

A132586


Numbers n such that sigma(n+1)n2 divides sigma(n)n1, where sigma(n) is sum of positive divisors of n and the ratio is greater than zero.


3




OFFSET

1,1


COMMENTS

The banal case of ratio equal to zero is excluded. In fact if n is a prime than sigma(n)n1=0. Therefore the ratio with sigma(n+1)n2 is equal to zero. Is this sequence finite?
a(6), if it exists, is larger than 10^13.  Giovanni Resta, Jul 13 2015


LINKS

Table of n, a(n) for n=1..5.


EXAMPLE

n=8 > sigma(8)=1+2+4+8 > sigma(n)n1=2+4=6.
n+1=9 > sigma(9)=1+3+9 > sigma(n+1)n2=3.
6/3 = 2 (integer >0)


MAPLE

with(numtheory); P:=proc(n) local a, i; for i from 1 by 1 to n do if sigma(i+1)i2>0 then a:=(sigma(i)i1)/(sigma(i+1)i2); if a>0 and trunc(a)=a then print(i); fi; fi; od; end: P(100000);


CROSSREFS

Cf. A002961, A058072, A058073, A132585.
Sequence in context: A105063 A166483 A274303 * A208400 A103953 A076444
Adjacent sequences: A132583 A132584 A132585 * A132587 A132588 A132589


KEYWORD

hard,more,nonn


AUTHOR

Paolo P. Lava and Giorgio Balzarotti, Aug 23 2007


EXTENSIONS

a(5) from Donovan Johnson, Aug 31 2008


STATUS

approved



