%I
%S 1,4,9,25,81,289,1681,3481,5041,7921,10201,17161,27889,29929,85849,
%T 146689,331776,458329,491401,552049,579121,597529,683929,703921,
%U 734449,786432,829921,1190281,1203409,1352569,1394761,1423249,1481089,1885129,2036329,2211169
%N Numbers k such that tau(sigma(tau(k))) = sigma(tau(sigma(k))), where tau is A000005 and sigma is A000203.
%C The squares of the terms of A053182 are a subset of this sequence. In fact, in general, if p is prime we have tau(p)=2 and tau(p^2)=3. Therefore tau(p^2)=3 > sigma(3)=4 > tau(4)=tau(2^2)=3 and if p belongs to A053182 we also have that sigma(p^2)=p^2+p+1 (prime) > tau(p^2+p+1)=2 > sigma(2)=3.
%p with(numtheory); P:=proc(q) local n;
%p for n from 1 to q do
%p if tau(sigma(tau(n)))=sigma(tau(sigma(n))) then print(n); fi;
%p od; end: P(10^6);
%o (PARI) s=[]; for(n=1, 2500000, if(sigma(sigma(sigma(n, 0)), 0) == sigma(sigma(sigma(n), 0)), s=concat(s, n))); s \\ _Colin Barker_, Feb 10 2014
%Y Cf. A000005, A000203, A053182.
%K nonn,changed
%O 1,2
%A _Paolo P. Lava_, Feb 10 2014
