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A056169
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Number of unitary prime divisors of n.
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66
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0, 1, 1, 0, 1, 2, 1, 0, 0, 2, 1, 1, 1, 2, 2, 0, 1, 1, 1, 1, 2, 2, 1, 1, 0, 2, 0, 1, 1, 3, 1, 0, 2, 2, 2, 0, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 0, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 0, 2, 3, 1, 1, 2, 3, 1, 0, 1, 2, 1, 1, 2, 3, 1, 1, 0, 2, 1, 2, 2, 2, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 0, 1, 3, 1, 1, 3
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OFFSET
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1,6
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COMMENTS
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The zeros of this sequences are the powerful numbers (A001694). There are no arbitrarily long subsequences with a given upper bound; for example, every sequence of 4 values includes one divisible by 2 but not 4, so there are no more than 3 consecutive zeros. Similarly, there can be no more than 23 consecutive values with none divisible by both 2 and 3 but neither 4 nor 9 (so a(n) >= 2), etc. In general, this gives an upper bound that is a (relatively) small multiple of the k-th primorial number (prime(k)#). One suspects that the actual upper bounds for such subsequences are quite a bit lower; e.g., Erdős conjectured that there are no three consecutive powerful numbers. - Franklin T. Adams-Watters, Aug 08 2006
In particular, for every A048670(k)*A002110(k) consecutive terms, at least one is greater than or equal to k. - Charlie Neder, Jan 03 2019
Following Catalan's conjecture (which became Mihăilescu's theorem in 2002), the first case of two consecutive zeros in this sequence is for a(8) and a(9), because 8 = 2^3 and 9 = 3^2, and there are no other consecutive zeros for consecutive powers. However, there are other pairs of consecutive zeros at powerful numbers (A001694, A060355). The next example is a(288) = a(289) = 0, because 288 = 2^5 * 3^2 and 289 = 17^2, then also a(675) and a(676). - Bernard Schott, Jan 06 2019
a(2k-1) is the number of primes p such that p || x + y and p^2 || x^(2k-1) + y^(2k-1) for some positive integers x and y. For any positive integers x, y and k > 1, there is no prime p such that p || x + y and p^2 || x^(2k) + y^(2k). - Jinyuan Wang, Apr 08 2020
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LINKS
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FORMULA
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A prime factor of n is unitary iff its exponent is 1 in prime factorization of n. In general, gcd(p, n/p) = 1 or = p.
Additive with a(p^e) = 1 if e = 1, 0 otherwise.
(End)
Sum_{k=1..n} a(k) ~ n * (log(log(n)) + B - C), where B is Mertens's constant (A077761) and C = Sum_{p prime} (1/p^2) = 0.452247... (A085548). - Amiram Eldar, Sep 28 2023
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EXAMPLE
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9 = 3^2 so a(9) = 0; 10 = 2 * 5 so a(10) = 2; 11 = 11^1 so a(11) = 1.
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MAPLE
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a:= n-> nops(select(i-> i[2]=1, ifactors(n)[2])):
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MATHEMATICA
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Join[{0}, Table[Count[Transpose[FactorInteger[n]][[2]], 1], {n, 2, 110}]] (* Harvey P. Dale, Mar 15 2012 *)
Table[DivisorSum[n, 1 &, And[PrimeQ@ #, CoprimeQ[#, n/#]] &], {n, 105}] (* Michael De Vlieger, Nov 28 2017 *)
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PROG
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(Haskell)
a056169 = length . filter (== 1) . a124010_row
(Python)
from sympy import factorint
def a(n):
f=factorint(n)
return 0 if n==1 else sum(1 for i in f if f[i]==1)
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CROSSREFS
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Cf. A001221, A001694, A002110, A034444, A056170, A055231, A076445, A162642, A275812, A295659, A295662, A295664, A001694.
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KEYWORD
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nonn,nice,easy
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AUTHOR
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STATUS
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approved
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