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A262262 Number of prime divisors p | n such that p^2 < n and p^2 does not divide n. 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,30
COMMENTS
a(n) = 0 if n is a prime power.
From Michael De Vlieger, Sep 17 2015: (Start)
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)
LINKS
EXAMPLE
a(6) = 1 because 4, 9 are squares of prime divisors of 6 and only 4 < 6 does not divide 6.
MATHEMATICA
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 *)
PROG
(PARI) a(n) = sumdiv(n, d, isprime(d) && (d^2 < n) && (n % d^2)); \\ Michel Marcus, Sep 17 2015
(PARI) a(n)=my(f=factor(n)[, 1]); sum(i=1, #f, f[i]^2<n && n%f[i]^2) \\ Charles R Greathouse IV, Sep 17 2015
CROSSREFS
Sequence in context: A359270 A318508 A100655 * A079275 A236314 A236322
KEYWORD
nonn,easy
AUTHOR
STATUS
approved

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Last modified March 28 05:39 EDT 2024. Contains 371235 sequences. (Running on oeis4.)