|
%I #14 Dec 22 2017 16:10:16
%S 1,2,2,4,3,5,3,6,4,6,3,9,3,6,6,10,5,11,5,12,8,8,5,15,6,7,7,11,4,14,4,
%T 13,7,9,8,20,6,11,9,19,6,18,6,15,13,11,6,24,6,12,9,12,4,17,8,16,9,8,4,
%U 25,4,8,13,19,11,20,7,17,12,19,7,32,7,13,15,19,11,21,7,29,11,11,5,29,11,11,9
%N a(1)=1. a(n) = Sum_{k|n} a(floor(sqrt(k))).
%e The divisors of 12 are 1,2,3,4,6,12. The floor of the square roots of these are 1,1,1,2,2,3. So a(12) = a(1) + a(1) + a(1) + a(2) + a(2) + a(3) = 1 + 1 + 1 + 2 + 2 + 2 = 9.
%p A133438 := proc(n) option remember ; local a,d; if n = 1 then 1; else a := 0 ; for d in numtheory[divisors](n) do a := a+ procnamefloor(sqrt(d))) ; od: RETURN(a) ; fi ; end:
%p seq(A133438(n),n=1..60) ; # _R. J. Mathar_, Jan 08 2008
%K nonn
%O 1,2
%A _Leroy Quet_, Nov 26 2007
%E More terms from _R. J. Mathar_ and _Matthew Conroy_, Jan 08 2008
|
