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.

A239207 a(n) = |{0 < k <= n: k*p(n)*q(n)*r(n) - 1 is prime}|, where p(.), q(.) and r(.) are given by A000041, A000009 and A047967 respectively.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 12 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1. If n > 1 is not equal to 25, then k*p(n)*q(n)*r(n) + 1 is prime for some k = 1, ..., n.
(ii) For any integer n > 1, there is a number k among 1, ..., n with k*p(n)*q(n) - 1 (or k*p(n)*q(n) + 1) prime.
(iii) For each n > 1, there is a positive integer k < n with k*p(n) + 1 (or k*q(n) + 1) prime. If n > 1, then k*p(n) - 1 is prime for some k = 1, ..., n. If n > 2, then k*q(n) - 1 is prime for some 0 < k < n.
We have verified that a(n) > 0 for all n = 2, ..., 83000.
See also A239209 and A239214 for related conjectures.

Examples

			a(2) = 1 since 2*p(2)*q(2)*r(2) - 1 = 2*2*1*1 - 1 = 3 is prime.
a(23) = 1 since 12*p(23)*q(23)*r(23) - 1 = 12*1255*104*1151 - 1 = 1802742239 is prime.
		

Crossrefs

Programs

  • Mathematica
    p[n_]:=PartitionsP[n]
    q[n_]:=PartitionsQ[n]
    f[n_]:=p[n]*q[n]*(p[n]-q[n])
    a[n_]:=Sum[If[PrimeQ[k*f[n]-1],1,0],{k,1,n}]
    Table[a[n],{n,1,80}]

A239214 a(n) = |{0 < k < n: p(k)*p(n)*(p(n)+1) - 1 is prime}|, where p(.) is the partition function (A000041).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 12 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1.
(ii) For each n = 2, 3, ... there is a positive integer k < n with p(k)*p(n)*(p(n)-1) + 1 prime. If n > 2, then p(k)*p(n)*(p(n)-1)-1 is prime for some 0 < k < n.
(iii) For any n > 1, there is a positive integer k < n with 2*p(k)*p(n)*A000009(n)*A047967(n) + 1 prime.
We have verified that a(n) > 0 for all n = 2, ..., 10^5.

Examples

			a(2) = 1 since p(1)*p(2)*(p(2)+1) - 1 = 1*2*3 - 1 = 5 is prime.
a(5) = 1 since p(3)*p(5)*(p(5)+1) - 1 = 3*7*8 - 1 = 167 is prime.
a(21) = 1 since p(10)*p(21)*(p(21)+1) - 1 = 42*792*793 - 1 = 26378351 is prime.
a(45) = 1 since p(20)*p(45)*(p(45)+1) - 1 = 627*89134*89135 - 1 = 4981489349429 is prime.
		

Crossrefs

Programs

  • Mathematica
    p[n_]:=PartitionsP[n]
    f[n_]:=p[n]*(p[n]+1)
    a[n_]:=Sum[If[PrimeQ[p[k]*f[n]-1],1,0],{k,1,n-1}]
    Table[a[n],{n,1,80}]

A239232 a(n) = |{0 < k <= n: p(n+k) + 1 is prime}|, where p(.) is the partition function (A000041).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 13 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 3.
(ii) If n > 15, then p(n+k) - 1 is prime for some k = 1, ..., n.
(iii) If n > 38, then p(n+k) is prime for some k = 1, ..., n.
The conjecture implies that there are infinitely many positive integers m with p(m) + 1 (or p(m) - 1, or p(m)) prime.

Examples

			a(4) = 1 since p(4+4) + 1 = 22 + 1 = 23 is prime.
a(8) = 2 since p(8+1) + 1 = 31 and p(8+2) + 1 = 43 are both prime.
a(11) = 1 since p(11+8) + 1 = 491 is prime.
		

Crossrefs

Programs

  • Mathematica
    p[n_]:=PartitionsP[n]
    a[n_]:=Sum[If[PrimeQ[p[n+k]+1],1,0],{k,1,n}]
    Table[a[n],{n,1,80}]
Showing 1-3 of 3 results.