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Antti Karttunen, <a href="/A318879/b318879.txt">Table of n, a(n) for n = 1..65537</a>
n = 105 has divisors [1, 3, 5, 7, 15, 21, 35, 105]. When A083254 is applied to them, we obtain [1, 1, 3, 5, 1, 3, 13, -9]. Summing the negative numbers present, and negating, we get a(105) = -(-9) = 9.
allocated for Antti Karttunen
a(n) = Sum_{d|n} [d-(2*phi(d)) > 0]*(d-(2*phi(d))).
0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 6, 0, 2, 0, 0, 0, 8, 0, 6, 0, 2, 0, 14, 0, 2, 0, 6, 0, 18, 0, 0, 0, 2, 0, 24, 0, 2, 0, 14, 0, 22, 0, 6, 0, 2, 0, 30, 0, 12, 0, 6, 0, 26, 0, 14, 0, 2, 0, 54, 0, 2, 0, 0, 0, 30, 0, 6, 0, 26, 0, 56, 0, 2, 0, 6, 0, 34, 0, 30, 0, 2, 0, 66, 0, 2, 0, 14, 0, 66, 0, 6, 0, 2, 0, 62, 0, 16, 0, 36, 0, 42, 0, 14, 9
1,6
(PARI) A318879(n) = sumdiv(n, d, d=d-(2*eulerphi(d)); (d>0)*d);
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Antti Karttunen, Sep 05 2018
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