A184395 a(n) = number of distinct values obtained when sigma is applied to the divisors of n.

1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 5, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 6, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 2, 10, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 7, 4, 7, 2, 6, 4, 8, 2, 12, 2, 4, 6, 6, 4, 8, 2, 10, 5, 4, 2, 12, 4, 4, 4, 8, 2, 12, 4, 6, 4, 4, 4, 12, 2, 6, 6, 9, 2, 8, 2, 8, 8, 4, 2, 12, 2, 8, 4, 10, 2, 8, 4

Offset: 1

Author

Jaroslav Krizek, Jan 12 2011

Keywords

• nonn

Comments

Sequence is not the same as A000005(n): a(66) = 7, A000005(66) = 8.

a(n) = number of numbers k <= sigma(n) such that k = sigma(d) for some divisor d of n, where sigma = A000203. - This is the original name of the sequence, except that I substituted "some divisor" for "any divisor". - Antti Karttunen, Aug 24 2017

Examples

			For n = 4, sigma(4) = 7, from numbers 1 - 7 there are three numbers k such that k = sigma(d) for any divisor d of n: 1 = sigma(1), 3 = sigma(2), 7 = sigma(4); a(4) = 3.
From _Antti Karttunen_, Aug 24 2017: (Start)
For n = 66, its 8 divisors are [1, 2, 3, 6, 11, 22, 33, 66]. When applying sigma to these, we obtain [1, 3, 4, 12, 12, 36, 48, 144], with one duplicate present, thus there are only 8-1 = 7 distinct values and a(66) = 7.
For n = 70, its 8 divisors are [1, 2, 5, 7, 10, 14, 35, 70]. When applying sigma to these, we obtain [1, 3, 6, 8, 18, 24, 48, 144], which are all unique values, thus a(70) = 8.
(End)
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A000203, A184396.

Programs

Mathematica

Table[Length[Union[DivisorSigma[1,Divisors[n]]]],{n,120}] (* Harvey P. Dale, Jun 20 2023 *)

PARI

A184395(n) = length(vecsort(apply(d->sigma(d),divisors(n)), , 8)); \\ Antti Karttunen, Aug 24 2017

Formula

a(n) = A000203(n) - A184396(n).

Extensions

Name changed, a(66) and a(70) corrected and more terms added by Antti Karttunen, Aug 24 2017