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a(n) = Sum_{j=1..n*(n-1)} (n*j mod (n+j)).
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%I #17 Feb 09 2021 22:54:55

%S 0,2,19,67,185,373,742,1249,2053,3111,4672,6467,9113,12164,16124,

%T 20862,26801,33376,41889,51089,62342,75007,89949,106152,125610,146699,

%U 170757,197305,227912,259643,297469,336895,381304,429869,483295,539575,603725,670931,745068,823421,910928

%N a(n) = Sum_{j=1..n*(n-1)} (n*j mod (n+j)).

%C Sum goes up to j = n*(n-1) because n*j == j-n*(n-1) (mod (n+j)) for j >= n*(n-1).

%H Robert Israel, <a href="/A309341/b309341.txt">Table of n, a(n) for n = 1..2000</a>

%H Robert Israel, <a href="/A309341/a309341.png">Plot of a(n)/n^4 for 100 <= n <= 2000</a>

%F Conjecture: a(n) ~ c*n^4 where c = 0.32246....

%e For k = 3, 1*3 = 3 == 3 (mod 4), 2*3 = 6 == 1 (mod 5), 3*3 = 9 == 3 (mod 6), 4*3 = 12 == 5 (mod 7), 5*3=15 == 7 (mod 8), 6*3 = 18 == 0 (mod 9), so a(3) = 3+1+3+5+7 = 19.

%p f:= k -> add((k*j) mod (k+j), j=1..k*(k-1)-1):

%p map(f, [$1..30]);

%K nonn

%O 1,2

%A _J. M. Bergot_ and _Robert Israel_, Jul 24 2019