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A238134
Number of primes p < n with q = floor((n-p)/4) and prime(q) - q + 1 both prime.
4
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 3, 3, 4, 4, 4, 3, 3, 3, 3, 4, 4, 4, 6, 5, 5, 5, 3, 4, 6, 6, 7, 6, 4, 4, 4, 4, 5, 5, 5, 5, 4, 4, 4, 4, 3, 3, 4, 4, 6, 6, 4, 5, 5, 5, 7, 6, 6, 6, 5, 5, 4, 4, 5, 5, 5, 5, 5, 6, 8, 8, 8, 7, 7, 7, 4, 4, 4, 4
OFFSET
1,11
COMMENTS
Conjecture: Let m > 0 and n > 2*m + 1 be integers. If m = 1 and 2 | n, or m = 3 and n is not congruent to 1 modulo 6, or m = 2, 4, 5, ..., then there is a prime p < n such that q = floor((n-p)/m) and prime(q) - q + 1 are both prime.
In the cases m = 1, 2, this gives refinements of Goldbach's conjecture and Lemoine's conjecture (see also A235189). For m > 2, the conjecture is completely new.
See also A238701 for a similar conjecture involving primes q with q^2 - 2 also prime.
LINKS
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.
EXAMPLE
a(29) = 3 since 7, floor((29-7)/4) = 5 and prime(5) - 5 + 1 = 11 - 4 = 7 are all prime; 17, floor((29-17)/4) = 3 and prime(3) - 3 + 1 = 5 - 2 = 3 are all prime; 19, floor((29-19)/4) = 2 and prime(2) - 2 + 1 = 3 - 1 = 2 are all prime.
MATHEMATICA
PQ[n_]:=PrimeQ[n]&&PrimeQ[Prime[n]-n+1]
a[n_]:=Sum[If[PQ[Floor[(n-Prime[k])/4]], 1, 0], {k, 1, PrimePi[n-1]}]
Table[a[n], {n, 1, 80}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 03 2014
STATUS
approved