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A343239 Irregular triangle read by rows giving the solutions x for x^2 == -5 (mod A343238(n)), for x from {0, 1, 2, ..., A343238(n)-1}, for n >= 1. 5
0, 1, 1, 2, 0, 1, 5, 3, 4, 2, 7, 5, 3, 11, 5, 10, 7, 11, 4, 10, 11, 17, 8, 15, 7, 20, 13, 16, 5, 25, 10, 25, 6, 35, 11, 17, 25, 31, 9, 34, 20, 25, 15, 31, 18, 29, 17, 32, 7, 47, 13, 45, 19, 42, 11, 25, 38, 52, 14, 53, 8, 31, 38, 61, 25, 45, 20, 61, 35, 47, 24, 59, 9, 77, 13, 16, 71, 74 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
COMMENTS
The length of row n is A343240(n).
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1468 (first 500 rows)
FORMULA
T(n, k) gives the solutions x from {0, 1, ..., A343238(n)-1} of the congruence x^2 + 5 == 0 (mod A343238(n)), for n >= 1.
EXAMPLE
The irregular triangle T with A(n) = A343238(n) begins:
n A(n) \ k 1 2 3 4 ...
---------------------------
1, 1: 0
2, 2: 1
3, 3: 1 2
4, 5: 0
5, 6: 1 5
6, 7: 3 4
7, 9: 2 7
8, 10: 5
9, 14: 3 11
10, 15: 5 10
11, 18: 7 11
12, 21: 4 10 11 17
13, 23: 8 15
14, 27: 7 20
15, 29: 13 16
16, 30: 5 25
17, 35: 10 25
18, 41: 6 35
19, 42: 11 17 25 31
20, 43: 9 34
...
PROG
(PARI) isok(k) = issquare(Mod(-5, k)); \\ A343238
lista(nn) = my(list = List()); for (n=1, nn, if (issquare(Mod(-5, n)), my(row = select(x->(Mod(x, n)^2 + 5 == 0), [0..n-1])); listput(list, row))); Vec(list); \\ Michel Marcus, Sep 17 2023
CROSSREFS
Sequence in context: A112340 A356264 A037186 * A004483 A197808 A085650
KEYWORD
nonn,tabf,easy
AUTHOR
Wolfdieter Lang, May 16 2021
STATUS
approved

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Last modified March 28 05:39 EDT 2024. Contains 371235 sequences. (Running on oeis4.)