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A262262
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Number of prime divisors p | n such that p^2 < n and p^2 does not divide n.
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1
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0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 3, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 2, 0, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 0, 1, 2, 0, 0, 1, 3, 0, 0, 0, 1, 1, 0, 1, 2, 0, 1, 0, 1, 0, 2, 1, 1, 1, 0, 0, 2, 1, 0, 1, 1, 1, 1, 0
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OFFSET
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1,30
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COMMENTS
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a(n) = 0 if n is a prime power.
Let k be a prime divisor p | n such that p^2 < n and d^2 does not divide n.
a(n) <= A262202(n), as any k = p^2 is a special case of a (prime or nonprime) divisor d of n where d is prime.
a(n) <= A010846(n), as any k is regular to n, i.e., k is a product less than n of the prime divisors of n.
a(n) <= A045763(n), as any k neither divides nor is coprime to n.
a(n) <= A243822(n), as any k is a "semidivisor" of n, i.e., k is a product less than n of the prime divisors of n that do not divide n.
(End)
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LINKS
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EXAMPLE
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a(6) = 1 because 4, 9 are squares of prime divisors of 6 and only 4 < 6 does not divide 6.
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MATHEMATICA
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f[n_] := Block[{d = First /@ FactorInteger@ n}, Select[d^2, And[Mod[n, #] != 0, # < n] &]]; Length@ f@ # & /@ Range@ 120 (* Michael De Vlieger, Sep 17 2015 *)
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PROG
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(PARI) a(n) = sumdiv(n, d, isprime(d) && (d^2 < n) && (n % d^2)); \\ Michel Marcus, Sep 17 2015
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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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