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Sum of remainders of mod(n, k), for 1 <= k <= sqrt(n).
1

%I #4 Nov 20 2013 16:34:57

%S 0,0,0,0,1,0,1,0,1,1,3,0,2,2,1,1,4,2,5,2,2,3,6,0,3,5,6,4,8,2,6,4,5,7,

%T 6,1,6,9,11,5,10,4,9,8,5,8,13,3,8,7,10,10,16,11,12,5,8,12,18,4,10,14,

%U 10,10,12,8,15,16,20,13,20,4,11,16,15,16,16,12,19,7,11,17,25,11,14,20,25

%N Sum of remainders of mod(n, k), for 1 <= k <= sqrt(n).

%C It appears, as one would expect, that a(n) is asymptotically 1/4 n.

%C 24 is the last n for which a(n) = 0.

%F a(n+1) = a(n) + floor(sqrt(n)) - A070039(n+1).

%t Total/@Table[Mod[n,k],{n,90},{k,Sqrt[n]}] (* _Harvey P. Dale_, Nov 20 2013 *)

%o (PARI) a(n) = sum(k=2,sqrtint(n),n%k)

%Y Cf. A004125, A070039.

%K nonn

%O 1,11

%A _Franklin T. Adams-Watters_, Apr 15 2010