Logo

Greetings from The On-Line Encyclopedia of Integer Sequences!

Hints

Search: id:A055210
Displaying 1-1 of 1 results found. page 1
     Format: long | short | internal | text      Sort: relevance | references | number      Highlight: on | off
%I A055210
%S A055210 1,1,1,3,1,1,1,3,7,1,1,3,1,1,1,11,1,7,1,3,1,1,1,3,21,1,7,3,1,1,1,11,1,
%T A055210 1,1,21,1,1,1,3,1,1,1,3,7,1,1,11,43,21,1,3,1,7,1,3,1,1,1,3,1,1,7,43,1,
%U A055210 1,1,3,1,1,1,21,1,1,21,3,1,1,1,11,61,1,1,3,1,1,1,3,1,7,1,3,1,1,1,11,1
%N A055210 Sum of totients of square divisors of n.
%F A055210 a(n) = Sum[Phi[d]; d is square and divides n]
%F A055210 Multiplicative with a(p^e) = (p^(e+1)+1)/(p+1) for even e and a(p^e) 
               = (p^e+1)/(p+1) for odd e. - Vladeta Jovovic (vladeta(AT)eunet.rs), 
               Dec 01 2001
%e A055210 n = 400: its square divisors are {1, 4, 16, 25, 100, 400}, their totients 
               are {1, 2, 8, 20, 40, 160} and the totient-sum over these divisors 
               is, so a(400) = 231. This value arises at special square-free multiples 
               of 400 (400 times 2, 3, 5, 6, 7, 10, 11, 13, 15, 17, 19, 21, 22, 
               23 etc.)
%e A055210 a(400) = a(2^4*5^2) = (2^5+1)/3*(5^3+1)/6 = 231.
%Y A055210 Sequence in context: A131270 A109223 A016466 this_sequence A082553 A143632 
               A130605
%Y A055210 Adjacent sequences: A055207 A055208 A055209 this_sequence A055211 A055212 
               A055213
%K A055210 nonn,mult
%O A055210 1,4
%A A055210 Labos E. (labos(AT)ana.sote.hu), Jun 19 2000

    
page 1

Search completed in 0.002 seconds

Lookup | Welcome | Find friends | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Puzzles | Hot | Classics
More pages | Superseeker | Maintained by N. J. A. Sloane (njas@research.att.com)

Last modified December 15 00:47 EST 2009. Contains 170825 sequences.


AT&T Labs Research