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A381348
Irregular triangle read by rows in which row n lists the largest subset of Z/nZ fixed by x -> x^2.
2
0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 3, 4, 0, 1, 2, 4, 0, 1, 0, 1, 4, 7, 0, 1, 5, 6, 0, 1, 3, 4, 5, 9, 0, 1, 4, 9, 0, 1, 3, 9, 0, 1, 2, 4, 7, 8, 9, 11, 0, 1, 6, 10, 0, 1, 0, 1, 0, 1, 4, 7, 9, 10, 13, 16, 0, 1, 4, 5, 6, 7, 9, 11, 16, 17, 0, 1, 5, 16, 0, 1, 4, 7, 9, 15, 16, 18
OFFSET
1,12
COMMENTS
Row n has length A277847(n).
Repeated squaring (application of f: x -> x^2) of the set of integers mod n results in a set that is fixed under f. Row n lists this set for modulus n. The number of applications of f to reach this fixed set is A309414(n).
Equivalently, row n lists the set of elements x such that, for some k, x^(2^k) == x (mod n), i.e. those x which are part of a cycle under x -> x^2 mod n.
For prime p of the form 4k + 3 (A002145), row p gives exactly the set of quadratic residues mod p. As such, row p has (p + 1) / 2 elements.
When n is a prime power (A000961) the product of row n (excluding 0) is equivalent to 1 mod n. Otherwise this product is equivalent to 0.
EXAMPLE
Triangle begins:
(mod 1) 0;
(mod 2) 0, 1;
(mod 3) 0, 1;
(mod 4) 0, 1;
(mod 5) 0, 1;
(mod 6) 0, 1, 3, 4;
(mod 7) 0, 1, 2, 4;
(mod 8) 0, 1;
(mod 9) 0, 1, 4, 7;
(mod 10) 0, 1, 5, 6;
(mod 11) 0, 1, 3, 4, 5, 9;
(mod 12) 0, 1, 4, 9;
(mod 13) 0, 1, 3, 9;
(mod 14) 0, 1, 2, 4, 7, 8, 9, 11;
(mod 15) 0, 1, 6, 10;
(mod 16) 0, 1;
(mod 17) 0, 1;
...
PROG
(Python)
def row(n):
l = set(range((n >> 1) + 1))
while True:
newl = {x**2 % n for x in l}
if newl == l: break
l = newl
return l
(PARI) row(n)=my(p=[0..n>>1], c=[0..n>>1]); until(p==c, p=c; c=Set([lift(Mod(v, n)^2)|v<-c])); return(c);
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Aloe Poliszuk, Feb 20 2025
STATUS
approved