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A329534
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Irregular triangle read by rows: for n >= 1 row n lists the k from [1, 2, ... , n] such that A002378(k-1) = (k-1)*k == 0 (mod n).
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1
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1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 3, 4, 6, 1, 7, 1, 8, 1, 9, 1, 5, 6, 10, 1, 11, 1, 4, 9, 12, 1, 13, 1, 7, 8, 14, 1, 6, 10, 15, 1, 16, 1, 17, 1, 9, 10, 18, 1, 19, 1, 5, 16, 20, 1, 7, 15, 21, 1, 11, 12, 22, 1, 23, 1, 9, 16, 24, 1, 25
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OFFSET
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1,3
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COMMENTS
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n-th row length gives 1 for n = 1, and 2^A001221(n) for n >= 2 , that is A034444(n). [Proof: Unique lifting theorem (e.g., Apostol, 5.30 (a), p.121) for this congruence, and only two solutions 1 and p for primes p. See also the Yuval Dekel, Sep 21 2003, comment in A034444. - Wolfdieter Lang, Feb 05 2020]
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REFERENCES
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Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1986.
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LINKS
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EXAMPLE
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The irregular triangle T(n,k) begins
n\k 1 2 3 4 ...
1: 1
2: 1 2
3: 1 3
4: 1 4
5: 1 5
6: 1 3 4 6
7: 1 7
8: 1 8
9: 1 9
10: 1 5 6 10
11: 1 11
12: 1 4 9 12
13: 1 13
14: 1 7 8 14
15: 1 6 10 15
16: 1 16
17: 1 17
18: 1 9 10 18
19: 1 19
20: 1 5 16 20
...
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MATHEMATICA
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Table[Select[Range@ n, Mod[-n + # (# - 1), n] == 0 &], {n, 25}] // Flatten (* Michael De Vlieger, Nov 18 2019 *)
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PROG
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(Magma) [[k: k in [1..n] | k^2 mod n eq k]: n in [1..38]];
(PARI) row(n) = select(x->(Mod(x, n) == Mod(x, n)^2), [1..n]); \\ Michel Marcus, Nov 19 2019
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CROSSREFS
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Cf. A000010, A000225, A000688, A000961, A001221, A006881, A006530, A007875, A020639, A024619, A034444, A077610, A135972, A309307.
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KEYWORD
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nonn,easy,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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