A238224 Number of pairs {j, k} with 0 < j < k <= n and k == 1 (mod j) such that pi(j*n) divides pi(k*n), where pi(.) is given by A000720.

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

Offset

1,4

Comments

Conjecture: a(n) > 0 for all n > 1.

This is a refinement of part (i) of the conjecture in A238165.

We have verified the conjecture for n up to 21500.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..2900

Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Example

a(18) = 1 since 6 == 1 (mod 1), and pi(1*18) = 7 divides pi(6*18) = 28.
a(50) = 1 since 7 == 1 (mod 3), and pi(3*50) = 35 divides pi(7*50) = 70.
a(379) = 1 since 353 == 1 (mod 4), and pi(4*379) = 240 divides pi(353*379) = 12480.

Mathematica

m[k_, j_, n_]:=Mod[PrimePi[k*n], PrimePi[j*n]]==0
a[n_]:=Sum[If[m[j*q+1, j, n], 1, 0], {j, 1, n-1}, {q, 1, (n-1)/j}]
Table[a[n], {n, 1, 80}]

Crossrefs

Cf. A000720, A237578, A237597, A237598, A238165.

Sequence in context: A361639 A055253 A103626 * A026268 A089258 A004065

Adjacent sequences: A238221 A238222 A238223 * A238225 A238226 A238227

Keyword

nonn

Author

Zhi-Wei Sun, Feb 20 2014

Status

approved