OFFSET
1,4
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3, 22, 25, 38, 101, 273.
(ii) Each n = 2, 3, ... can be written as k + m with k > 0 and m > 0 such that 6*k - 1, 6*k + 1 and prime(prime(m)) + 2 are all prime.
(iii) Any integer n > 5 can be written as k + m with k > 0 and m > 0 such that phi(k) - 1, phi(k) + 1 and prime(prime(m)) + 2 are all prime, where phi(.) is Euler's totient function.
(iv) If n > 2 is neither 10 nor 31, then n can be written as k + m with k > 0 and m > 0 such that prime(k) + 2 and prime(prime(prime(m))) + 2 are both prime.
(v) If n > 1 is not equal to 133, then n can be written as k + m with k > 0 and m > 0 such that 6*k - 1, 6*k + 1 and prime(prime(prime(m))) + 2 are all prime.
Clearly, each part of the conjecture implies the twin prime conjecture.
We have verified part (i) for n up to 10^9. See the comments in A237348 for an extension of this part.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Andrei-Lucian Dragoi, The "Vertical" Generalization of the Binary Goldbach's Conjecture as Applied on "Iterative" Primes with (Recursive) Prime Indexes (i-primeths), Journal of Advances in Mathematics and Computer Science (2017), Vol. 25, No. 2, pp. 1-32.
Zhi-Wei Sun, Unification of Goldbach's conjecture and the twin prime conjecture, a message to Number Theory List, Jan. 29, 2014.
Zhi-Wei Sun, Super Twin Prime Conjecture, a message to Number Theory List, Feb. 6, 2014.
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014
EXAMPLE
a(3) = 1 since 3 = 2 + 1 with prime(2) + 2 = 3 + 2 = 5 and prime(prime(1)) + 2 = prime(2) + 2 = 5 both prime.
a(22) = 1 since 22 = 20 + 2 with prime(20) + 2 = 71 + 2 = 73 and prime(prime(2)) + 2 = prime(3) + 2 = 5 + 2 = 7 both prime.
a(25) = 1 since 25 = 2 + 23 with prime(2) + 2 = 3 + 2 = 5 and prime(prime(23)) + 2 = prime(83) + 2 = 431 + 2 = 433 both prime.
a(38) = 1 since 38 = 35 + 3 with prime(35) + 2 = 149 + 2 = 151 and prime(prime(3)) + 2 = prime(5) + 2 = 11 + 2 = 13 both prime.
a(101) = 1 since 101 = 98 + 3 with prime(98) + 2 = 521 + 2 = 523 and prime(prime(3)) + 2 = prime(5) + 2 = 11 + 2 = 13 both prime.
a(273) = 1 since 273 = 2 + 271 with prime(2) + 2 = 3 + 2 = 5 and prime(prime(271)) + 2 = prime(1741) + 2 = 14867 + 2 = 14869 both prime.
MATHEMATICA
pq[n_]:=PrimeQ[Prime[n]+2]
PQ[n_]:=PrimeQ[Prime[Prime[n]]+2]
a[n_]:=Sum[If[pq[k]&&PQ[n-k], 1, 0], {k, 1, n-1}]
Table[a[n], {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 05 2014
STATUS
approved