cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A234694 a(n) = |{0 < k < n: p = k + prime(n-k) and prime(p) - p + 1 are both prime}|.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 29 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 9. Also, for any integer n > 51 there is a positive integer k < n such that p = k + prime(n-k) and prime(p) + p + 1 are both prime.
(ii) If n > 9 (or n > 21), then there is a positive integer k < n such that m - 1 and prime(m) + m (or prime(m) - m, resp.) are both prime, where m = k + prime(n-k).
(iii) If n > 483, then for some 0 < k < n both prime(m) + m and prime(m) - m are prime, where m = k + prime(n-k).
(iv) If n > 3, then there is a positive integer k < n such that prime(k + prime(n-k)) + 2 is prime.
Clearly, part (i) of the conjecture implies that there are infinitely many primes p with prime(p) - p + 1 (or prime(p) + p + 1) also prime.
See A234695 for primes p with prime(p) - p + 1 also prime.

Examples

			a(5) = 1 since 2 + prime(3) = 7 and prime(7) - 6 = 11 are both prime.
a(25) = 1 since 20 + prime(5) = 31 and prime(31) - 30 = 97 are both prime.
a(27) = 1 since 18 + prime(9) = 41 and prime(41) - 40 = 139 are both prime.
a(45) = 1 since 6 + prime(39) = 173 and prime(173) - 172 = 859 are both prime.
a(49) = 1 since 26 + prime(23) = 109 and prime(109) - 108 = 491 are both prime.

Links

Crossrefs

Cf. A000040, A014688, A014689, A014692, A064269, A064270, A232861, A233150, A233183, A233206, A233296, A234695.

Programs

  • Mathematica
    f[n_,k_]:=k+Prime[n-k]
q[n_,k_]:=PrimeQ[f[n,k]]&&PrimeQ[Prime[f[n,k]]-f[n,k]+1]
a[n_]:=Sum[If[q[n,k],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

A233183 Number of ways to write n = k + m with 0 < k < m such that C(2*k, k) + prime(m) is prime.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 1, 3, 1, 2, 2, 3, 3, 2, 4, 4, 3, 7, 3, 4, 4, 4, 5, 2, 3, 5, 5, 3, 7, 7, 6, 2, 5, 3, 7, 6, 9, 6, 5, 5, 6, 8, 6, 6, 2, 12, 6, 7, 6, 9, 4, 5, 7, 5, 3, 7, 8, 8, 6, 5, 7, 9, 10, 4, 9, 6, 7, 7, 8, 6, 10, 8, 6, 6, 8, 5, 5, 10, 8, 10, 5, 9, 8, 15, 8, 12, 3, 12, 9, 10, 9, 10, 5, 11, 12, 8, 3, 12, 12, 8
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 05 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 2.
We have verified this for n up to 10^8.

Examples

			a(6) = 2 since 6 = 1 + 5 = 2 + 4 with C(2*1, 1) + prime(5) = C(2*2, 2) + prime(4) = 13 prime.
a(9) = 1 since 9 = 2 + 7 with C(2*2, 2) + prime(7) = 6 + 17 = 23 prime.

Links

Crossrefs

Cf. A000040, A000984, A231201, A233150.

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[Binomial[2k,k]+Prime[n-k]],1,0],{k,1,(n-1)/2}]
Table[a[n],{n,1,100}]

A233204 Number of ways to write n = k + m with 0 < k < m such that 2^k * prime(m) + 3 is prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 05 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 3. Also, any integer n > 2 can be written as k + m (0 < k <= m) with 2^k*prime(m) - 3 prime.
(ii) Any integer n > 6 can be written as k + m (0 < k < m) with prime(k) + 6 and prime(m) + 6 both prime. Each integer n > 4 can be written as k + m (0 < k < m) with prime(k) + 2 and prime(m) + 6 both prime. Also, for every integer n > 3 not among 11, 21, 32, 49, 171, there is a positive integer k < n with prime(k) + 2 and prime(n-k) + 2 both prime.

Links

Crossrefs

Cf. A000040, A000079, A231201, A231561, A233150, A233183.

Programs

  • Maple
    a(6) = 1 since 6 = 2 + 4 with 2^2*prime(4) + 3 = 4*7 + 3 = 31 prime.
a(22) = 1 since 22 = 1 + 21 with 2^1*prime(21) + 3 = 2*73 + 3 = 149 prime.
  • Mathematica
    a[n_]:=Sum[If[PrimeQ[2^k*Prime[n-k]+3],1,0],{k,1,(n-1)/2}]
Table[a[n],{n,1,100}]

A233206 Number of ways to write n = k + m (0 < k <= m) with k! + prime(m) prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 05 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
We have verified this for n up to 10^7. For n = 1356199, the least positive integer k with k! + prime(n-k) prime is 4496. For n = 7212995, the smallest positive integer k with k! + prime(n-k) prime is 4507.

Examples

			a(6) = 1 since 6 = 3 + 3 with 3! + prime(3) = 6 + 5 = 11 prime.
a(11) = 1 since 11 = 4 + 7 with 4! + prime(7) = 24 + 17 = 41 prime.

Links

Crossrefs

Cf. A000040, A000142, A231201, A231516, A231557, A231561, A231631, A233150, A233183, A233204.

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[k!+Prime[n-k]],1,0],{k,1,n/2}]
Table[a[n],{n,1,100}]

A233296 a(n) = |{0 < k < n: k*prime(n-k) + 1 is prime}|.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 2, 2, 2, 3, 3, 4, 2, 2, 5, 4, 1, 6, 1, 2, 5, 4, 4, 4, 3, 3, 2, 4, 6, 5, 4, 5, 6, 7, 7, 6, 5, 8, 6, 4, 5, 7, 8, 4, 6, 7, 6, 10, 7, 4, 8, 11, 9, 11, 6, 5, 5, 8, 8, 10, 5, 7, 10, 10, 11, 7, 6, 6, 12, 6, 10, 11, 6, 11, 7, 11, 8, 12, 7, 7, 9, 13, 9, 10, 14, 7, 13, 8, 10, 11, 9, 14, 10, 14, 17, 14, 13, 8, 12, 12
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 07 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1. Similarly, for any integer n > 2, k*prime(n-k) - 1 (or k^2*prime(n-k) - 1) is prime for some 0 < k < n.
(ii) Let n > 3 be an integer. Then k + prime(n-k) is prime for some 0 < k < n. Also, if n is not equal to 13, then k^2 + prime(n-k)^2 is prime for some 0 < k < n.

Examples

			a(17) = 1 since 17 = 14 + 3 with 14*prime(3) + 1 = 14*5 + 1 = 71 prime.
a(19) = 1 since 19 = 18 + 1 with 18*prime(1) + 1 = 18*2 + 1 = 37 prime.

Links

Crossrefs

Cf. A000040, A232465, A232502, A233150, A233183, A233204, A233206.

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[k*Prime[n-k]+1],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

A233439 a(n) = |{0 < k < n: prime(k)^2 + 4*prime(n-k)^2 is prime}|.

Original entry on oeis.org

0, 0, 0, 1, 2, 1, 2, 1, 3, 4, 4, 8, 4, 6, 3, 1, 7, 3, 8, 5, 2, 9, 2, 11, 8, 7, 5, 4, 8, 7, 8, 8, 8, 7, 5, 9, 5, 10, 9, 7, 13, 9, 11, 10, 14, 5, 11, 10, 10, 11, 12, 7, 13, 10, 10, 8, 15, 11, 12, 11, 13, 14, 6, 12, 11, 22, 21, 5, 15, 7, 13, 15, 17, 15, 10, 16, 11, 13, 14, 12, 17, 12, 16, 16, 19, 22, 17, 12, 19, 17, 19, 17, 16, 17, 18, 20, 19, 17, 10, 16
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 09 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 3.
(ii) For any integer n > 10, prime(j)^3 + 2*prime(n-j)^2 is prime for some 0 < j < n, and prime(k)^3 + 2*prime(n-k)^3 is prime for some 0 < k < n.
(iii) If n > 5, then prime(k)^3 + 2*p(n-k)^3 is prime for some 0 < k < n, where p(.) is the partition function (A000041). If n > 2, then prime(k)^3 + 2*q(n-k)^3 is prime for some 0 < k < n, where q(.) is the strict partition function (A000009).

Examples

			a(4) = 1 since prime(3)^2 + 4*prime(1)^2 = 5^2 + 4*2^2 = 41 is prime.
a(6) = 1 since prime(5)^2 + 4*prime(1)^2 = 11^2 + 4*2^2 = 137 is prime.
a(8) = 1 since prime(3)^2 + 4*prime(5)^2 = 5^2 + 4*11^2 = 509 is prime.
a(16) = 1 since prime(6)^2 + 4*prime(10)^2 = 13^2 + 4*29^2 = 3533 is prime.

Links

Crossrefs

Cf. A000040, A002313, A220413, A232174, A232269, A232465, A232502, A233150, A233204, A233206, A233296, A233307, A233346.

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[Prime[k]^2+4*Prime[n-k]^2],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]

A233529 a(n) = |{0 < k <= n/2: prime(k)*prime(n-k) - 6 is prime}|.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 1, 1, 2, 1, 3, 4, 1, 4, 5, 1, 5, 3, 2, 1, 2, 5, 5, 4, 5, 6, 5, 5, 4, 8, 5, 7, 4, 3, 6, 6, 4, 8, 6, 7, 7, 8, 7, 5, 5, 5, 7, 8, 6, 13, 9, 5, 3, 9, 6, 8, 11, 5, 9, 9, 10, 8, 9, 14, 9, 10, 13, 11, 6, 9, 12, 10, 12, 14, 10, 12, 7, 13, 9, 7, 7, 15, 12, 6, 10, 11, 12, 12, 9, 18, 15, 14, 11, 10, 10, 8, 13, 21, 9, 14
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 11 2013

Keywords

Comments

Conjectures:
(i) a(n) > 0 for all n > 5. Also, for any n > 5, 2*prime(k)*prime(n-k) - 3 is prime for some 0 < k < n.
(ii) For any n > 1 not among 3, 9, 13, 26, there is a positive integer k < n with prime(k)*prime(n-k) - 2 prime. For any n > 2 not among 8, 23, 33, there is a positive integer k < n with prime(k)*prime(n-k) - 4 prime.

Examples

			a(8) = 1 since prime(4)*prime(4) - 6 = 7*7 - 6 = 43 is prime.
a(10) = 1 since prime(3)*prime(7) - 6 = 5*17 - 6 = 79 is prime.
a(16) = 1 since prime(3)*prime(13) - 6 = 5*41 - 6 = 199 is prime.
a(20) = 1 since prime(7)*prime(13) - 6 = 17*41 - 6 = 691 is prime.

Links

Crossrefs

Cf. A000040, A232465, A232502, A232861, A233150, A233204, A233206, A233439.

Programs

  • Mathematica
    PQ[n_]:=n>0&&PrimeQ[n]
a[n_]:=Sum[If[PQ[Prime[k]*Prime[n-k]-6],1,0],{k,1,n/2}]
Table[a[n],{n,1,100}]
Showing 1-7 of 7 results.

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