

A184395


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


5



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
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OFFSET

1,2


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


LINKS

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


FORMULA

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


EXAMPLE

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 81 = 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)


PROG

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


CROSSREFS

Cf. A000005, A000203, A184396.
Sequence in context: A335519 A167447 A134687 * A329484 A179941 A179942
Adjacent sequences: A184392 A184393 A184394 * A184396 A184397 A184398


KEYWORD

nonn


AUTHOR

Jaroslav Krizek, Jan 12 2011


EXTENSIONS

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


STATUS

approved



