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A276935
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Number of distinct prime factors prime(k) of n such that prime(k)^k, but not prime(k)^(k+1) is a divisor of n.
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4
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0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0
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OFFSET
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1,18
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LINKS
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FORMULA
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a(1) = 0, for n > 1, a(n) = a(A028234(n)) + [A067029(n) = A055396(n)], where [] is Iverson bracket, giving 1 as its result when the stated equivalence is true and 0 otherwise.
Additive with a(p^e) = 1 if e = primepi(p), and 0 otherwise.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{k>=1} (1/prime(k)^k - 1/prime(k)^(k+1)) = 0.33083690651252383414... . (End)
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EXAMPLE
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For n = 12 = 2*2*3 = prime(1)^2 * prime(2)^1, neither of the prime factors satisfies the condition, thus a(12) = 0.
For n = 18 = 2*3*3 = prime(1)^1 * prime(2)^2, both prime factors satisfy the condition, thus a(18) = 1+1 = 2.
For n = 750 = 2*3*5*5*5 = prime(1)^1 * prime(2)^1 * prime(3)^3, only the prime factors 2 and 5 satisfy the condition, thus a(750) = 1+1 = 2.
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MATHEMATICA
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f[p_, e_] := If[PrimePi[p] == e, 1, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 30 2023 *)
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PROG
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(PARI) a(n) = {my(f = factor(n)); sum(i = 1, #f~, primepi(f[i, 1]) == f[i, 2]); } \\ Amiram Eldar, Sep 30 2023
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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