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A322582
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a(n) = n - A003958(n), where A003958 is fully multiplicative with a(p) = (p-1).
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15
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0, 1, 1, 3, 1, 4, 1, 7, 5, 6, 1, 10, 1, 8, 7, 15, 1, 14, 1, 16, 9, 12, 1, 22, 9, 14, 19, 22, 1, 22, 1, 31, 13, 18, 11, 32, 1, 20, 15, 36, 1, 30, 1, 34, 29, 24, 1, 46, 13, 34, 19, 40, 1, 46, 15, 50, 21, 30, 1, 52, 1, 32, 39, 63, 17, 46, 1, 52, 25, 46, 1, 68, 1, 38, 43, 58, 17, 54, 1, 76, 65, 42, 1, 72, 21, 44, 31, 78, 1
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OFFSET
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1,4
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COMMENTS
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a(p*(n/p)) - (n/p) = (p-1)*a(n/p) holds for all prime divisors p of n, which can be seen by expanding the left hand side as p*(n/p) - A003958(p*(n/p)) - (n/p) = (p-1)*(n/p) - (p-1)*A003958(n/p) = (p-1)*((n/p) - A003958(n/p)) = (p-1)*a(n/p). This shows that this sequence gives a lower limit for arithmetic derivative (A003415) in the same way as A348507 gives an upper limit for it. - Antti Karttunen, Nov 07 2021
With n = Product_{i=1..k} p_i the prime factorization of n, if one constructs for each i a test with a probability of success equal to 1/p_i, and if the tests are independent, then a(n)/n is the probability that at least one of the k tests succeeds. - Luc Rousseau, Jan 14 2023
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LINKS
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FORMULA
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(End)
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MATHEMATICA
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a[1] = 0; a[n_] := n - Times @@ ((First[#] - 1)^Last[#] & /@ FactorInteger[n]); Array[a, 60] (* Amiram Eldar, Dec 17 2018 *)
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PROG
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(PARI)
A003958(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1]--); factorback(f); };
(PARI)
A020639(n) = if(1==n, n, (factor(n)[1, 1]));
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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