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 A087726 Number of elements X in the matrix ring M_2(Z_n) such that X^2 == 0 mod n. 2
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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 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 *) PROG (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; a

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Last modified July 25 11:31 EDT 2021. Contains 346289 sequences. (Running on oeis4.)