cp's OEIS Frontend

A082903 Highest power of two that divides the sum of divisors of n.

%I A082903 #51 Jul 02 2022 13:15:10
%S A082903 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 A082903 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 A082903 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 A082903 Highest power of two that divides the sum of divisors of n.
%C A082903 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)). (The original, equivalent definition for this sequence).
%C A082903 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 2-adic 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(p-1). Applying the well-known LTE Lemma (see Hossein link) Case 4 arrives at v_2(p^(k(e+1))-1)-v_2(p-1) = 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
%C A082903 Also the highest power of two that divides the sum of odd divisors of n. - _Antti Karttunen_, Mar 27 2022
%H A082903 Robert Israel, <a href="/A082903/b082903.txt">Table of n, a(n) for n = 1..10000</a>
%H A082903 Amir Hossein, <a href="http://www.artofproblemsolving.com/community/c6h393335">Lifting The Exponent Lemma (Containing PDF file)</a>
%H A082903 Jon Maiga, <a href="http://sequencedb.net/s/A082903">Computer-generated formulas for A082903</a>, Sequence Machine.
%F A082903 From _Antti Karttunen_, Mar 27 2022: (Start)
%F A082903 (Some of these formulas were found by Sequence Machine.)
%F A082903 a(n) = a(A000265(n)) = a(2*n).
%F A082903 a(n) = A006519(A000593(n)) = A006519(A000203(n)) = A000203(n) / A161942(n).
%F A082903 a(n) = 2^(A286357(n)-1).
%F A082903 (End)
%F A082903 a(n) = 2^A336937(n). - _Chai Wah Wu_, Jul 02 2022
%p A082903 seq(2^min(n, padic:-ordp(numtheory:-sigma(n),2)), n=1..100); # _Robert Israel_, Oct 23 2016
%t A082903 Array[2^IntegerExponent[DivisorSigma[1, #], 2] &, 97] (* _Michael De Vlieger_, Apr 03 2022 *)
%o A082903 (PARI) a(n) = gcd(2^n, sigma(n)); \\ _Michel Marcus_, Oct 15 2016
%o A082903 (PARI) A082903(n) = (2^valuation(sigma(n), 2)); \\ _Antti Karttunen_, Mar 27 2022
%o A082903 (Python)
%o A082903 from sympy import divisor_sigma
%o A082903 def A082903(n): return 1<<(~(m:=int(divisor_sigma(n))) & m-1).bit_length() # _Chai Wah Wu_, Jul 02 2022
%Y A082903 Cf. A000203, A000265, A000593, A001157, A001158, A006519, A007814, A082902, A161942, A286357.
%Y A082903 Cf. A028982 (positions of 1's).
%K A082903 nonn,mult
%O A082903 1,3
%A A082903 _Labos Elemer_, Apr 22 2003
%E A082903 Name replaced with a simpler one and the original definition moved to the Comments section by _Antti Karttunen_, Apr 03 2022