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 1-6 of 6 results.

A237496 Number of ordered ways to write n = k + m (0 < k <= m) with pi(k) + pi(m) - 2 prime, where pi(.) is given by A000720.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 08 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 5.
(ii) Any integer n > 23 can be written as k + m (k > 0 and m > 0) with pi(k) + pi(m) prime. Also, each integer n > 25 can be written as k + m (k > 0 and m > 0) with pi(k) + pi(m) - 1 prime.

Examples

			a(6) = 1 since 6 = 3 + 3 with pi(3) + pi(3) - 2 = 2 + 2 - 2 = 2 prime.
a(17) = 1 since 17 = 2 + 15 with pi(2) + pi(15) - 2 = 1 + 6 - 2 = 5 prime.
a(99) = 1 since 99 = 1 + 98 with pi(1) + pi(98) - 2 = 0 + 25 - 2 = 23 prime.
		

Crossrefs

Programs

  • Mathematica
    PQ[n_]:=n>0&&PrimeQ[n]
    p[k_,m_]:=PQ[PrimePi[k]+PrimePi[m]-2]
    a[n_]:=Sum[If[p[k,n-k],1,0],{k,1,n/2}]
    Table[a[n],{n,1,80}]

A232502 Number of ways to write n = k + m (0 < k < m) with 2*prime(m) - prime(k) prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Nov 24 2013

Keywords

Comments

Note that prime(k), prime(m), 2*prime(m) - prime(k) form a three-term arithmetic progression. It is known that there are infinitely many nontrivial three-term arithmetic progressions whose terms are all prime.
Conjecture: (i) a(n) > 0 for all n > 4, and a(n) = 1 only for n = 5, 6, 7, 9, 11, 14, 17, 21, 34, 43.
(ii) Any integer n > 4, can be written as k + m (0 < k < m) with 2*prime(m) + prime(k) prime.

Examples

			a(17) = 1 since 2*prime(10) - prime(7) = 2*29 - 17 = 41 is prime.
a(21) = 1 since 2*prime(19) - prime(2) = 2*67 - 3 = 131 is prime.
a(34) = 1 since 2*prime(24) - prime(10) = 2*89 - 29 = 149 is prime.
a(43) = 1 since 2*prime(28) - prime(15) = 2*107 - 47 = 167 is prime.
		

Crossrefs

Programs

  • Mathematica
    a[n_]:=Sum[If[PrimeQ[2*Prime[n-k]-Prime[k]],1,0],{k,1,(n-1)/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.
		

Crossrefs

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.
		

Crossrefs

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.
		

Crossrefs

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}]

A238585 Number of primes p < n with prime(p)^2 + (prime(n)-1)^2 prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 01 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 unless n divides 6, and a(n) = 1 only for n = 4, 5, 7, 10, 11, 12, 19, 21, 22, 31, 42, 44.
(ii) If n > 2 is not equal to 9, then prime(n)^2 + (prime(p) - 1)^2 is prime for some prime p < n.
(iii) For n > 3, there is a prime p < n with prime(p) + prime(n) + 1 prime. If n > 9 is not equal to 18, then prime(p)^2 + prime(n)^2 - 1 is prime for some prime p < n.

Examples

			a(7) = 1 since 3 and prime(3)^2 + (prime(7)-1)^2 = 5^2 + 16^2 = 281 are both prime.
a(44) = 1 since 23 and prime(23)^2 + (prime(44)-1)^2 = 83^2 + 192^2 = 43753 are both prime.
		

Crossrefs

Programs

  • Mathematica
    p[n_,k_]:=PrimeQ[k]&&PrimeQ[Prime[k]^2+(Prime[n]-1)^2]
    a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
    Table[a[n],{n,1,80}]
Showing 1-6 of 6 results.