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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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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). 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
| J. S. McCranie, A study of hyperperfect numbers, J. Int. Seqs. Vol. 3 (2000) #P00.1.3
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FORMULA
| a(n) = A175505(A003627(n)). [From Jaroslav Krizek (jaroslav.krizek(AT)atlas.cz), Aug 01 2010]
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EXAMPLE
| a(6) = 19 because for A003627(6) = 29 holds: B(29) = A053818(29) / A023896(29) = 7714 / 406 = 19. Cf. A179871-A179891, A003627, A007645. [From Jaroslav Krizek (jaroslav.krizek(AT)atlas.cz), Aug 01 2010]
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MATHEMATICA
| Select[Range[500], PrimeQ[(3# + 1)/2] &] (* From Harvey P. Dale, Jan 15 2011 *)
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PROG
| (MAGMA) [ n: n in [1..400 by 2] | IsPrime((3*n+1) div 2) ];
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CROSSREFS
| Cf. A038536 and A045309.
Sequence in context: A039957 A079422 A194442 * A191151 A194440 A189364
Adjacent sequences: A034931 A034932 A034933 * A034935 A034936 A034937
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KEYWORD
| nonn
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AUTHOR
| Jud McCranie (JudMcCranie(AT)ugaalum.uga.edu)
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EXTENSIONS
| Corrected by Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Mar 24 2010
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