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A238881
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Number of odd primes p < 2*n with prime(n*(p+1)/2) + n*(p+1)/2 prime.
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3
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0, 1, 1, 0, 0, 2, 1, 3, 1, 2, 2, 5, 2, 3, 2, 4, 3, 5, 2, 3, 2, 4, 1, 8, 4, 4, 4, 3, 2, 5, 2, 4, 4, 4, 4, 8, 2, 4, 2, 5, 4, 6, 3, 2, 6, 6, 3, 11, 6, 10, 4, 8, 2, 11, 4, 7, 4, 7, 2, 12, 4, 6, 2, 6, 3, 8, 3, 5, 8, 12, 6, 12, 4, 15, 8, 11, 5, 12, 2, 11
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OFFSET
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1,6
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COMMENTS
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Conjecture: (i) a(n) > 0 for all n > 5, and a(n) = 1 only for n = 2, 3, 7, 9, 23. Moreover, for any r = 1,-1 and n > 5*(2+r) there is a positive integer k < n such that 2*k+r and prime(k*n)+k*n are both prime.
(ii) If n > 1 is not equal to 13, then prime(k*n) - k*n is prime for some k = 1, ..., n.
This conjecture implies that there are infinitely many positive integers m with prime(m) + m (or prime(m) - m) prime.
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REFERENCES
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Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28--Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169--187. (See Conjecture 3.21(i) and note that the typo 2k+1 there should be 2k-1.)
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LINKS
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EXAMPLE
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a(7) = 1 since 11 and prime(7*(11+1)/2) + 7*(11+1)/2 = prime(42) + 42 = 181 + 42 = 223 are both prime.
a(23) = 1 since 7 and prime(23*(7+1)/2) + 23*(7+1)/2 = prime(92) + 92 = 479 + 92 = 571 are both prime.
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MATHEMATICA
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PQ[n_]:=PrimeQ[Prime[n]+n]
p[k_, n_]:=PQ[(Prime[k]+1)/2*n]
a[n_]:=Sum[If[p[k, n], 1, 0], {k, 2, PrimePi[2n-1]}]
Table[a[n], {n, 1, 80}]
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PROG
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(PARI) a(n) = {my(nb = 0); forprime(p=3, 2*n, if (isprime(prime(n*(p+1)/2) + n*(p+1)/2), nb++); ); nb; } \\ Michel Marcus, Sep 21 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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