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a(n) = 1 + sum of the anti-divisors of n.
6

%I #7 Dec 25 2017 13:54:25

%S 1,1,3,4,6,5,11,9,9,15,13,14,20,17,19,15,29,29,19,25,23,37,35,24,40,

%T 25,43,47,25,37,43,59,49,31,53,33,51,71,53,56,42,67,57,41,87,59,61,57,

%U 73,81,43,95,89,53,75,57,75,97,91,108,58,79,113,47,85

%N a(n) = 1 + sum of the anti-divisors of n.

%C See A066272 for definition of anti-divisor.

%H Jon Perry, <a href="http://www.users.globalnet.co.uk/~perry/maths">Anti-divisors</a> [Broken link]

%H Jon Perry, <a href="/A066272/a066272a.html">The Anti-divisor</a> [Cached copy]

%H Jon Perry, <a href="/A066272/a066272.html">The Anti-divisor: Even More Anti-Divisors</a> [Cached copy]

%F a(n) = A066417(n) + 1.

%e Consider n = 18: 2n-1, 2n, 2n+1 are 35, 36, 37 with odd divisors > 1 {5,7,35}, {3,9}, {37} respectively and quotients {7, 5, 1}, {12, 4}, {1}; so the anti-divisors of 18 are 4, 5, 7, 12. Therefore a(18) = 1 + 28 = 29.

%Y Cf. A066417, A066241, A066452.

%K nonn

%O 1,3

%A _Jon Perry_, Dec 28 2001