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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.
2
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
1,4
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.
LINKS
EXAMPLE
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.
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}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 19 2014
STATUS
approved