A069521

[Davide Rotondo]

A069521
Smallest multiple of n with digit sum = 2, or 0 if no such number exists, e.g., a(3k)=0.
2, 2, 0, 20, 20, 0, 1001, 200, 0, 20, 11, 0, 1001, 10010, 0, 2000, 100000001, 0, 1000000001, 20, 0, 110, 100000000001, 0, 200, 10010, 0, 100100, 100000000000001, 0, 0, 20000, 0, 1000000010, 10010, 0, 0, 10000000010, 0, 200, 0, 0, 0, 1100, 0,...

1 2
2 2
3 0
4 20
5 20
6 0
7 1001
8 200
9 0
10 20
11 11
12 0
13 1001
14 10010
15 0
16 2000
17 100000001
18 0
19 1000000001
20 20
21 0
22 110
23 100000000001
24 0
25 200
26 10010
27 0
28 100100
29 100000000000001
30 0
31 0
32 20000
33 0
34 1000000010
35 10010
36 0
37 0
38 10000000010
39 0
40 200
41 0
42 0
43 0
44 1100
45 0
46 1000000000010
47 100000000000000000000001
48 0
49 1000000000000000000001
50 200
51 0
52 100100
53 0
54 0
55 110
56 1001000
57 0
58 1000000000000010
59 100000000000000000000000000001
60 0
61 1000000000000000000000000000001
62 0
63 0
64 200000
65 10010
66 0
67 0
68 10000000100
69 0
70 10010
71 0
72 0
73 10001
74 0
75 0
76 100000000100
77 1001
78 0
79 0
80 2000
81 0
82 0
83 0
84 0
85 1000000010
86 0
87 0
88 11000
89 10000000000000000000001
90 0
91 1001
92 10000000000100
93 0
94 1000000000000000000000010
95 10000000010
96 0
97 1000000000000000000000000000000000000000000000001
98 10000000000000000000010
99 0
100 200

and

A186635
Primes p such that the decimal expansion of 1/p has a periodic part of odd length.
2, 3, 5, 31, 37, 41, 43, 53, 67, 71, 79, 83, 107, 151, 163, 173, 191, 199, 227, 239, 271, 277, 283, 307, 311, 317, 347, 359, 397, 431, 439, 443, 467, 479, 523, 547, 563, 587, 599, 613, 631, 643, 683, 719, 733, 751, 757, 773, 787, 797, 827, 839, 853, 883, 907, 911, 919, 947, 991, 1013, 1031, 1039, 1093, 1123, 1151, 1163, 1187

It seems to be a match between these two sequences, what do you think?

See you soon

All the best

[Gareth McCaughan]

On 24/08/2025 15:44, Davide Rotondo wrote:

A069521
Smallest multiple of n with digit sum = 2, or 0 if no such number exists, e.g., a(3k)=0.
2, 2, 0, 20, 20, 0, 1001, 200, 0, 20, 11, 0, 1001, 10010, 0, 2000, 100000001, 0, 1000000001, 20, 0, 110, 100000000001, 0, 200, 10010, 0, 100100, 100000000000001, 0, 0, 20000, 0, 1000000010, 10010, 0, 0, 10000000010, 0, 200, 0, 0, 0, 1100, 0,...

...

A186635
Primes p such that the decimal expansion of 1/p has a periodic part of odd length.
2, 3, 5, 31, 37, 41, 43, 53, 67, 71, 79, 83, 107, 151, 163, 173, 191, 199, 227, 239, 271, 277, 283, 307, 311, 317, 347, 359, 397, 431, 439, 443, 467, 479, 523, 547, 563, 587, 599, 613, 631, 643, 683, 719, 733, 751, 757, 773, 787, 797, 827, 839, 853, 883, 907, 911, 919, 947, 991, 1013, 1031, 1039, 1093, 1123, 1151, 1163, 1187

It seems to be a match between these two sequences, what do you think?

A number has digit sum 2 exactly when it's twice a power of 10 (the only primes dividing these are 2 and 5) or it's 10^a+10^b for distinct a,b. So a prime p divides some such number precisely when it divides some 10^c+1; in other words, if -1 is a power of 10; in other words, if the order of 10 mod p is even.

The length of the periodic part of the base-10 expansion of 1/p is precisely the order of 10 mod p.

So, for p other than 2 or 5, A069521(p)=0 iff p is in A186635.

--
g

[Davide Rotondo]

Thanks 🙏 

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