A329534 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).

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

Offset

1,3

Comments

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]

References

Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1986.

Links

Table of n, a(n) for n=1..69.

Example

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
...

Mathematica

Table[Select[Range@ n, Mod[-n + # (# - 1), n] == 0 &], {n, 25}] // Flatten (* Michael De Vlieger, Nov 18 2019 *)

Prog

(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

Crossrefs

Cf. A000010, A000225, A000688, A000961, A001221, A006881, A006530, A007875, A020639, A024619, A034444, A077610, A135972, A309307.

Sequence in context: A222266 A077609 A077610 * A317746 A364449 A228179

Adjacent sequences: A329531 A329532 A329533 * A329535 A329536 A329537

Keyword

nonn,easy,tabf

Author

Juri-Stepan Gerasimov, Nov 15 2019

Extensions

Edited by Wolfdieter Lang, Feb 05 2020

Status

approved