%I
%S 1,1,4,1,2,4,8,1,1,2,4,4,2,8,8,1,2,1,4,2,32,4,8,4,1,2,8,8,2,8,32,1,16,
%T 2,16,1,2,4,8,2,2,32,4,4,2,8,16,4,1,1,8,2,2,8,8,8,16,2,4,8,2,32,8,1,4,
%U 16,4,2,32,16,8,1,2,2,4,4,32,8,16,2,1,2,4,32,4,4,8,4,2,2,16,8,128,16,8,4,2
%N a(n) = gcd(2^n, sigma_1(n)) = gcd(A000079(n), A000203(n)) also a(n) = gcd(2^n, sigma_3(n)) = gcd(A000079(n), A001158(n)).
%C a(n) = gcd(2^n, sigma_k(n)) when k is an odd positive integer. Proof: It suffices to show that v_2(sigma_k(n)) does not depend on k, where v_2(n) is the 2adic valuation of n. Since v_2(ab) = v_2(a)+v_2(b) and sigma_k(n) is an arithmetic function, we need only prove it for n=p^e with p prime. If p is 2 or e is even, sigma_k(p^e) is odd, so we can disregard those cases. Otherwise, we sum the geometric series to obtain v_2(sigma_k(p^e)) = v_2(p^(k(e+1))1)v_2(p1). Applying the wellknown LTE Lemma (see Hossein link) Case 4 arrives at v_2(p^(k(e+1))1)v_2(p1) = v_2(p+1)+v_2(k(e+1))1. But v_2(k(e+1)) = v_2(k)+v_2(e+1), and k is odd, so we conclude that v_2(sigma_k(p^e)) = v_2(p+1)+v_2(e+1)1, a result independent of k.  _Rafay A. Ashary_, Oct 15 2016
%H Robert Israel, <a href="/A082903/b082903.txt">Table of n, a(n) for n = 1..10000</a>
%H Amir Hossein, <a href="http://www.artofproblemsolving.com/community/c6h393335">Lifting The Exponent Lemma (Containing PDF file)</a>
%p seq(2^min(n, padic:ordp(numtheory:sigma(n),2)), n=1..100); # _Robert Israel_, Oct 23 2016
%o (PARI) a(n) = gcd(2^n, sigma(n)); \\ _Michel Marcus_, Oct 15 2016
%Y Cf. A000203, A001157, A001158, A082902.
%K nonn,mult
%O 1,3
%A _Labos Elemer_, Apr 22 2003
