A046071 -id:A046071 - cp's OEIS frontend

A307550 Irregular array of distinct terms read by rows: for n > 0, row n = [r(1),...,r(k)] with r(1) = n^2 (mod prime(n)), r(2) = r(1)^2 (mod prime(n)), ..., r(k) = r(k-1)^2 (mod prime(n)), where r(k) is the last term of the cycle.

1, 1, 4, 1, 2, 4, 3, 9, 4, 5, 10, 9, 3, 15, 4, 16, 1, 7, 11, 12, 6, 13, 8, 18, 2, 4, 16, 3, 9, 13, 24, 25, 16, 28, 9, 19, 20, 33, 16, 34, 9, 7, 12, 5, 25, 10, 18, 37, 16, 24, 17, 31, 15, 10, 14, 37, 6, 36, 27, 24, 12, 3, 9, 34, 28, 32, 44, 28, 42, 15, 13, 10

Offset: 1

Michel Lagneau, Apr 14 2019

Consider the map of quadratic residues x -> x^2 (mod prime(n)) with the initial term x = r(1) = n^2 (mod prime(n)) needed to reach the end of the cycle. Row n contains all distinct quadratic residues r(i) such that r(i) = r(j)^2 (mod prime(n)) for some i, j.

The corresponding row lengths are given by the sequence {b(n)} = {1, 1, 2, 2, 4, 3, 4, 2, 10, 4, 4, 6, 6, 6, 11, 12, 28, 5, 10, 3, 4, 12, 20, 12, 5, 21, ...} with b(n) = A307551(n) + 1. We observe the following property: if prime(n) = 2p + 1 with p prime, b(n) = p - 1 if 2 is a primitive root mod p; that is, p is in A001122 (see A141305). Example: b(17) = 28 because prime(17) = 59 = 2*29 + 1 with 28 = 29 - 1, and 2 is a primitive root mod 29.

			Row 5 = [3, 9, 4, 5] because prime(5) = 11, and 3 = 5^2 (mod 11), 9 = 3^2 (mod 11), 4 = 9^2 (mod 11) and 5 = 4^2 (mod 11).
Irregular array starts:
  [1];
  [1];
  [4, 1];
  [2, 4];
  [3, 9, 4, 5];
  [10, 9, 3];
  [15, 4, 16, 1];
   ...

Cf. A000040, A001122, A000224, A046071, A063987, A096008, A141305, A307551.

nn:=30:T:=array(1..280):j:=0 :
for n from 1 to nn do:
p:=ithprime(n):lst0:={}:lst1:={}:ii:=0:r:=n:
for k from 1 to 10^6 while(ii=0) do:
  r1:=irem(r^2,p):lst0:=lst0 union {r1}:j:=j+1:T[j]:=r1:
      if lst0=lst1
       then
        ii:=1:
        else
        r:=r1:lst1:=lst0:
      fi:
     od:
   if lst0 intersect {r1} = {r1}
    then
    j:=j-1:else fi:
od:
print(T):

s[n_] := Module[{p = Prime[n]}, f[x_] := Mod[x^2, p]; Most[NestWhileList[f, f[n], Unequal, All]]]; seq = {}; Do[AppendTo[seq, s[n]], {n, 20}]; seq // Flatten (* Amiram Eldar, Jul 05 2019 *)

A307551 Number of iterations of the map of quadratic residues x -> x^2 (mod prime(n)) with the initial term x = n^2 (mod prime(n)) needed to reach the end of the cycle.

Original entry on oeis.org

0, 0, 1, 1, 3, 2, 3, 1, 9, 3, 3, 5, 5, 5, 10, 11, 27, 4, 9, 2, 3, 11, 19, 11, 4, 20, 7, 51, 17, 2, 5, 11, 9, 10, 35, 19, 11, 5, 81, 13, 10, 3, 35, 6, 21, 29, 11, 35, 27, 18, 27, 7, 5, 99, 7, 129, 65, 35, 10, 2, 22, 9, 23, 19, 13, 38, 19, 8, 171, 27, 13, 177, 59

Offset: 1

Michel Lagneau, Apr 14 2019

nonn

Let L(n) be the number of elements in row n of A307550. Then a(n) = L(n) - 1.

			a(5) = 3 because prime(5) = 11, and 5^2 (mod 11) = 3 -> 3^2 (mod 11) = 9 ->  9^2 (mod 11) = 4 -> 4^2 (mod 11) = 5 with 3 iterations, where 5 is the last term of the cycle.

Cf. A000040, A000224, A046071, A063987, A096008, A307550.

nn:=100:T:=array(1..3000):j:=0 :
for n from 1 to nn do:
p:=ithprime(n):lst0:={}:lst1:={}:ii:=0:r:=n:
for k from 1 to 10^6 while(ii=0) do:
  r1:=irem(r^2,p):lst0:=lst0 union {r1}:j:=j+1:T[j]:=r1:
      if lst0=lst1
       then
        ii:=1: printf(`%d, `,nops(lst0)-1):
        else
        r:=r1:lst1:=lst0:
      fi:
     od:
   if lst0 intersect {r1} = {r1}
    then
    j:=j-1:else fi:
od:

a[n_] := Module[{p = Prime[n]}, f[x_] := Mod[x^2, p]; Length[NestWhileList[f, f[n], Unequal, All]] - 2]; Array[a, 100] (* Amiram Eldar, Jul 05 2019 *)

A309680 The smallest nonsquare nonzero integer that is a quadratic residue modulo n, or 0 if no such integer exists.

Original entry on oeis.org

0, 0, 0, 0, 0, 3, 2, 0, 7, 5, 3, 0, 3, 2, 6, 0, 2, 7, 5, 5, 7, 3, 2, 12, 6, 3, 7, 8, 5, 6, 2, 17, 3, 2, 11, 13, 3, 5, 3, 20, 2, 7, 6, 5, 10, 2, 2, 33, 2, 6, 13, 12, 6, 7, 5, 8, 6, 5, 3, 21, 3, 2, 7, 17, 10, 3, 6, 8, 3, 11, 2, 28, 2, 3, 6, 5, 11, 3, 2, 20, 7

Offset: 1

John Prosser, Aug 12 2019

nonn

			For n=5, the nonzero quadratic residues modulo 5 are 1 and 4. Since these are both squares we have a(5) = 0.
For n=6, the nonzero quadratic residues modulo 6 are 1,3, and 4. Since 3 is not a square we have a(6) = 3.
For n=10, the nonzero quadratic residues modulo 10 are 1,4,5,6,9. Since 5 is the least nonsquare value, we have a(10) = 5.

A330404 is an alternate version.

Cf. A046071, A053760, A057126.

a[n_] := SelectFirst[ Union@ Mod[Range[n-1]^2, n], ! IntegerQ@ Sqrt@ # &, 0]; Array[a, 81] (* Giovanni Resta, Aug 13 2019 *)

a(n) = my(v=select(x->issquare(x), vector(n-1, k, Mod(k,n)), 1), y = select(x->!issquare(x), Vec(v))); if (#y, y[1], 0); \\ Michel Marcus, Aug 16 2019

a(n) = 2 for n in A057126 and n > 2. - Michel Marcus, Aug 24 2019

cp's OEIS Frontend

A307550 Irregular array of distinct terms read by rows: for n > 0, row n = [r(1),...,r(k)] with r(1) = n^2 (mod prime(n)), r(2) = r(1)^2 (mod prime(n)), ..., r(k) = r(k-1)^2 (mod prime(n)), where r(k) is the last term of the cycle.

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A307551 Number of iterations of the map of quadratic residues x -> x^2 (mod prime(n)) with the initial term x = n^2 (mod prime(n)) needed to reach the end of the cycle.

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Maple

Mathematica

A309680 The smallest nonsquare nonzero integer that is a quadratic residue modulo n, or 0 if no such integer exists.

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