A060594 -id:A060594 - cp's OEIS frontend

A228179 Irregular table where the n-th row consists of the square roots of 1 in Z_n.

1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 3, 5, 7, 1, 8, 1, 9, 1, 10, 1, 5, 7, 11, 1, 12, 1, 13, 1, 4, 11, 14, 1, 7, 9, 15, 1, 16, 1, 17, 1, 18, 1, 9, 11, 19, 1, 8, 13, 20, 1, 21, 1, 22, 1, 5, 7, 11, 13, 17, 19, 23, 1, 24, 1, 25, 1, 26, 1, 13, 15, 27, 1, 28, 1, 11

Offset: 2

Tom Edgar, Aug 20 2013

Each 1 starts a new row.
This is a subsequence of A020652.
Row n has A060594(n) entries.
Each row forms a subgroup of the multiplicative group of units of Z_n.

			The table starts out as follows:
  1
  1 2
  1 3
  1 4
  1 5
  1 6
  1 3 5 7
  1 8
  1 9
  1 10
  1 5 7 11
  ...

Alois P. Heinz, Rows n = 2..2000 of irregular triangle, flattened

Cf. A060594, A038566, A020652, A082568.
Cf. A070667 (second column), A358016 (second-last column).
Cf. A277776 (nontrivial square roots of 1).

T:= n-> seq(`if`(k&^2 mod n=1, k, NULL), k=1..n-1):
seq(T(n), n=2..50);  # Alois P. Heinz, Aug 20 2013

Flatten[Table[Position[Mod[Range[n]^2, n], 1], {n, 2, 50}]] (* T. D. Noe, Aug 20 2013 *)

from itertools import chain, count, islice
from sympy.ntheory import sqrt_mod_iter
def A228179_gen(): # generator of terms
    return chain.from_iterable((sorted(sqrt_mod_iter(1,n)) for n in count(2)))
A228179_list = list(islice(A228179_gen(),30)) # Chai Wah Wu, Oct 26 2022

[[i for i in [1..k-1] if (i*i).mod(k)==1] for k in [2..n]] #changing n gives you the table up to the n-th row.

A327924 Square array read by ascending antidiagonals: T(m,n) is the number of non-isomorphic groups G such that G is the semidirect product of C_m and C_n, where C_m is a normal subgroup of G and C_n is a subgroup of G.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 3, 1, 2, 1, 1, 1, 4, 2, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 4, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 4, 1, 3, 1, 2, 1, 1, 1, 4, 1, 3, 1, 4, 1, 2, 1, 2, 1, 1, 1

Offset: 1

Jianing Song, Sep 30 2019

The semidirect product of C_m and C_n has group representation G = , where r is any number such that r^n == 1 (mod m). Two groups G =  and G' =  are isomorphic if and only if there exists some k, gcd(k,n) = 1 such that r^k == s (mod m), in which case f(x^i*y^j) = x^i*y^(k*j) is an isomorphic mapping from G to G'.
Given m, T(m,n) only depends on the value of gcd(n,psi(m)), psi = A002322 (Carmichael lambda). So each row is periodic with period psi(m). See A327925 for an alternative version.
Every number k occurs in the table. By Dirichlet's theorem on arithmetic progressions, there exists a prime p such that p == 1 (mod 2^(k-1)), then T(p,2^(k-1)) = d(gcd(2^(k-1),p-1)) = k (see the formula below). For example, T(5,4) = 3, T(17,8) = 4, T(17,16) = 5, T(97,32) = 6, T(193,64) = 7, ...
Row m and Row m' are the same if and only if (Z/mZ)* = (Z/m'Z)*, where (Z/mZ)* is the multiplicative group of integers modulo m. The if part is clear; for the only if part, note that the two sequences {(number of x in (Z/mZ)* such that x^n = 1)}{n>=1} and {T(m,n)}{n>=1} determine each other, and the structure of a finite abelian group G is uniquely determined by the sequence {(number of x in G such that x^n = 1)}{n>=1}. - _Jianing Song, May 16 2022

			  m/n  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
   1   1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1
   2   1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1
   3   1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2
   4   1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2
   5   1  2  1  3  1  2  1  3  1  2  1  3  1  2  1  3  1  2  1  3
   6   1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2  1  2
   7   1  2  2  2  1  4  1  2  2  2  1  4  1  2  2  2  1  4  1  2
   8   1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4
   9   1  2  2  2  1  4  1  2  2  2  1  4  1  2  2  2  1  4  1  2
  10   1  2  1  3  1  2  1  3  1  2  1  3  1  2  1  3  1  2  1  3
  11   1  2  1  2  2  2  1  2  1  4  1  2  1  2  2  2  1  2  1  4
  12   1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4  1  4
  13   1  2  2  3  1  4  1  3  2  2  1  6  1  2  2  3  1  4  1  3
  14   1  2  2  2  1  4  1  2  2  2  1  4  1  2  2  2  1  4  1  2
  15   1  4  1  6  1  4  1  6  1  4  1  6  1  4  1  6  1  4  1  6
  16   1  4  1  6  1  4  1  6  1  4  1  6  1  4  1  6  1  4  1  6
  17   1  2  1  3  1  2  1  4  1  2  1  3  1  2  1  5  1  2  1  3
  18   1  2  2  2  1  4  1  2  2  2  1  4  1  2  2  2  1  4  1  2
  19   1  2  2  2  1  4  1  2  3  2  1  4  1  2  2  2  1  6  1  2
  20   1  4  1  6  1  4  1  6  1  4  1  6  1  4  1  6  1  4  1  6
Example shows that T(16,4) = 6: The semidirect product of C_16 and C_4 has group representation G = <x, y|x^16 = y^4 = 1, yxy^(-1) = x^r>, where r = 1, 3, 5, 7, 9, 11, 13, 15. Since 3^3 == 11 (mod 16), 5^3 == 13 (mod 16), <x, y|x^16 = y^4 = 1, yxy^(-1) = x^3> and <x, y|x^16 = y^4 = 1, yxy^(-1) = x^11> are isomorphic, <x, y|x^16 = y^4 = 1, yxy^(-1) = x^5> and <x, y|x^16 = y^4 = 1, yxy^(-1) = x^13> are isomorphic, giving a total of 6 non-isomorphic groups.

Jianing Song, Table of n, a(n) for n = 1..5050 (the first 100 ascending diagonals)
Math Overflow, When are two semidirect products of two cyclic groups isomorphic.
Jianing Song, An equivalent formula.

Cf. A327925, A002322, A000010, A000005.

Cf. also A060594, A060839, A073103, A319099, A319100, A319101, A247257.

numord(n,q) = my(v=divisors(q),r=znstar(n)[2]); sum(i=1,#v,prod(j=1,#r,gcd(v[i],r[j]))*moebius(q/v[i]))
T(m,n) = my(u=divisors(n)); sum(i=1,#u,numord(m,u[i])/eulerphi(u[i]))

T(m,n) = Sum_{d|n} (number of elements x such that ord(x,m) = d)/phi(d), where ord(x,m) is the multiplicative order of x modulo m, phi = A000010. There is a version to compute the terms more conveniently, see the links section. [Proof: let (Z/mZ)* denote the multiplicative group modulo m. For every d|n, the elements in (Z/mZ)* having order d are put into different equivalence classes, where each class is of the form {a^k: gcd(k,n)=1}. The size of each equivalence class is the number of different residues modulo d of the numbers that are coprime to n, which is phi(d). - Jianing Song, Sep 17 2022]
Equivalently, T(m,n) = Sum_{d|gcd(n,psi(m))} (number of elements x such that ord(x,m) = d)/phi(d). - Jianing Song, May 16 2022 [This is because the order of elements in (Z/mZ)* must divide psi(m). - Jianing Song, Sep 17 2022]
Let U(m,q) be the number of solutions to x^q == 1 (mod m):
T(m,1) = U(m,1) = 1;
T(m,2) = U(m,2) = A060594(m);
T(m,3) = (1/2)*U(m,3) + (1/2)*U(m,1) = (1/2)*A060839(m) + 1/2;
T(m,4) = (1/2)*U(m,4) + (1/2)*U(m,2) = (1/2)*A073103(m) + 1/2;
T(m,5) = (1/4)*U(m,5) + (3/4)*U(m,1) = (1/4)*A319099(m) + 3/4;
T(m,6) = (1/2)*U(m,6) + (1/2)*U(m,2) = (1/2)*A319100(m) + 1/2;
T(m,7) = (1/6)*U(m,7) + (5/6)*U(m,1) = (1/6)*A319101(m) + 5/6;
T(m,8) = (1/4)*U(m,8) + (1/4)*U(m,4) + (1/2)*U(m,2) = (1/4)*A247257(m) + (1/4)*A073103(m) + (1/2)*A060594(m);
T(m,9) = (1/6)*U(m,9) + (1/3)*U(m,3) + (1/2)*U(m,1);
T(m,10) = (1/4)*U(m,10) + (3/4)*U(m,2).
For odd primes p, T(p^e,n) = d(gcd(n,(p-1)*p^(e-1))), d = A000005; for e >= 3, T(2^e,n) = 2*(min{v2(n),e-2}+1) for even n and 1 for odd n, where v2 is the 2-adic valuation.

A087688 a(n) = number of solutions to x^3 - x == 0 (mod n).

Original entry on oeis.org

1, 2, 3, 3, 3, 6, 3, 5, 3, 6, 3, 9, 3, 6, 9, 5, 3, 6, 3, 9, 9, 6, 3, 15, 3, 6, 3, 9, 3, 18, 3, 5, 9, 6, 9, 9, 3, 6, 9, 15, 3, 18, 3, 9, 9, 6, 3, 15, 3, 6, 9, 9, 3, 6, 9, 15, 9, 6, 3, 27, 3, 6, 9, 5, 9, 18, 3, 9, 9, 18

Offset: 1

Yuval Dekel (dekelyuval(AT)hotmail.com), Sep 27 2003

Shadow transform of A007531. - Michel Marcus, Jun 06 2013

a(n) = 3 iff n belongs to (A061345 \ {1}) Union {4}. - Bernard Schott, Sep 16 2019

Eric M. Schmidt, Table of n, a(n) for n = 1..10000
Lorenz Halbeisen and Norbert Hungerbuehler, Number theoretic aspects of a combinatorial function, Notes on Number Theory and Discrete Mathematics 5(4) (1999), 138-150; see Definition 7 for the shadow transform.
N. J. A. Sloane, Transforms.

Cf. A034444, A224516, A060594.

A087688 := proc(n) local a,x ; a := 0 ; for x from 0 to n-1 do if (x*(x^2-1)) mod n = 0 then a := a+1 ; end if; end do; a ; end proc:
seq(A087688(n),n=1..70) ; # R. J. Mathar, Jan 07 2011

nsols[n_]:=Length[Select[Range[0,n-1],Mod[#^3-#,n]==0&]]; nsols/@Range[80]  (* Harvey P. Dale, Mar 22 2011 *)
f[2, e_] := Which[e == 1, 2, e == 2, 3, e >= 3, 5]; f[p_, e_] := 3; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 19 2020 *)

a(n)=if(n%2,3^omega(n),my(v=valuation(n,2));3^omega(n>>v)*[2,3,5][min(3,v)]) \\ Charles R Greathouse IV, Mar 22 2011

Multiplicative with a(p^e) = 3 for p an odd prime, a(2^1) = 2, a(2^2) = 3, a(2^e) = 5 for e >= 3. - Eric M. Schmidt, Apr 08 2013

A087780 Number of non-congruent solutions to x^2 == 2 mod n.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 2, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 2, 0, 0, 2, 2, 0, 0, 0, 0, 2, 0, 0

Offset: 1

Yuval Dekel (dekelyuval(AT)hotmail.com), Oct 06 2003

Eric M. Schmidt, Table of n, a(n) for n = 1..10000

Cf. A038873, A057126, A060594.

Cf. A013661, A196525.

f[2, e_] := Boole[e == 1]; f[p_, e_] := If[MemberQ[{1, 7}, Mod[p, 8]], 2, 0]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 19 2020 *)

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,1] == 2, (f[i,2] == 1), if(f[i,1]%8 == 1 || f[i,1]%8 == 7, 2, 0)));} \\ Amiram Eldar, Nov 21 2023

def A087780(n) :
    res = 1
    for (p, m) in factor(n) :
        if p % 8 in [1, 7] : res *= 2
        elif not (p==2 and m==1) : return 0
    return res
# Eric M. Schmidt, Apr 20 2013

Multiplicative with a(p^m) = 2 for p == 1, 7 (mod 8); a(p^m) = 0 for p == 3, 5 (mod 8); a(2^1) = 1; a(2^m) = 0 for m > 1. - Eric M. Schmidt, Apr 20 2013

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log(sqrt(2)+1)/(sqrt(2)*zeta(2)) = A196525/A013661 = 0.37887551404073012021... . - Amiram Eldar, Nov 21 2023

More terms from David Wasserman, Jun 17 2005

A155828 Number of integers k in {1,2,3,..,n} such that kn+1 is a square.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 3, 3, 1, 1, 1, 3, 3, 1, 1, 7, 1, 1, 1, 3, 1, 3, 1, 3, 3, 1, 3, 3, 1, 1, 3, 7, 1, 3, 1, 3, 3, 1, 1, 7, 1, 1, 3, 3, 1, 1, 3, 7, 3, 1, 1, 7, 1, 1, 3, 3, 3, 3, 1, 3, 3, 3, 1, 7, 1, 1, 3, 3, 3, 3, 1, 7, 1, 1, 1, 7, 3, 1, 3, 7, 1, 3, 3, 3, 3, 1, 3, 7, 1, 1, 3, 3, 1, 3, 1, 7, 7

Offset: 1

John W. Layman, Jan 28 2009

nonn

Conjecture: All terms a(n) are of the form 2^m - 1. This has been verified up to n = 1000.
It appears that the terms of this sequence are exactly one less than those of A060594, indicating that the terms are related to the square roots of unity mod n. - John W. Layman, Feb 04 2009
The conjecture that a(n) is of the form 2^m - 1 holds trivially. In fact, the conjecture can be restated as follows: For any positive integer n, the solutions of the congruence x^2 = 1 (mod n) with 1 <= x <= n is a power of two. By the Chinese remainder theorem, this reduces to the case when n is a prime power. It is well known that the solution of x^2 = 1 (mod p^a) is two when p is an odd prime (one may obtain this by induction on a). As for the number of solutions of the congruence x^2 = 1 (mod 2^a), it equals 1, 2, 4 according to whether a=1, 2, or a>2. - Zhi-Wei Sun, Feb 11 2009

			1*8 + 1 = 9, 3*8 + 1 = 25, 6*8 + 1 = 49, whereas other values are not square, so a(8) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A060594.

a(n) = sum(k=1, n, issquare(k*n+1)); \\ Michel Marcus, Jul 10 2014

a(n)=if(n>2, 2^#znstar(n)[3]-1) \\ Charles R Greathouse IV, Jun 18 2025

a(n) = A060594(n) - 1. - after Zhi-Wei Sun, Michel Marcus, Jul 10 2014

A211415 Number of ordered triples (w,x,y) with all terms in {-n,...,0,...,n} and w^2 + x*y = 1.

Original entry on oeis.org

0, 12, 20, 36, 52, 68, 84, 100, 132, 148, 164, 180, 212, 228, 244, 276, 308, 324, 340, 356, 388, 420, 436, 452, 516, 532, 548, 564, 596, 612, 644, 660, 692, 724, 740, 772, 804, 820, 836, 868, 932, 948, 980, 996, 1028, 1060, 1076, 1092, 1156, 1172

Offset: 0

Clark Kimberling, Apr 09 2012

nonn

For n>0, a(n)+2 is the number of self-inverse 2 X 2 matrices with all terms in {-n,...,0,...,n} (see A211416).

For a guide to related sequences, see A211422.

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Cf. A060594, A211416, A211422.

a:=[]; for n in [0..50] do Append(~a,#[:w,x,y in [-n..n]|w^2 + x*y eq 1]); end for; a; // Marius A. Burtea, Jan 22 2020

t[n_] := t[n] = Flatten[Table[w^2 + x*y - 1, {w, -n, n}, {x, -n, n}, {y, -n, n}]]
c[n_] := Count[t[n], 0]
t = Table[c[n], {n, 0, 20}]  (* A211415 *)
t + 2                        (* A211416 *)
(t + 2)/2    (* integers *)
t/4          (* integers *)
(t/4 - 1)/4  (* integers for n>1 *)

For n>0, a(n+1) - a(n) = 8*A060594(n+1). - Pontus von Brömssen, Jan 22 2020

A329152 a(n) = Sum_{i=1..n-1} Sum_{j=1..i-1} [1 == i*j (mod n)], where [] is the Iverson bracket.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 2, 0, 2, 1, 4, 0, 5, 2, 2, 2, 7, 2, 8, 2, 4, 4, 10, 0, 9, 5, 8, 4, 13, 2, 14, 6, 8, 7, 10, 4, 17, 8, 10, 4, 19, 4, 20, 8, 10, 10, 22, 4, 20, 9, 14, 10, 25, 8, 18, 8, 16, 13, 28, 4, 29, 14, 16, 14, 22, 8, 32, 14, 20, 10, 34, 8, 35, 17, 18, 16, 28, 10, 38, 12

Offset: 1

Torlach Rush, Feb 25 2020

In other words, a(n) is the number of choices for k, 1 <= k <= n, so that k*n + 1 has the form a*b for 1 < a < b < n.
Solutions exist only when 1 <= k <= n - 3.
Any odd number greater than 3 has at least one solution, 2*ceiling(n/2).
Observations:
- Unique values of a(n) exist, 6 = a(32), 24 = a(240), 126 = a(512), 144 = a(1320), ..., and n mod 8 = 0.
- If a(n) is an odd prime then a(n) = a(2*n) OR a(n) = a(n/2). For example, 1013 = a(2029) = a(2197) = a(4058) = a(4394).

			a(1)=0 because there is no solution to k*1 + 1 = a*b, 1 < a < b < n, 1 <= k < n.
a(5)=1 because 1 == 3*2 (mod 5).
a(7)=2 because 1 == 4*2 == 5*3 (mod 7).
a(11)=4 because 1 == 4*3 == 6*2 == 8*7 == 9*5 (mod 11).
a(13)=5 because 1 == 7*2 == 8*5 == 9*3 == 10*4 = 11*6 (mod 13).
a(32)=6 because 1 == 11*3 == 13*5 == 23*7 == 25*9 == 27*19 == 29*21 (mod 32).

David Broadhurst, Table of n, a(n) for n = 1..10000

Cf. A000010, A060594, A343292.

Array[Sum[Sum[Boole[Mod[i j, #] == 1], {j, i - 1}], {i, # - 1}] &, 80] (* Michael De Vlieger, Mar 15 2020 *)

a(n) = {my(x=0); for (i = 1, n - 1, for (ii = 1, i - 1, if(1 == ((ii*i) % n), x++))); return(x)}

a(n)=sum(i=2,n-2,gcd(i,n)==1&&(1/i)%n>i); \\ David Broadhurst, Feb 20 2025

a(n)=eulerphi(n)/2-2^(#znstar(n)[3]-1); \\ David Broadhurst, Feb 21 2025

a(n) = (A000010(n) - A060594(n))/2 = n - A343292(n). - Ridouane Oudra, May 04 2025

A073503 Numbers n such that the number of solutions to x^4 == 1 (mod n) is twice the number of solutions of x^2 == 1 (mod n).

Original entry on oeis.org

5, 10, 13, 15, 16, 17, 20, 25, 26, 29, 30, 32, 34, 35, 37, 39, 40, 41, 45, 48, 50, 51, 52, 53, 55, 58, 60, 61, 64, 68, 70, 73, 74, 75, 78, 82, 87, 89, 90, 91, 95, 96, 97, 100, 101, 102, 104, 105, 106, 109, 110, 111, 112, 113, 115, 116, 117, 119, 120, 122, 123, 125

Offset: 1

Benoit Cloitre, Aug 19 2002

Conjectures: 2n > a(n) or 2n < a(n) for infinitely many values of n and abs(a(n)-2n) < sqrt(n) for n > 45. a(n)=2n for n = 318, 338, 350, 488, 490, 492, 494,...

Numbers divisible by 16 which have no prime factors = 1 mod 4, together with numbers not divisible by 16 which have exactly one prime factor = 1 mod 4. This refutes the conjectures. - Charles R Greathouse IV, Apr 16 2012

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A060594, A073103.

Select[Range[3, 125], Length[Reduce[x^4 - 1 == 0, x, Modulus -> #]] == 2*Length[Reduce[x^2 - 1 == 0, x, Modulus -> #]] &] (* Jayanta Basu, Jul 01 2013 *)

is(n)=my(v=factor(n)[,1]%4, s=sum(i=1,#v,v[i]==1), e=valuation(n, 2)); s==(e<4) \\ Charles R Greathouse IV, Apr 16 2012

a(n) seems to be asymptotic to 2n.

A081754 Numbers n such that the number of noncongruent solutions to x^(2^m) == 1 (mod n) is the same for any m>=1.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 14, 18, 19, 21, 22, 23, 24, 27, 28, 31, 33, 36, 38, 42, 43, 44, 46, 47, 49, 54, 56, 57, 59, 62, 63, 66, 67, 69, 71, 72, 76, 77, 79, 81, 83, 84, 86, 88, 92, 93, 94, 98, 99, 103, 107, 108, 114, 118, 121, 124, 126, 127, 129, 131, 132, 133, 134

Offset: 1

Benoit Cloitre, Apr 08 2003

nonn

Numbers n such that the multiplicative group of residues modulo n does not contain C4 as a subgroup. Equivalently, numbers not divisible by 16 or by any primes of the form 4k+1. - Ivan Neretin, Aug 02 2016
From Jianing Song, Oct 18 2021: (Start)
Numbers k such that psi(k) = A002322(k) is not divisible by 4.
Numbers k such that there are an odd number of coprime squares modulo k, i.e., numbers k such that A046073(k) is odd. (End)

Ivan Neretin, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Modulo Multiplication Group.

Cf. A060594, A002322, A046073.

filter:= n -> n mod 16 <> 0 and not member(1,numtheory:-factorset(n) mod 4):
select(filter, [$1..1000]); # Robert Israel, Aug 02 2016

Select[Range@135, ! Divisible[#, 16] && FreeQ[Mod[FactorInteger[3 #][[All, 1]], 4], 1] &] (* Ivan Neretin, Aug 02 2016 *)

isA081754(n) = if(n>2, znstar(n)[2][1]%4==2, 1) \\ Jianing Song, Oct 18 2021

A157223 Number of primitive inequivalent oblique sublattices of centered rectangular lattice of index n.

Original entry on oeis.org

0, 1, 1, 2, 2, 5, 3, 4, 5, 8, 5, 10, 6, 11, 10, 10, 8, 17, 9, 16, 14, 17, 11, 20, 14, 20, 17, 22, 14, 34, 15, 22, 22, 26, 22, 34, 18, 29, 26, 32, 20, 46, 21, 34, 34, 35, 23, 44, 27, 44, 34, 40, 26, 53, 34, 44, 38, 44, 29, 68, 30, 47, 46, 46, 40, 70, 33, 52, 46

Offset: 1

N. J. A. Sloane, Feb 25 2009

nonn

John S. Rutherford, Sublattice enumeration. IV. Equivalence classes of plane sublattices by parent Patterson symmetry and colour lattice group type, Acta Cryst. (2009). A65, 156-163. [See Table 4.]

Cf. A060594 (primitive mirror-symmetric sublattices), A145390 (all mirror-symmetric sublattices), A145391 (all sublattices), A001615, A304182.

a(n) = (A001615(n) - A060594(n))/2. - Andrey Zabolotskiy, May 09 2018

New name and more terms from Andrey Zabolotskiy, May 09 2018

cp's OEIS Frontend

A228179 Irregular table where the n-th row consists of the square roots of 1 in Z_n.

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A087688 a(n) = number of solutions to x^3 - x == 0 (mod n).

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A087780 Number of non-congruent solutions to x^2 == 2 mod n.

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A211415 Number of ordered triples (w,x,y) with all terms in {-n,...,0,...,n} and w^2 + x*y = 1.

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A329152 a(n) = Sum_{i=1..n-1} Sum_{j=1..i-1} [1 == i*j (mod n)], where [] is the Iverson bracket.

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