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 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)). 2
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS 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 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 Amir Hossein, Lifting The Exponent Lemma (Containing PDF file) MAPLE seq(2^min(n, padic:-ordp(numtheory:-sigma(n), 2)), n=1..100); # Robert Israel, Oct 23 2016 PROG (PARI) a(n) = gcd(2^n, sigma(n)); \\ Michel Marcus, Oct 15 2016 CROSSREFS Cf. A000203, A001157, A001158, A082902. Sequence in context: A014571 A324466 A152523 * A258770 A225815 A154589 Adjacent sequences:  A082900 A082901 A082902 * A082904 A082905 A082906 KEYWORD nonn,mult AUTHOR Labos Elemer, Apr 22 2003 STATUS approved

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Last modified October 16 13:32 EDT 2019. Contains 328093 sequences. (Running on oeis4.)