A238277 -id:A238277 - cp's OEIS frontend

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}|.

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}]

A238281 a(n) = |{0 < k < n: the two intervals (kn, (k+1)n) and ((k+1)n, (k+2)n) contain the same number of primes}|.

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 3, 1, 5, 2, 4, 4, 8, 3, 7, 4, 4, 4, 2, 3, 7, 3, 10, 4, 12, 7, 7, 15, 7, 9, 8, 5, 8, 9, 11, 8, 8, 10, 8, 4, 10, 10, 10, 11, 7, 10, 8, 11, 8, 8, 9, 9, 8, 11, 7, 8, 13, 10, 8, 14, 13, 4, 14, 8, 11, 12, 14, 12, 8, 10, 16, 12, 16, 12, 14, 19, 11, 14, 8, 9

Offset: 1

Zhi-Wei Sun, Feb 22 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 1. Moreover, if n > 1 is not equal to 8, then there is a positive integer k < n with 2*k + 1 prime such that the two intervals ((k-1)*n, k*n) and (k*n, (k+1)*n) contain the same number of primes.

(ii) For any integer n > 4, there is a positive integer k < prime(n) such that all the three intervals (k*n, (k+1)*n), ((k+1)*n, (k+2)*n), ((k+2)*n, (k+3)*n) contain the same number of primes, i.e., pi(k*n), pi((k+1)*n), pi((k+2)*n), pi((k+3)*n) form a 4-term arithmetic progression.

			a(8) = 1 since each of the two intervals (7*8, 8*8) and (8*8, 9*8) contains exactly two primes.

Zhi-Wei Sun, Table of n, a(n) for n = 1..5000
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Cf. A000040, A000720, A237578, A238224, A238277, A238278.

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

A238289 Least positive integer k such that prime(kn), prime((k+1)n) and prime((k+2)*n) form an arithmetic progression, or 0 if such a number k does not exist.

Original entry on oeis.org

2, 2, 17, 4, 1, 1, 59, 3, 56, 1, 39, 10, 9, 130, 2, 18, 11, 7, 80, 67, 2, 19, 27, 17, 92, 73, 180, 65, 5, 110, 282, 4, 6, 8, 16, 2, 23, 198, 20, 3, 99, 83, 217, 13, 110, 28, 16, 6, 5, 3, 144, 31, 9, 187, 176, 145, 75, 11, 43, 424, 4, 54, 272, 8, 26, 131, 123, 107, 8, 4, 52, 9, 127, 84, 264, 33, 145, 663, 16, 285

Offset: 1

Zhi-Wei Sun, Feb 22 2014

nonn

Conjecture: (i) We always have 0 < a(n) < 3*prime(n) + 9.
(ii) For any integer n > 3, there is a positive integer k < 5*n^3 such that prime(k*n), prime((k+1)*n), prime((k+2)*n), and prime((k+3)*n) form a 4-term arithmetic progression.
(iii) In general, for each m = 3, 4, ..., if n is sufficiently large then there is a positive integer k = O(n^(m-1)) such that prime((k+j)*n) (j = 0, ..., m-1) form an arithmetic progression.
The conjecture is a refinement of the Green-Tao theorem.

			a(2) = 2 since prime(2*2) = 7, prime(3*2) = 13 and prime(4*2) = 19 form a 3-term arithmetic progression, but prime(1*2) = 3,  prime(2*2) = 7 and prime(3*2) = 13 do not form a 3-term arithmetic progression.

Zhi-Wei Sun, Table of n, a(n) for n = 1..8000
B. Green and T. Tao, The primes contain arbitrarily long arithmetic progressions, Annals of Math. 167(2008), 481-547.
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Cf. A000040, A238277, A238278, A238281.

d[k_,n_]:=Prime[(k+1)*n]-Prime[k*n]
Do[Do[If[d[k,n]==d[k+1,n],Print[n," ",k];Goto[aa]],{k,1,3*Prime[n]+8}];
Print[n," ",0];Label[aa];Continue,{n,1,80}]

okpr(p, q, r) = (q - p) == (r - q);
a(n) = {k = 1; while(! okpr(prime(k*n), prime((k+1)*n), prime((k+2)*n)), k++); k;} \\ Michel Marcus, Aug 28 2014

A239328 Number of primes p < n with pi(p*n) - pi((p-1)n) prime, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Mar 16 2014

nonn

Conjecture: a(n) > 0 for all n > 3. Also, for any integer n > 4, there is a prime p < n with pi((p+1)*n) - pi(p*n) prime.

We have verified that a(n) > 0 for all n = 4, ..., 7*10^5.

			a(4) = 1 since 2 and pi(2*4) - pi(1*4) = 4 - 2 = 2 are both prime.
a(5) = 1 since 3 and pi(3*5) - pi(2*5) = 6 - 4 = 2 are both prime.
a(17) = 1 since 11 and pi(11*17) - pi(10*17) = 42 - 39 = 3 are both prime.
a(47) = 1 since 37 ad pi(37*47) - pi(36*47) = 270 - 263 = 7 are both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..5000
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Cf. A000040, A000720, A237578, A238277, A238278, A238281, A239330.

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

A230022 a(n) = |{the number of primes in the interval (kn, (k+1)n]: k = 0, 1, ..., n-1}|.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 4, 3, 4, 5, 5, 4, 5, 4, 5, 5, 6, 7, 6, 5, 6, 6, 6, 5, 5, 7, 6, 6, 7, 7, 6, 6, 7, 7, 8, 9, 8, 9, 9, 8, 8, 8, 9, 8, 9, 9, 8, 10, 10, 9, 10, 9, 10, 10, 10, 10, 11, 10, 10, 9, 10, 9, 11, 10, 11, 11, 11, 11, 11, 11, 11, 10, 12, 11, 10, 11, 12, 13, 11

Offset: 1

Zhi-Wei Sun, Feb 23 2014

nonn

Conjecture: (i) a(n) is at least sqrt(n-1) for each n > 0, and equality holds only when n is 2 or 26.
(ii) The sequence contains all positive integers.
We have verified part (i) of the conjecture for n up to 10000.

			a(1) = 1 since the interval (0,1*1] contains no prime, and the set {0} has cardinaly 1.
a(3) = 2 since the intervals (0, 1*3], (1*3, 2*3], (2*3, 3*3] contain exactly 2, 1, 1 primes respectively, and the set {2, 1, 1} has cardinality 2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..4000
Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Cf. A000040, A000720, A238277, A238278, A238281.

d[k_,n_]:=PrimePi[(k+1)*n]-PrimePi[k*n]
a[n_]:=Length[Union[Table[d[k,n],{k,0,n-1}]]]
Table[a[n],{n,1,80}]

cp's OEIS Frontend

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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A238281 a(n) = |{0 < k < n: the two intervals (k*n, (k+1)*n) and ((k+1)*n, (k+2)*n) contain the same number of primes}|.

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A238289 Least positive integer k such that prime(k*n), prime((k+1)*n) and prime((k+2)*n) form an arithmetic progression, or 0 if such a number k does not exist.

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A239328 Number of primes p < n with pi(p*n) - pi((p-1)n) prime, where pi(x) denotes the number of primes not exceeding x.

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A230022 a(n) = |{the number of primes in the interval (k*n, (k+1)*n]: k = 0, 1, ..., n-1}|.

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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}|.

A238281 a(n) = |{0 < k < n: the two intervals (kn, (k+1)n) and ((k+1)n, (k+2)n) contain the same number of primes}|.

A238289 Least positive integer k such that prime(kn), prime((k+1)n) and prime((k+2)*n) form an arithmetic progression, or 0 if such a number k does not exist.

A230022 a(n) = |{the number of primes in the interval (kn, (k+1)n]: k = 0, 1, ..., n-1}|.