login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A233204
Number of ways to write n = k + m with 0 < k < m such that 2^k * prime(m) + 3 is prime.
7
0, 0, 0, 1, 2, 1, 3, 1, 4, 3, 2, 2, 2, 4, 3, 2, 6, 3, 2, 1, 8, 1, 2, 2, 4, 7, 2, 5, 6, 8, 5, 4, 4, 8, 3, 5, 2, 7, 5, 8, 5, 3, 4, 4, 4, 8, 6, 2, 4, 3, 7, 7, 3, 4, 7, 5, 3, 4, 6, 8, 4, 2, 6, 6, 4, 7, 7, 5, 7, 7, 6, 6, 2, 7, 8, 7, 7, 5, 11, 3, 4, 8, 2, 7, 8, 6, 9, 7, 6, 10, 11, 4, 5, 8, 4, 8, 8, 6, 7, 9
OFFSET
1,5
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 3. Also, any integer n > 2 can be written as k + m (0 < k <= m) with 2^k*prime(m) - 3 prime.
(ii) Any integer n > 6 can be written as k + m (0 < k < m) with prime(k) + 6 and prime(m) + 6 both prime. Each integer n > 4 can be written as k + m (0 < k < m) with prime(k) + 2 and prime(m) + 6 both prime. Also, for every integer n > 3 not among 11, 21, 32, 49, 171, there is a positive integer k < n with prime(k) + 2 and prime(n-k) + 2 both prime.
LINKS
MAPLE
a(6) = 1 since 6 = 2 + 4 with 2^2*prime(4) + 3 = 4*7 + 3 = 31 prime.
a(22) = 1 since 22 = 1 + 21 with 2^1*prime(21) + 3 = 2*73 + 3 = 149 prime.
MATHEMATICA
a[n_]:=Sum[If[PrimeQ[2^k*Prime[n-k]+3], 1, 0], {k, 1, (n-1)/2}]
Table[a[n], {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 05 2013
STATUS
approved