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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.
4
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
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
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 20 2014
STATUS
approved