[seqfan] cototient divisibility in A196736

[Richard Mathar]

I have problems reproducing http://oeis.org/A196736 .
According to my implementation, 24 and 30 are the first terms
that should be missing from the sequence, not 10 and 14 .

The sequence k(n,x) sorted along rows n>= appears to be (with n after the colon, 
see also A197099)
1,0,0,0,0,0,0,0,0,0,:1
1,1,0,0,0,0,0,0,0,0,:2
2,1,0,0,0,0,0,0,0,0,:3
2,1,1,0,0,0,0,0,0,0,:4
4,1,0,0,0,0,0,0,0,0,:5
2,3,0,1,0,0,0,0,0,0,:6
6,1,0,0,0,0,0,0,0,0,:7
4,2,1,1,0,0,0,0,0,0,:8
6,2,1,0,0,0,0,0,0,0,:9
4,5,0,1,0,0,0,0,0,0,:10
10,1,0,0,0,0,0,0,0,0,:11
4,4,2,1,0,1,0,0,0,0,:12
12,1,0,0,0,0,0,0,0,0,:13
6,7,0,1,0,0,0,0,0,0,:14
8,6,0,1,0,0,0,0,0,0,:15
8,4,2,1,1,0,0,0,0,0,:16
16,1,0,0,0,0,0,0,0,0,:17
6,8,1,2,0,1,0,0,0,0,:18
18,1,0,0,0,0,0,0,0,0,:19
8,6,4,1,0,1,0,0,0,0,:20
12,8,0,1,0,0,0,0,0,0,:21
10,11,0,1,0,0,0,0,0,0,:22
22,1,0,0,0,0,0,0,0,0,:23
8,8,2,4,0,1,0,1,0,0,:24
20,4,1,0,0,0,0,0,0,0,:25

So the complementary row sums n-sum_{j>=1} k(n,j) are
(with cototient(n), then n after the colon):
0,0,0,0,0,0,0,0,0,0,:0 1
1,0,0,0,0,0,0,0,0,0,:1 2
1,0,0,0,0,0,0,0,0,0,:1 3
2,1,0,0,0,0,0,0,0,0,:2 4
1,0,0,0,0,0,0,0,0,0,:1 5
4,1,1,0,0,0,0,0,0,0,:4 6
1,0,0,0,0,0,0,0,0,0,:1 7
4,2,1,0,0,0,0,0,0,0,:4 8
3,1,0,0,0,0,0,0,0,0,:3 9
6,1,1,0,0,0,0,0,0,0,:6 10
1,0,0,0,0,0,0,0,0,0,:1 11
8,4,2,1,1,0,0,0,0,0,:8 12
1,0,0,0,0,0,0,0,0,0,:1 13
8,1,1,0,0,0,0,0,0,0,:8 14
7,1,1,0,0,0,0,0,0,0,:7 15
8,4,2,1,0,0,0,0,0,0,:8 16
1,0,0,0,0,0,0,0,0,0,:1 17
12,4,3,1,1,0,0,0,0,0,:12 18
1,0,0,0,0,0,0,0,0,0,:1 19
12,6,2,1,1,0,0,0,0,0,:12 20
9,1,1,0,0,0,0,0,0,0,:9 21
12,1,1,0,0,0,0,0,0,0,:12 22
1,0,0,0,0,0,0,0,0,0,:1 23
16,8,6,2,2,1,1,0,0,0,:16 24
5,1,0,0,0,0,0,0,0,0,:5 25
So the nonzero numbers in the table (left from the colon) divide the cototient(n)
(right after the colon) until we reach 24. At n=24, the 6 does not divide the 16.

Where is the error?
RJM