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A236531 a(n) = |{0 < k < n: {6*k -1 , 6*k + 1} and {prime(n-k), prime(n-k) + 2} are both twin prime pairs}|. 10
0, 0, 1, 2, 2, 2, 2, 3, 2, 3, 1, 4, 2, 3, 4, 1, 3, 2, 3, 5, 2, 4, 3, 2, 4, 1, 5, 4, 3, 5, 3, 3, 4, 3, 7, 5, 4, 7, 1, 7, 1, 5, 8, 3, 8, 5, 5, 5, 3, 9, 6, 6, 7, 4, 6, 3, 5, 8, 6, 7, 5, 6, 4, 5, 7, 7, 6, 5, 4, 4, 6, 5, 7, 6, 9, 3, 5, 5, 5, 6, 5, 8, 5, 5, 6, 5, 7, 4, 5, 10, 3, 7, 5, 6, 3, 4, 7, 5, 6, 6 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 2.
(ii) If n > 3 is neither 11 nor 125, then n can be written as k + m with k > 0 and m > 0 such that 6*k - 1, 6*k + 1, prime(m) + 2 and 3*prime(m) - 10 are all prime.
(iii) Any integer n > 458 can be written as p + q with q > 0 such that {p, p + 2} and {prime(q), prime(q) + 2} are both twin prime pairs.
This is much stronger than the twin prime conjecture. We have verified part (i) of the conjecture for n up to 2*10^7.
LINKS
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014
EXAMPLE
a(11) = 1 since {6*1 - 1, 6*1 + 1} = {5, 7} and {prime(10), prime(10) + 2} = {29, 31} are both twin prime pairs.
a(16) = 1 since {6*3 - 1, 6*3 + 1} = {17, 19} and {prime(13), prime(13) + 2} = {41, 43} are both twin prime pairs.
MATHEMATICA
p[n_]:=PrimeQ[6n-1]&&PrimeQ[6n+1]
q[n_]:=PrimeQ[Prime[n]+2]
a[n_]:=Sum[If[p[k]&&q[n-k], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 100}]
CROSSREFS
Sequence in context: A199800 A338094 A165035 * A332334 A217403 A081309
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 27 2014
STATUS
approved

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Last modified May 10 16:57 EDT 2024. Contains 372388 sequences. (Running on oeis4.)