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.

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.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

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}]

Crossrefs

Cf. A000040, A000203, A002372, A002375, A232270, A233544, A233654, A233793.

Sequence in context: A364027 A308967 A241844 * A133232 A137152 A328902

Adjacent sequences: A233861 A233862 A233863 * A233865 A233866 A233867

Keyword

nonn

Author

Zhi-Wei Sun, Dec 16 2013

Status

approved