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 1-4 of 4 results.

A238645 Number of odd primes p < 2*n such that the number of squarefree numbers among 1, ..., ((p-1)/2)*n is prime.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 02 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 4, 7, 14, 19, 29.
This is an analog of the conjecture in A237578 for squarefree numbers. We have verified it for n up to 20000.
See also A238646 for a similar conjecture involving squarefree numbers.

Examples

			a(4) = 1 since 3 is prime and there are exactly 3 squarefree numbers among 1, ..., (3-1)/2*4 (namely, 1, 2, 3).
a(14) = 1 since 5 and 17 are both prime, and there are exactly 17 squarefree numbers among 1, ..., (5-1)/2*14.
a(19) = 1 since 3 and 13 are both prime, and there are exactly 13 squarefree numbers among 1, ..., (3-1)/2*19 (namely, 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19).
a(29) = 1 since 41 and 353 are both prime, and there are exactly 353 squarefree numbers among 1, ..., (41-1)/2*29 = 580.

Links

Crossrefs

Cf. A000040, A005117, A013928, A237578, A237615, A238646.

Programs

  • Mathematica
    s[n_]:=Sum[If[SquareFreeQ[k],1,0],{k,1,n}]
a[n_]:=Sum[If[PrimeQ[s[(Prime[k]-1)/2*n]],1,0],{k,2,PrimePi[2n-1]}]
Table[a[n],{n,1,80}]

A238165 Number of pairs {j, k} with 0 < j < k <= n such that pi(j*n) divides pi(k*n), where pi(.) is given by A000720.

Original entry on oeis.org

0, 1, 1, 2, 3, 2, 1, 5, 5, 5, 3, 5, 12, 5, 5, 7, 3, 2, 12, 7, 8, 9, 9, 6, 6, 11, 9, 12, 9, 15, 12, 12, 13, 7, 16, 12, 18, 15, 16, 11, 8, 8, 13, 15, 20, 13, 7, 15, 13, 7, 18, 7, 18, 15, 11, 15, 15, 12, 15, 17, 6, 18, 17, 16, 11, 15, 9, 18, 15, 13
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 19 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1.
(ii) For any integer n > 4, the sequence pi(k*n)^(1/k) (k = 1, ..., n) is strictly decreasing.
See also A238224 for a refinement of part (i) of this conjecture.

Examples

			a(5) = 3 since pi(1*5) = 3 divides both pi(3*5) = 6 and pi(5*5) = 9, and pi(2*5) = 4 divides pi(4*5) = 8.
a(7) = 1 since pi(1*7) = 4 divides pi(3*7) = 8.

Links

Crossrefs

Cf. A000720, A237578, A237597, A237598, A237612, A237614, A237615, A238224.

Programs

  • Mathematica
    m[k_,j_]:=Mod[PrimePi[k],PrimePi[j]]==0
a[n_]:=Sum[If[m[k*n,j*n],1,0],{k,2,n},{j,1,k-1}]
Do[Print[n," ",a[n]],{n,1,70}]

A238568 a(n) = |{0 < k < n: n^2 - pi(k*n) is prime}|, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 28 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 8, 10, 24, 41.
(ii) For any integer n > 6, there is a positive integer k < n with n^2 + pi(k*n) - 1 prime.
(iii) If n > 2, then pi(n^2) - pi(k*n) is prime for some 0 < k < n. If n > 1, then pi(n^2) + pi(k*n) - 1 is prime for some 0 < k < n.

Examples

			a(2) = 1 since 2^2 - pi(1*2) = 4 - 1 = 3 is prime.
a(3) = 1 since 3^2 - pi(1*3) = 9 - 2 = 7 is prime.
a(4) = 1 since 4^2 - pi(3*4) = 16 - 5 = 11 is prime.
a(8) = 1 since 8^2 - pi(4*8) = 64 - 11 = 53 is prime.
a(10) = 1 since 10^2 - pi(6*10) = 100 - 17 = 83 is prime.
a(24) = 1 since 24^2 - pi(14*24) = 576 - 67 = 509 is prime.
a(41) = 1 since 41^2 - pi(10*41) = 1681 - 80 = 1601 is prime.

Links

Crossrefs

Cf. A000040, A000720, A237578, A237615, A237712, A238570.

Programs

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

A238703 a(n) = |{0 < k < n: floor(k*n/3) is prime}|.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 03 2014

Keywords

Comments

Conjecture: If n > m > 0 with n not divisible by m, then floor(k*n/m) is prime for some 0 < k < n.

Examples

			a(4) = 1 since floor(2*4/3) = 2 is prime.
If p is a prime, then a(3*p) = 1 since floor(k*3p/3) = k*p is prime only for k = 1. If m > 1 is composite, then a(3*m) = 0 since floor(k*3m/3) = k*m is composite for all k > 0.

Links

Crossrefs

Cf. A000040, A237578, A237615, A238645, A238701.

Programs

  • Mathematica
    p[n_,k_]:=PrimeQ[Floor[k*n/3]]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
Table[a[n],{n,1,80}]
Showing 1-4 of 4 results.

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