A238701 -id:A238701 - cp's OEIS frontend

A238134 Number of primes p < n with q = floor((n-p)/4) and prime(q) - q + 1 both prime.

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

Zhi-Wei Sun, Mar 03 2014

nonn

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.

			 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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Cf. A000040, A045917, A046927, A234694, A234695, A235189, A237705, A238701.

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}]

A238732 Number of primes p < n with floor((n-p)/3) a square.

Original entry on oeis.org

0, 0, 1, 2, 2, 3, 3, 3, 2, 2, 1, 2, 1, 3, 4, 4, 3, 3, 3, 3, 3, 2, 2, 3, 3, 2, 2, 1, 2, 4, 5, 5, 4, 4, 3, 3, 1, 2, 2, 3, 3, 4, 3, 4, 4, 4, 2, 3, 2, 4, 5, 4, 3, 3, 5, 4, 4, 3, 3, 4, 4, 3, 3, 3, 4, 4, 3, 3, 3, 3, 4, 5, 4, 4, 4, 3, 3, 4, 5, 6

Offset: 1

Zhi-Wei Sun, Mar 03 2014

nonn

Conjecture: (i) For any integers m > 2 and n > 2, there is a prime p < n such that floor((n-p)/m) is a square.
(ii) For any integer k > 2, if m and n are sufficiently large, then there is a prime p < n such that floor((n-p)/m) is a k-th power. In particular, for any integer n > 2, there is a prime p < n such that floor((n-p)/5) is a cube.
We have verified part (i) for m = 3, ..., 10 and n = 3, ..., 10^6.

			a(3) = 1 since 2 is prime with floor((3-2)/3) = 0^2.
a(11) = 1 since 7 is prime with floor((11-7)/3) = 1^2.
a(13) = 1 since 13 is prime with floor((13-11)/3) = 0^2.
a(28) = 1 since 23 is prime with floor((28-23)/3) = 1^2.
a(37) = 1 since 23 is prime with floor((37-23)/3) = 2^2.
a(173) = 1 since 97 is prime with floor((173-97)/3) = 5^2.
It seems that a(n) = 1 only for n = 3, 11, 13, 28, 37, 173.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Cf. A000040, A000290, A237706, A238701, A238134.

SQ[n_]:=IntegerQ[Sqrt[n]]
s[n_,k_]:=SQ[Floor[(n-Prime[k])/3]]
a[n_]:=Sum[If[s[n,k],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]

A238703 a(n) = |{0 < k < n: floor(k*n/3) is prime}|.

Original entry on oeis.org

0, 0, 1, 1, 2, 1, 3, 3, 1, 3, 4, 0, 4, 2, 1, 3, 5, 0, 4, 4, 1, 4, 5, 0, 3, 4, 0, 3, 6, 0, 5, 4, 1, 6, 6, 0, 7, 4, 1, 5, 4, 0, 7, 6, 0, 8, 5, 0, 8, 7, 1, 6, 7, 0, 9, 9, 1, 9, 8, 0, 6, 7, 0, 7, 12, 0, 9, 7, 1, 11, 10, 0, 6, 8, 0, 7, 9, 0, 7, 12

Offset: 1

Zhi-Wei Sun, Mar 03 2014

nonn

Conjecture: If n > m > 0 with n not divisible by m, then floor(k*n/m) is prime for some 0 < k < n.

			a(4) = 1 since floor(2*4/3) = 2 is prime.
If p is a prime, then a(3*p) = 1 since floor(k*3p/3) = k*p is prime only for k = 1. If m > 1 is composite, then a(3*m) = 0 since floor(k*3m/3) = k*m is composite for all k > 0.

Zhi-Wei Sun, Table of n, a(n) for n = 1..6000
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Cf. A000040, A237578, A237615, A238645, A238701.

p[n_,k_]:=PrimeQ[Floor[k*n/3]]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
Table[a[n],{n,1,80}]

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A238134 Number of primes p < n with q = floor((n-p)/4) and prime(q) - q + 1 both prime.

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A238732 Number of primes p < n with floor((n-p)/3) a square.

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A238703 a(n) = |{0 < k < n: floor(k*n/3) is prime}|.

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