)). - Jaroslav Krizek, Aug 01 2010
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EXAMPLE
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a(6) = 19 because for A003627(6) = 29 holds: B(29) = A053818(29) / A023896(29) = 7714 / 406 = 19. Cf. A179871-A179891, A003627, A007645. - Jaroslav Krizek, Aug 01 2010
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MATHEMATICA
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Select[Range[500], PrimeQ[(3# + 1)/2] &] (* Harvey P. Dale, Jan 15 2011 *)
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PROG
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(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
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CROSSREFS
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Cf. A038536, A045309.
Sequence in context: A194442 A220522 A220526 * A191151 A220494 A194440
Adjacent sequences: A034931 A034932 A034933 * A034935 A034936 A034937
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KEYWORD
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nonn
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AUTHOR
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Jud McCranie
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EXTENSIONS
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Corrected by Vincenzo Librandi, Mar 24 2010
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STATUS
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approved
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