|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
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
|
|
|
FORMULA
|
|
|
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.
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)
|
|
MATHEMATICA
|
Table[Length[Union[DivisorSigma[1, Divisors[n]]]], {n, 120}] (* Harvey P. Dale, Jun 20 2023 *)
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
Name changed, a(66) and a(70) corrected and more terms added by Antti Karttunen, Aug 24 2017
|
|
STATUS
|
approved
|
|
|
|