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 1-3 of 3 results.

A218829 Number of ordered ways to write n = k + m with k > 0 and m > 0 such that prime(k) + 2 and prime(prime(m)) + 2 are both prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 05 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3, 22, 25, 38, 101, 273.
(ii) Each n = 2, 3, ... can be written as k + m with k > 0 and m > 0 such that 6*k - 1, 6*k + 1 and prime(prime(m)) + 2 are all prime.
(iii) Any integer n > 5 can be written as k + m with k > 0 and m > 0 such that phi(k) - 1, phi(k) + 1 and prime(prime(m)) + 2 are all prime, where phi(.) is Euler's totient function.
(iv) If n > 2 is neither 10 nor 31, then n can be written as k + m with k > 0 and m > 0 such that prime(k) + 2 and prime(prime(prime(m))) + 2 are both prime.
(v) If n > 1 is not equal to 133, then n can be written as k + m with k > 0 and m > 0 such that 6*k - 1, 6*k + 1 and prime(prime(prime(m))) + 2 are all prime.
Clearly, each part of the conjecture implies the twin prime conjecture.
We have verified part (i) for n up to 10^9. See the comments in A237348 for an extension of this part.

Examples

			a(3) = 1 since 3 = 2 + 1 with prime(2) + 2 = 3 + 2 = 5 and prime(prime(1)) + 2 = prime(2) + 2 = 5 both prime.
a(22) = 1 since 22 = 20 + 2 with prime(20) + 2 = 71 + 2 = 73 and prime(prime(2)) + 2 = prime(3) + 2 = 5 + 2 = 7 both prime.
a(25) = 1 since 25 = 2 + 23 with prime(2) + 2 = 3 + 2 = 5 and prime(prime(23)) + 2 = prime(83) + 2 = 431 + 2 = 433 both prime.
a(38) = 1 since 38 = 35 + 3 with prime(35) + 2 = 149 + 2 = 151 and prime(prime(3)) + 2 = prime(5) + 2 = 11 + 2 = 13 both prime.
a(101) = 1 since 101 = 98 + 3 with prime(98) + 2 = 521 + 2 = 523 and prime(prime(3)) + 2 = prime(5) + 2 = 11 + 2 = 13 both prime.
a(273) = 1 since 273 = 2 + 271 with prime(2) + 2 = 3 + 2 = 5 and prime(prime(271)) + 2 = prime(1741) + 2 = 14867 + 2 = 14869 both prime.
		

Crossrefs

Programs

  • Mathematica
    pq[n_]:=PrimeQ[Prime[n]+2]
    PQ[n_]:=PrimeQ[Prime[Prime[n]]+2]
    a[n_]:=Sum[If[pq[k]&&PQ[n-k],1,0],{k,1,n-1}]
    Table[a[n],{n,1,80}]

A237413 Number of ways to write n = k + m with k > 0 and m > 0 such that p(k)^2 - 2, p(m)^2 - 2 and p(p(m))^2 - 2 are all prime, where p(j) denotes the j-th prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 07 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This conjecture was motivated by the "Super Twin Prime Conjecture".
See A237414 for primes q with q^2 - 2 and p(q)^2 - 2 both prime.

Examples

			a(7) = 1 since 7 = 6 + 1 with p(6)^2 - 2 = 13^2 - 2 = 167, p(1)^2 - 2 = 2^2 - 2 = 2 and p(p(1))^2 - 2 = p(2)^2 - 2 = 3^2 - 2 = 7 are all prime.
a(516) = 1 since 516 = 473 + 43 with p(473)^2 - 2 = 3359^2 - 2 = 11282879, p(43)^2 - 2 = 191^2 - 2 = 36479 and p(p(43))^2 - 2 = p(191)^2 - 2 = 1153^2 - 2 = 1329407 all prime.
		

Crossrefs

Programs

  • Mathematica
    pq[k_]:=PrimeQ[Prime[k]^2-2]
    a[n_]:=Sum[If[pq[k]&&pq[n-k]&&pq[Prime[n-k]],1,0],{k,1,n-1}]
    Table[a[n],{n,1,80}]

A237367 Number of ordered ways to write n = k + m with k > 0 and m > 0 such that 2*k - 1, prime(k)^2 - 2 and prime(m)^2 - 2 are all prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 07 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 2, and a(n) = 1 only for n = 3, 73, 81, 534.
(ii) Any integer n > 2 can be written as k + m with k > 0 and m > 0 such that 2*k - 1, prime(k) + k*(k-1) and prime(m) + m*(m-1) are all prime.
(iii) Every n = 9, 10, ... can be written as k + m with k > 0 and m > 0 such that 6*k - 1, prime(k) + 2*k and prime(m) + 2*m are all prime.
Clearly, part (i) of this conjecture implies that there are infinitely many primes p with p^2 - 2 also prime. Similar comments apply to parts (ii) and (iii).

Examples

			a(3) = 1 since 3 = 2 + 1 with 2*2 - 1 = 3, prime(2)^2 - 2 = 3^2 - 2 = 7 and prime(1)^2 - 2 = 2^2 - 2 = 2 all prime.
a(73) = 1 since 73 = 55 + 18 with 2*55 - 1 = 109, prime(55)^2 - 2 = 257^2 - 2 = 66047 and prime(18)^2 - 2 = 61^2 - 2 = 3719 all prime.
a(81) = 1 since 81 = 34 + 47 with 2*34 - 1 = 67, prime(34)^2 - 2 = 139^2 - 2 = 19319 and prime(47)^2 - 2 = 211^2 - 2 = 44519 all prime.
a(534) = 1 since 534 = 100 + 434 with 2*100 - 1 = 199, prime(100)^2 - 2 = 541^2 - 2 = 292679 and prime(434)^2 - 2 = 3023^2 - 2 = 9138527 all prime.
		

Crossrefs

Programs

  • Mathematica
    pq[k_]:=PrimeQ[Prime[k]^2-2]
    a[n_]:=Sum[If[PrimeQ[2k-1]&&pq[k]&&pq[n-k],1,0],{k,1,n-1}]
    Table[a[n],{n,1,80}]
Showing 1-3 of 3 results.