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%I
%S 1,1,5,10,30,26,91,84,159,140,385,196,650,406,620,680,1496,654,2109,
%T 1080,1806,1650,3795,1544,4150,2756,4365,3164,7714,2360,9455,5456,
%U 7370,6256,9940,5196,16206,8778,12324,8560,22140,6972,25585
%N Sum_{k=1..n, gcd(n,k) = 1} k^2.
%C Equals row sums of triangle A143612 [From _Gary W. Adamson_, Aug 27 2008]
%C a(n) = A175505(n) * A023896(n) / A175506(n). For number n >= 1 holds B(n) = a(n) / A023896(n) = A175505(n) / A175506(n), where B(n) = antiharmonic mean of numbers k such that GCD(k, n) = 1 for k < n. Cf. A179871, A179872, A179873, A179874, A179875, A179876, A179877, A179878, A179879, A179880, A179882, A179883, A179884, A179885, A179886, A179887, A179890, A179891, A007645, A003627, A034934. [From _Jaroslav Krizek_, Aug 01 2010]
%D T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 48, problem 15, the function phi_2(n).
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 199, #2.
%H P. G. Brown, <a href="http://www.jstor.org/stable/3621931">Some comments on inverse arithmetic functions</a>, Math. Gaz. 89 (516) (2005) 403-408.
%F If n = p_1^e_1 * ... *p_r^e_r then a(n) = n^2*phi(n)/3 + (-1)^r*p_1*..._p_r*phi(n)/6.
%F a(n) = n^2*A000010(n)/3 + n*A023900(n)/6, n>1. [Brown]
%F a(n) = A000010(n)/3 * (n^2 + (-1)^A001221(n)*A007947(n)/2)) for n>=2. [From _Jaroslav Krizek_, Aug 24 2010]
%t f[n_] := Plus @@ (Select[ Range@n, GCD[ #, n] == 1 &]^2); Array[f, 43] [From _Robert G. Wilson v_, Jul 01 2010]
%Y Cf. A023896, A053819, A053820.
%Y A143612 [From _Gary W. Adamson_, Aug 27 2008]
%K nonn
%O 1,3
%A _N. J. A. Sloane_, Apr 07 2000
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