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A082903
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Highest power of two that divides the sum of divisors of n.
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6
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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
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OFFSET
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1,3
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COMMENTS
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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).
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
Also the highest power of two that divides the sum of odd divisors of n. - Antti Karttunen, Mar 27 2022
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LINKS
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FORMULA
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(Some of these formulas were found by Sequence Machine.)
(End)
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MAPLE
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seq(2^min(n, padic:-ordp(numtheory:-sigma(n), 2)), n=1..100); # Robert Israel, Oct 23 2016
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MATHEMATICA
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Array[2^IntegerExponent[DivisorSigma[1, #], 2] &, 97] (* Michael De Vlieger, Apr 03 2022 *)
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PROG
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(Python)
from sympy import divisor_sigma
def A082903(n): return 1<<(~(m:=int(divisor_sigma(n))) & m-1).bit_length() # Chai Wah Wu, Jul 02 2022
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CROSSREFS
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KEYWORD
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nonn,mult
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AUTHOR
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EXTENSIONS
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Name replaced with a simpler one and the original definition moved to the Comments section by Antti Karttunen, Apr 03 2022
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STATUS
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approved
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