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

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

Offset: 1

Author

Labos Elemer, Apr 22 2003

Keywords

• nonn
• mult

Comments

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

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Amir Hossein, Lifting The Exponent Lemma (Containing PDF file)
• Jon Maiga, Computer-generated formulas for A082903, Sequence Machine.

Crossrefs

Cf. A000203, A000265, A000593, A001157, A001158, A006519, A007814, A082902, A161942, A286357.

Cf. A028982 (positions of 1's).

Programs

Maple

seq(2^min(n, padic:-ordp(numtheory:-sigma(n),2)), n=1..100); # Robert Israel, Oct 23 2016

Mathematica

Array[2^IntegerExponent[DivisorSigma[1, #], 2] &, 97] (* Michael De Vlieger, Apr 03 2022 *)

PARI

a(n) = gcd(2^n, sigma(n)); \\ Michel Marcus, Oct 15 2016

PARI

A082903(n) = (2^valuation(sigma(n), 2)); \\ Antti Karttunen, Mar 27 2022

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

Formula

From Antti Karttunen, Mar 27 2022: (Start)

(Some of these formulas were found by Sequence Machine.)

a(n) = a(A000265(n)) = a(2*n).

a(n) = A006519(A000593(n)) = A006519(A000203(n)) = A000203(n) / A161942(n).

a(n) = 2^(A286357(n)-1).

(End)

a(n) = 2^A336937(n). - Chai Wah Wu, Jul 02 2022

Extensions

Name replaced with a simpler one and the original definition moved to the Comments section by Antti Karttunen, Apr 03 2022