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.

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A238902 a(n) = |{0 < k <= n: pi(pi(k*n)) is a square}|, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 06 2014

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) For every n = 1, 2, 3, ..., there exists a positive integer k <= (n+1)/2 such that pi(pi(k*n)) is a triangular number.
We have verified parts (i) and (ii) for n up to 2*10^5 and 10^5 respectively.
See A239884 for a sequence related to part (i) of the conjecture.

Examples

			a(8)    = 1 since pi(pi(3*8)) = pi(pi(24)) = pi(9) = 2^2.
a(434)  = 1 since pi(pi(297*434)) = pi(pi(128898)) = pi(12064) = 38^2.
a(1042) = 1 since pi(pi(698*1042)) = pi(pi(727316)) = pi(58590) = 77^2.
a(9143) = 1 since pi(pi(8514*9143)) = pi(pi(77843502)) = pi(4550901) = 565^2.
a(48044)  > 0 since pi(pi(18332*48044))  = pi(45075237)  = 1650^2.
a(52158)  > 0 since pi(pi(27976*52158))  = pi(72792062)  = 2067^2.
a(78563)  > 0 since pi(pi(26031*78563))  = pi(100326489) = 2404^2.
a(98213)  > 0 since pi(pi(37308*98213))  = pi(174740922) = 3123^2.
 a(141589) > 0 since pi(pi(42375*141589)) = pi(279538049)= 3899^2.
a(154473) > 0 since pi(pi(42954*154473)) = pi(307695484) = 4080^2.
a(195387) > 0 since pi(pi(60161*195387)) = pi(530982180) = 5282^2.

Links

Crossrefs

Cf. A000040, A000217, A000290, A000720, A237598, A237840, A238504, A239884.

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
p[k_,n_]:=SQ[PrimePi[PrimePi[k*n]]]
a[n_]:=Sum[If[p[k,n],1,0],{k,1,n}]
Table[a[n],{n,1,80}]
  • PARI
    {a(n) = sum( k=1, n, issquare( primepi( primepi( k*n))))}; /* Michael Somos, Mar 10 2014 */

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}]
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