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

Author

Zhi-Wei Sun, Feb 20 2014

Keywords

• nonn

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.

Examples

			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.
		

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

Crossrefs

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

Programs

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}]