A054030 - OEIS

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A054030

Sigma(n)/n for n such that sigma(n) is divisible by n.

1, 2, 2, 3, 2, 3, 2, 4, 4, 3, 4, 4, 2, 4, 4, 3, 4, 3, 2, 5, 5, 4, 3, 4, 2, 4, 4, 5, 4, 5, 5, 4, 5, 5, 4, 4, 4, 5, 4, 4, 2, 5, 4, 5, 6, 5, 5, 5, 5, 5, 5, 6, 5, 5, 4, 5, 6, 5, 4, 4, 5, 4, 5, 4, 6, 6, 6, 6, 6, 6, 6, 6, 5, 6, 6, 5, 6, 5, 6, 6, 5, 4, 4, 5, 4, 4, 5, 6, 5, 5, 4, 6, 4, 4, 6, 5, 6, 6, 6, 6, 6, 6, 6, 5, 6

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

The graph supports the conjecture that all numbers except 2 appear only a finite number of times. Sequences A000396, A005820, A027687, A046060 and A046061 give the n for which the abundancy sigma(n)/n is 2, 3, 4, 5 and 6, respectively. See A134639 for the number of n having abundancy greater than 2. - T. D. Noe, Nov 04 2007

LINKS

T. D. Noe, Table of n, a(n) for n = 1..1600 (using Flammenkamp's data)

Achim Flammenkamp, The Multiply Perfect Numbers Page

Eric Weisstein's World of Mathematics, Abundancy

FORMULA

a(n) = sigma(A007691(n))/A007691(n)

MAPLE

with(numtheory): for i while i < 33000 do

if sigma(i) mod i = 0 then print(sigma(i)/i) fi od;

PROG

(PARI) for(n=1, 1e7, if(denominator(k=sigma(n, -1))==1, print1(k", "))) \\ Charles R Greathouse IV, Mar 09 2014

CROSSREFS