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D. H. Lehmer shows that tau(n) == n*sigma_9(n) (mod 7), so a(n) is an integer for all n. Furthermore, if n == 3, 5, 6 (mod 7) then tau(n) == n*sigma_9(n) (mod 49). See the Wikipedia link below. It seems that the latter congruence also holds for most of the other numbers. Among the 571 numbers in [1, 1000] congruent to 0, 1, 2, 4 modulo 7, tau(n) == n*sigma_9(n) holds for 311 n's, and among the 5715 numbers in [1, 10000] congruent to 0, 1, 2, 4 modulo 7, the congruence holds for 3492 n's.
It seems that 150 divides a(n) for all n. There are no counterexamples for n <= 10000.
Number of n's in [2, N] which satisfy the higher-order congruence tau(n) == n*sigma_9(n) (mod 7^e) but not tau(n) == n*sigma_9(n) (mod 7^(e+1)):
N = 1000:
e | n == 3, 5, 6 (mod 7) | n == 0, 1, 2, 4 (mod 7) | total
---+----------------------+-------------------------+-------
1 | 0 | 260 | 260
---+----------------------+-------------------------+-------
2 | 358 | 80 | 438
---+----------------------+-------------------------+-------
3 | 45 | 195 | 240
---+----------------------+-------------------------+-------
4 | 24 | 28 | 52
---+----------------------+-------------------------+-------
5 | 2 | 5 | 7
---+----------------------+-------------------------+-------
6 | 0 | 2* | 2
* n = 686, 942.
N = 10000:
e | n == 3, 5, 6 (mod 7) | n == 0, 1, 2, 4 (mod 7) | total
---+----------------------+-------------------------+-------
1 | 0 | 2223 | 2223
---+----------------------+-------------------------+-------
2 | 3368 | 728 | 4096
---+----------------------+-------------------------+-------
3 | 466 | 2280 | 2746
---+----------------------+-------------------------+-------
4 | 397 | 384 | 781
---+----------------------+-------------------------+-------
5 | 46 | 87 | 133
---+----------------------+-------------------------+-------
6 | 6 | 12 | 18
---+----------------------+-------------------------+-------
7 | 2** | 0 | 2
** n = 5185, 9021.
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