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A233654
|{prime p < n: n - p = sigma(k) for some k > 0}|, where sigma(k) is the sum of all (positive) divisors of k.
5
0, 0, 1, 1, 1, 3, 1, 3, 3, 3, 3, 2, 2, 4, 4, 3, 5, 4, 4, 6, 4, 3, 5, 3, 6, 5, 5, 1, 4, 4, 7, 5, 6, 4, 8, 3, 8, 5, 5, 2, 9, 5, 10, 8, 8, 4, 10, 3, 11, 6, 10, 2, 8, 4, 11, 5, 8, 3, 12, 5, 13, 7, 7, 3, 13, 3, 14, 7, 7, 5, 12, 3, 14, 9, 11, 6, 12, 2, 16, 7, 11, 5, 12, 3, 18, 8, 12, 2, 11, 3, 19, 6, 11, 4, 13, 4, 17, 8, 10, 6
OFFSET
1,6
COMMENTS
Conjecture: (i) Let n > 1 be an integer. Then we have a(2*n) > 0. Also, 2*n + 1 can be written as p + sigma(k), where p is a Sophie Germain prime and k is a positive integer.
(ii) Each odd number greater than one can be written as sigma(k^2) + phi(m), where k and m are positive integers, and phi(.) is Euler's totient function.
That a(2*n+1) > 0 for n > 1 is a consequence of Goldbach's conjecture, for, if 2*n = p + q with p and q both prime, then 2*n + 1 = p + sigma(q) = q + sigma(p).
EXAMPLE
a(3) = 1 since 3 = 2 + 1 = 2 + sigma(1) with 2 prime.
a(7) = 1 since 7 = 3 + 4 = 3 + sigma(3) with 3 prime.
a(10) = 3 since 10 = 2 + 8 = 2 + sigma(7) with 2 prime, 10 = 3 + 7 = 3 + sigma(4) with 3 prime, and 10 = 7 + 3 = 7 + sigma(2) with 7 prime.
a(13) = 2 since 13 = 5 + 8 = 5 + sigma(7) with 5 prime, and 13 = 7 + 6 = 7 + sigma(5) with 7 prime.
a(28) = 1 since 28 = 13 + 15 = 13 + sigma(8) with 13 prime.
a(36) = 3 since 36 = 5 + 31 = 5 + sigma(16) = 5 + sigma(25) with 5 prime, 36 = 23 + 13 = 23 + sigma(9) with 23 prime, and 36 = 29 + 7 = 29 + sigma(4) with 29 prime.
a(148) = 1 since 148 = 109 + 39 = 109 + sigma(18) with 109 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[n-Part[S[n], i]], 1, 0], {i, 1, Length[S[n]]}]
Table[a[n], {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 14 2013
STATUS
approved