a(3) = 0 since for j=1, k=2, 1*3+2 = 5 and 2*3+1 = 7, which have distinct squarefree kernels.
a(4) = 0 since {j,k} is {1,2}, {1,3}, and {2,3} are such that rad(j*n+k) != rad(k*n+j).
a(5) = 1 since for j=1, k=3, 1*5+3 = 8 and 3*5+1 = 16, and 8 and 16 have the same squarefree kernel 2. This is true for no other pairs 0 < j < k < 5.
a(11) = 4 since there are 4 pairs {j,k} such that rad(j*n+k) = rad(k*n+j). These are as follows:
{1,4}: 1*11+4 = 15, 4*11+1 = 45, both with squarefree kernel 15;
{1,9}: 1*11+9 = 20, 9*11+1 = 100, both with squarefree kernel 10;
{2,8}: 2*11+8 = 30, 8*11+2 = 90, both with squarefree kernel 30;
{3,7}: 3*11+7 = 40, 7*11+3 = 80, both with squarefree kernel 10.
Table of n, a(n) for select n, listing solutions {j,k} such that rad(j*n+k) = rad(k*n+j):
n a(n) Solutions {j,k}
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5 1 {1,3}.
7 1 {1,5}.
9 1 {1,7}.
11 4 {1,4}, {1,9}, {2,8}, {3,7}.
13 4 {1,11}, {4,11}, {5,7}, {5,10}.
15 1 {1,13}, {9,12}.
17 2 {1,15}, {5,11}.
19 7 {1,5}, {1,17}, {2,7}, {2,10},
{3,15}, {4,14}, {7,11}.
20 2 {2,8}, {4,16}.
21 4 {1,19}, {2,13}, {3,9}, {6,18}.
22 3 {4,10}, {8,20}, {10,20}.
23 3 {1,21}, {5,17}, {7,15}.
25 3 {1,14}, {1,23}, {7,17}.
27 1 {1,25}.
29 20 {1,6}, {1,11}, {1,13}, {1,27}, {2,5},
{2,12}, {2,22}, {2,26}, {3,18}, {4,10},
{4,24}, {5,23}, {6,15}, {7,13}, {8,20},
{9,19}, {10,25}, {11,17}, {13,15}, {14,26}.
