A237705 -id:A237705 - cp's OEIS frontend

A237706 Number of primes p < n with pi(n-p) a square, where pi(.) is given by A000720.

0, 0, 1, 2, 1, 1, 1, 1, 2, 2, 2, 4, 3, 3, 3, 1, 1, 2, 2, 3, 3, 1, 1, 2, 3, 4, 4, 4, 4, 6, 5, 4, 4, 2, 2, 3, 3, 5, 5, 3, 3, 3, 3, 4, 4, 3, 3, 3, 3, 3, 3, 2, 2, 4, 5, 5, 5, 4, 4, 7, 6, 5, 5, 4, 4, 5, 5, 7, 7, 5

Offset: 1

Zhi-Wei Sun, Feb 11 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3, 5, 6, 7, 8, 16, 17, 22, 23, 148, 149.
(ii) For any integer n > 2, there is a prime p < n with pi(n-p) a triangular number.
We have verified that a(n) > 0 for every n = 3, ..., 1.5*10^7. See A237710 for the least prime p < n with pi(n-p) a square.
See also A237705, A237720 and A237721 for similar conjectures.

			a(8) = 1 since 7 is prime with pi(8-7) = 0^2.
a(16) = 1 since 7 is prime with pi(16-7) = 2^2.
a(149) = 1 since 139 is prime with pi(149-139) = pi(10) = 2^2.
a(637) = 2 since 409 is prime with pi(637-409) = pi(228) = 7^2, and 613 is prime with pi(637-613) = pi(24) = 3^2.

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

Cf. A000040, A000217, A000290, A000720, A237598, A237612, A237705, A237710, A237720, A237721.

SQ[n_]:=IntegerQ[Sqrt[n]]
q[n_]:=SQ[PrimePi[n]]
a[n_]:=Sum[If[q[n-Prime[k]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,70}]

A237768 Number of primes p < n with pi(n-p) a Sophie Germain prime, where pi(.) is given by A000720.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Feb 13 2014

nonn

Conjecture: a(n) > 0 for all n > 4, and a(n) = 1 only for n = 5, 12, 20, 21, 26, 27, 30, 31, 32, 60, 61, 68, 69, 80, 81.
This is stronger than part (i) of the conjecture in A237705.
We have verified that a(n) > 0 for all n = 5, ..., 2*10^7.

			a(5) = 1 since 2, pi(5-2) = pi(3) = 2 and 2*2 + 1 = 5 are all prime.
a(12) = 1 since 7, pi(12-7) = pi(5) = 3 and 2*3 + 1 = 7 are all prime.
a(81) = 1 since 47, pi(81-47) = pi(34) = 11 and 2*11 + 1 = 23 are all prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000040, A000720, A005384, A237284, A237705, A237706.

sg[n_]:=PrimeQ[n]&&PrimeQ[2n+1]
a[n_]:=Sum[If[sg[PrimePi[n-Prime[k]]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]

A237769 Number of primes p < n with pi(n-p) - 1 and pi(n-p) + 1 both prime, where pi(.) is given by A000720.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 3, 2, 2, 3, 3, 3, 4, 3, 4, 4, 2, 2, 2, 2, 4, 4, 2, 2, 2, 2, 3, 3, 1, 1, 2, 2, 3, 4, 3, 3, 4, 3, 5, 5, 3, 3, 2, 2, 5, 5, 3, 3, 3, 3, 5, 5, 2, 2, 3, 3, 3, 4, 2, 2, 6, 6, 9, 8, 4, 4, 3, 3, 6, 6, 5, 5, 4, 4, 7

Offset: 1

Zhi-Wei Sun, Feb 13 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 8, and a(n) = 1 only for n = 9, 34, 35.
(ii) For any integer n > 4, there is a prime p < n such that 3*pi(n-p) - 1, 3*pi(n-p) + 1 and 3*pi(n-p) + 5 are all prime. Also, for each integer n > 8, there is a prime p < n such that 3*pi(n-p) - 1, 3*pi(n-p) + 1 and 3*pi(n-p) - 5 are all prime.
(iii) For any integer n > 6, there is a prime p < n such that phi(n-p) - 1 and phi(n-p) + 1 are both prime, where phi(.) is Euler's totient function.

			a(9) = 1 since 2, pi(9-2) - 1 = 3 and pi(9-2) + 1 = 5 are all prime.
a(34) = 1 since 19, pi(34-19) - 1 = pi(15) - 1 = 5 and pi(34-19) + 1 = pi(15) + 1 = 7 are all prime.
a(35) = 1 since 19, pi(35-19) - 1 = pi(16) - 1 = 5 and pi(35-19) + 1 = pi(16) + 1 = 7 are all prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000010, A000040, A000720, A001359, A006512, A014574, A022004, A022005, A237705, A237706, A237768.

TQ[n_]:=PrimeQ[n-1]&&PrimeQ[n+1]
a[n_]:=Sum[If[TQ[PrimePi[n-Prime[k]]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]

A238278 a(n) = |{0 < k < n: the number of primes in the interval ((k-1)n, kn] and the number of primes in the interval (kn, (k+1)n] are both prime}|.

Original entry on oeis.org

0, 0, 0, 1, 1, 3, 3, 2, 7, 6, 8, 4, 9, 4, 9, 8, 1, 1, 3, 5, 4, 6, 3, 4, 4, 6, 3, 11, 8, 8, 7, 7, 12, 9, 4, 8, 9, 12, 8, 12, 8, 7, 6, 7, 7, 9, 4, 8, 9, 11, 5, 6, 3, 11, 2, 5, 14, 8, 8, 11, 2, 1, 11, 4, 6, 4, 5, 4, 1, 9, 5, 2, 10, 5, 4, 9, 10, 11, 6, 7

Offset: 1

Zhi-Wei Sun, Feb 22 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 3.
(ii) For any integer n > 3, there is a prime p < n such that the number of primes in the interval ((p-1)*n, p*n) is a prime.
We have verified part (i) for n up to 150000.
See also A238277 and A238281 for related conjectures.

			a(17) = 1 since the interval (9*17, 10*17] contains exactly 3 primes with 3 prime, and the interval (10*17, 11*17] contains exactly 3 primes with 3 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..3500
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Cf. A000040, A237578, A237643, A237705, A238224, A238277, A238281.

d[k_,n_]:=PrimePi[k*n]-PrimePi[(k-1)n]
a[n_]:=Sum[If[PrimeQ[d[k,n]]&&PrimeQ[d[k+1,n]],1,0],{k,1,n-1}]
Table[a[n],{n,1,80}]

A237815 Number of primes p < n such that the number of Sophie Germain primes among 1, ..., n-p is a Sophie Germain prime.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Feb 13 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 4.
(ii) For any integer n > 11, there is a prime p < n such that the number of Sophie Germain primes among 1, ..., n-p is a square.
See also A237817 for a similar conjecture involving twin primes.

			a(5) = 1 since there are exactly two Sophie Germain primes not exceeding 5-2 = 3, and 2 is a Sophie Germain prime.

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

Cf. A000040, A005384, A000290, A237705, A237706, A237768, A237817.

sg[n_]:=PrimeQ[n]&&PrimeQ[2n+1]
sum[n_]:=Sum[If[PrimeQ[2Prime[k]+1],1,0],{k,1,PrimePi[n]}]
a[n_]:=Sum[If[sg[sum[n-Prime[k]]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]

A237817 Number of primes p < n such that r = |{q <= n-p: q and q + 2 are both prime}| and r + 2 are both prime.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 2, 3, 3, 4, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 8, 7, 6, 6, 5, 5, 5, 5, 5, 5, 6, 6, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 4, 4, 4

Offset: 1

Zhi-Wei Sun, Feb 13 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 12.
(ii) For any integer n > 2, there is a prime p < n such that r = |{q <= n-p: q and q + 2 are both prime}| is a square.
See also A237815 for a similar conjecture involving Sophie Germain primes.

			a(13) = 1 since {q <= 13 - 2: q and q + 2 are both prime} = {3, 5, 11} has cardinality 3, and {3, 3 + 2} is a twin prime pair.

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

Cf. A000040, A000290, A001359, A006512, A237705, A237706, A237769, A237815.

TQ[n_]:=PrimeQ[n]&&PrimeQ[n+2]
sum[n_]:=Sum[If[PrimeQ[Prime[k]+2],1,0],{k,1,PrimePi[n]}]
a[n_]:=Sum[If[TQ[sum[n-Prime[k]]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]

A237720 Number of primes p <= (n+1)/2 with floor( sqrt(n-p) ) prime.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 3, 2, 2, 2, 2, 1, 1, 2, 2, 2, 3, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 4, 4, 3, 4, 3, 4, 4, 4, 3, 4, 3, 3, 4, 5, 4, 5, 4, 5, 6, 6, 5, 6, 7, 8, 8, 8, 7, 7, 5, 6, 5, 5

Offset: 1

Zhi-Wei Sun, Feb 12 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 5, and a(n) = 1 only for n = 6, 23, 24, 111, 112, ..., 120.
(ii) For any integer n > 2, there is a prime p < n with floor(sqrt(n+p)) prime.
Note that floor(sqrt(n)) is the number of squares among 1, ..., n.
See also A237705, A237706 and A237721 for similar conjectures.

			a(6) = 1 since 2 and floor(sqrt(6-2)) = 2 are both prime.
a(23) = 1 since 11 and floor(sqrt(23-11)) = 3 are both prime.
a(24) = 1 since 11 and floor(sqrt(24-11)) = 3 are both prime.
a(27) = 2 since 2 and floor(sqrt(27-2)) = 5 are both prime, and 13 and floor(sqrt(27-13)) = 3 are both prime.
a(n) = 1 for n = 111, ..., 116 since 53 and floor(sqrt(n-53)) = 7 are both prime.
a(n) = 1 for n = 117, 118, 119, 120 since 59 and floor(sqrt(n-59)) = 7 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000040, A000290, A237706, A237710, A237721.

q[n_]:=PrimeQ[Floor[Sqrt[n]]]
a[n_]:=Sum[If[q[n-Prime[k]],1,0],{k,1,PrimePi[(n+1)/2]}]
Table[a[n],{n,1,70}]

A237721 Number of primes p <= n with floor( sqrt(n-p) ) a square.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Feb 12 2014

nonn

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 9, 10, 11, 12, 15, 16, 17.
We have verified this for n up to 10^6.
See also A237705, A237706 and A237720 for similar conjectures.

			a(2) = 1 since 2 is prime with floor(sqrt(2-2)) = 0^2.
a(3) = 2 since 2 is prime with floor(sqrt(3-2)) = 1^2, and 3 is prime with floor(sqrt(3-3)) = 0^2.
a(9) = a(10) = 1 since 7 is prime with floor(sqrt(9-7)) = floor(sqrt(10-7)) = 1^2.
a(11) = 1 since 11 is prime with floor(sqrt(11-11)) = 0^2.
a(12) = 1 since 11 is prime with floor(sqrt(12-11)) = 1^2.
a(15) = a(16) = 1 since 13 is prime with floor(sqrt(15-13)) = floor(sqrt(16-13)) = 1^2.
a(17) = 1 since 17 is prime with floor(sqrt(17-17)) = 0^2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000040, A000290, A237705, A237706, A237710, A237720.

SQ[n_]:=IntegerQ[Sqrt[n]]
q[n_]:=SQ[Floor[Sqrt[n]]]
a[n_]:=Sum[If[q[n-Prime[k]],1,0],{k,1,PrimePi[n]}]
Table[a[n],{n,1,70}]

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

Original entry on oeis.org

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}]

A238701 Number of primes p < n with q = floor((n-p)/4) and q^2 - 2 both prime.

Original entry on oeis.org

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, 5, 6, 6, 5, 5, 5, 5, 3, 4, 5, 5

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 with q = floor((n-p)/m) and q^2 - 2 both prime.

In the case m = 1, this is a refinement of Goldbach's conjecture. In the case m = 2, this is stronger than Lemoine's conjecture (cf. A046927). The conjecture for m > 2 seems completely new. We view the conjecture as a natural extension of Goldbach's conjecture.

			a(11) = 2 since 2, floor((11-2)/4)= 2 and 2^2 - 2 are all prime, and 3, floor((11-3)/4) = 2 and 2^2 - 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, A049002, A062326, A230502, A237413, A237705.

PQ[n_]:=PrimeQ[n]&&PrimeQ[n^2-2]
p[n_,k_]:=PQ[Floor[(n-Prime[k])/4]]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]

cp's OEIS Frontend

A237706 Number of primes p < n with pi(n-p) a square, where pi(.) is given by A000720.

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A237768 Number of primes p < n with pi(n-p) a Sophie Germain prime, where pi(.) is given by A000720.

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A237769 Number of primes p < n with pi(n-p) - 1 and pi(n-p) + 1 both prime, where pi(.) is given by A000720.

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A238278 a(n) = |{0 < k < n: the number of primes in the interval ((k-1)*n, k*n] and the number of primes in the interval (k*n, (k+1)*n] are both prime}|.

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A237815 Number of primes p < n such that the number of Sophie Germain primes among 1, ..., n-p is a Sophie Germain prime.

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A237817 Number of primes p < n such that r = |{q <= n-p: q and q + 2 are both prime}| and r + 2 are both prime.

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A237720 Number of primes p <= (n+1)/2 with floor( sqrt(n-p) ) prime.

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A237721 Number of primes p <= n with floor( sqrt(n-p) ) a square.

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A238278 a(n) = |{0 < k < n: the number of primes in the interval ((k-1)n, kn] and the number of primes in the interval (kn, (k+1)n] are both prime}|.