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%I #5 Aug 01 2020 14:36:48
%S 2,5,15,14,7,25,7,36,30,26,15,68,2,21,125,78,9,94,7,90,52,33,38,208,
%T 15,11,74,69,3,227,2,166,66,33,129,276,2,25,101,228,7,115,8,76,329,47,
%U 28,482,19,65,40,50,7,248,99,234,46,34,20,572,2,5,426,356,16
%N Number of k = x*y such that phi(k) = n*(phi(x) + phi(y)).
%C If phi(x*y) = n*(phi(x) + phi(y)) and phi(x) <= phi(y), then phi(x) <= 2*n and phi(y) <= n*phi(x).
%C a(n) >= 1 because (k, x, y) = (4*n^2, 2*n, 2*n) is a solution.
%C If gcd(n, 6) = 1, then a(n) >= 2 because (k, x, y) = (12*n^2, 3*n, 4*n) is also a solution. Note that a(n) = 2 when n = 1, 13, 31, 37, 61, 73, 97, 103, 149, 151, 157, 181, ...
%C Conjecture: a(n) > 2 if n is composite.
%H Zhang Sibao and Xi Xiaozhong, <a href="http://www.cqvip.com/QK/96482X/201601/668542873.html">Positive integer solutions on phi(ab) = k*(phi(a) + phi(b))</a>
%p a(2) = 5 because k = 16, 24, 36, 40 and 60 satisfy the equation.
%o (PARI) f(n) = floor(n*exp(Euler)*log(log(n^2))+2.5*n/log(log(n^2)));
%o a(n) = {if(n==1, return(2)); my(t, v=List([])); for(x=1, f(2*n), if((t=eulerphi(x)) <= 2*n, for(y=1, f(t=n*t), if(eulerphi(x*y) == t+n*eulerphi(y), listput(v, x*y))))); #Set(v); }
%Y Cf. A000010, A057635, A336384, A336710.
%K nonn
%O 1,1
%A _Jinyuan Wang_, Aug 01 2020