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A234694 a(n) = |{0 < k < n: p = k + prime(n-k) and prime(p) - p + 1 are both prime}|. 22
0, 1, 0, 2, 1, 2, 1, 0, 0, 2, 2, 4, 1, 1, 2, 4, 2, 1, 1, 2, 3, 3, 2, 3, 1, 1, 1, 3, 5, 4, 3, 4, 3, 3, 3, 2, 4, 3, 2, 5, 4, 4, 4, 1, 1, 5, 4, 2, 1, 2, 5, 5, 2, 3, 4, 2, 3, 5, 7, 7, 6, 2, 5, 6, 2, 5, 4, 4, 7, 6, 6, 5, 4, 8, 7, 4, 5, 3, 5, 7, 3, 5, 4, 7, 6, 7, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 9. Also, for any integer n > 51 there is a positive integer k < n such that p = k + prime(n-k) and prime(p) + p + 1 are both prime.
(ii) If n > 9 (or n > 21), then there is a positive integer k < n such that m - 1 and prime(m) + m (or prime(m) - m, resp.) are both prime, where m = k + prime(n-k).
(iii) If n > 483, then for some 0 < k < n both prime(m) + m and prime(m) - m are prime, where m = k + prime(n-k).
(iv) If n > 3, then there is a positive integer k < n such that prime(k + prime(n-k)) + 2 is prime.
Clearly, part (i) of the conjecture implies that there are infinitely many primes p with prime(p) - p + 1 (or prime(p) + p + 1) also prime.
See A234695 for primes p with prime(p) - p + 1 also prime.
LINKS
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014
EXAMPLE
a(5) = 1 since 2 + prime(3) = 7 and prime(7) - 6 = 11 are both prime.
a(25) = 1 since 20 + prime(5) = 31 and prime(31) - 30 = 97 are both prime.
a(27) = 1 since 18 + prime(9) = 41 and prime(41) - 40 = 139 are both prime.
a(45) = 1 since 6 + prime(39) = 173 and prime(173) - 172 = 859 are both prime.
a(49) = 1 since 26 + prime(23) = 109 and prime(109) - 108 = 491 are both prime.
MATHEMATICA
f[n_, k_]:=k+Prime[n-k]
q[n_, k_]:=PrimeQ[f[n, k]]&&PrimeQ[Prime[f[n, k]]-f[n, k]+1]
a[n_]:=Sum[If[q[n, k], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
CROSSREFS
Sequence in context: A105661 A082451 A121362 * A091704 A175799 A123739
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 29 2013
STATUS
approved

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Last modified April 26 21:53 EDT 2024. Contains 372004 sequences. (Running on oeis4.)