A087726 Number of elements X in the matrix ring M_2(Z_n) such that X^2 == 0 mod n.

1, 4, 9, 28, 25, 36, 49, 112, 153, 100, 121, 252, 169, 196, 225, 640, 289, 612, 361, 700, 441, 484, 529, 1008, 1225, 676, 1377, 1372, 841, 900, 961, 2560, 1089, 1156, 1225, 4284, 1369, 1444, 1521, 2800, 1681, 1764, 1849, 3388, 3825, 2116, 2209, 5760, 4753, 4900, 2601, 4732

Offset: 1

Author

Yuval Dekel (dekelyuval(AT)hotmail.com), Sep 28 2003

Keywords

• mult
• nonn

Comments

Conjecture: a(n)=n^2 if and only if n is squarefree. [Ben Branman, Mar 22 2013]

Preceding conjecture is true in the case where n is squarefree. - Eric M. Schmidt, Mar 23 2013

It appears that a(p^k) = (1+3*p^2 + 2*k*(p^2-1) + (-1)^k*(p^2-1))*p^(2*k-2)/4 for primes p. Since the sequence is multiplicative, this would imply the conjecture. - Robert Israel, Jun 10 2015

A proof of the formula for k=1 can be done easily (see pdf). - Manfred Scheucher, Jun 10 2015

Links

• Manfred Scheucher, Table of n, a(n) for n = 1..1000
• Manfred Scheucher, A proof of the formula for k=1

Crossrefs

Cf. A066907, A000188.

Programs

C

#include
#include
int main(int argc,char** argv)
{
  long ct = 0;
  int n = atoi(argv[1]);
  int a,b,c,d;
  for(a=0;aManfred Scheucher, Jun 09 2015 */

Maple

f:= proc(n)
  local tot, S, a, mult, sa, d, ad, g, cands;
  tot:= 0;
  S:= ListTools:-Classify(t -> t^2 mod n, [$0..n-1]);
  for a in numtheory:-divisors(n) do
    mult:= numtheory:-phi(n/a);
    sa:= a^2 mod n;
    for d in S[sa] do
       g:= igcd(a+d,n);
       cands:= [seq(i*n/g, i=0..g-1)];
       tot:= tot + mult * numboccur(sa,[seq(seq(s*t,s=cands),t=cands)] mod n);
    od
  od;
  tot
end proc:
map(f, [$1..100]); # Robert Israel, Jun 09 2015

Mathematica

a[m_] := Count[Table[Mod[MatrixPower[Partition[IntegerDigits[n, m, 4], 2], 2], m] == {{0, 0}, {0, 0}}, {n, 0, m^4 - 1}], True]; Table[a[n], {n,2,30}] (* Ben Branman, Mar 22 2013 *)

Extensions

More terms from Ben Branman, Mar 22 2013

More terms from Manfred Scheucher, Jun 09 2015