Sigma_2(n)

Fri Mar 26 11:33:14 CET 2004

```The sequence is 1,4,9,16,20,25,36,49,50,64,81,100,117
Sigma_2(20)= 546
Sigma(20) = 42
and Sigma_2(20)/Sigma(20) =13 then 20 is candidate
Look at A020487 and A046870

%I A020487
%S A020487 1,4,9,16,20,25,36,49,50,64,81,100,117,121,144,169,180,196,200,225,242,
%T A020487 256,289,324,325,361,400,441,450,468,484,500,529,576,578,605,625,650,676,
%U A020487 729,784,800,841,900,961,968,980,1024,1025,1058,1089,1156,1225,1280,1296
%N A020487 sigma_1(n) divides sigma_2(n).
%Y A020487 Sequence in context: A066213 A010441 A046871 this_sequence A046870 A010461 A010421
%Y A020487 Adjacent sequences: A020484 A020485 A020486 this_sequence A020488 A020489 A020490
%K A020487 nonn
%O A020487 1,2
%A A020487 David W. Wilson (davidwwilson(AT)comcast.net)

Thanks

Mohammed BOUAYOUN

> I knew the following sequence doesn't exist on OEIS.
>    a(n) : 1,4,9,16,25,36,49,50,64,81,100,117,
>         Sigma_2(n)/Sigma(n)=integer
>    example :
>         Sigma_2(4)/Sigma(4)=(1+4+16)/(1+2+4)=3 , so 4 appears on a(n).
>    It is easy but interesting.

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