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A233864
a(n) = |{0 < m < 2*n: m = sigma(k) for some k > 0, and 2*n - 1 - m and 2*n - 1 + m are both prime}|, where sigma(k) is the sum of all (positive) divisors of k.
3
0, 0, 0, 1, 1, 2, 1, 2, 3, 1, 1, 3, 3, 3, 3, 2, 4, 5, 3, 4, 4, 4, 4, 4, 3, 5, 4, 5, 4, 5, 3, 4, 7, 4, 5, 6, 4, 8, 8, 4, 4, 4, 7, 5, 6, 5, 6, 8, 4, 6, 8, 6, 7, 6, 6, 5, 5, 9, 7, 9, 7, 6, 8, 7, 7, 8, 6, 9, 9, 6, 6, 12, 9, 6, 10, 8, 9, 12, 7, 7, 11, 5, 10, 9, 9, 10, 7, 11, 8, 9, 6, 8, 14, 10, 8, 8, 10, 12, 9, 6
OFFSET
1,6
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 3.
(ii) For any even number 2*n > 0, 2*n + sigma(k) is prime for some 0 < k < 2*n.
See also A233793 for a related conjecture.
Clearly part (i) of the conjecture implies Goldbach's conjecture for even numbers 2*(2*n - 1) with n > 3; we have verified part (i) for n up to 10^8. Concerning part (ii), we remark that 1024 is the unique positive integer k < 1134 with 1134 + sigma(k) prime, and that sigma(1024) = 2047 > 1134.
EXAMPLE
a(7) = 1 since sigma(5) = 6, and 2*7 - 1 - 6 = 7 and 2*7 - 1 + 6 = 19 are both prime.
a(10) = 1 since sigma(6) = sigma(11) = 12, and 2*10 - 1 - 12 = 7 and 2*10 - 1 + 12 = 31 are both prime.
a(11) = 1 since sigma(7) = 8, and 2*11 - 1 - 8 = 13 and 2*11 - 1 + 8 = 29 are both prime.
MATHEMATICA
f[n_]:=Sum[If[Mod[n, d]==0, d, 0], {d, 1, n}]
S[n_]:=Union[Table[f[j], {j, 1, n}]]
PQ[n_]:=n>0&&PrimeQ[n]
a[n_]:=Sum[If[PQ[2n-1-Part[S[2n-1], i]]&&PQ[2n-1+Part[S[2n-1], i]], 1, 0], {i, 1, Length[S[2n-1]]}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 16 2013
STATUS
approved