A235872 Number of solutions to the equation x^2=0 in the ring of Gaussian integers modulo n.
1, 2, 1, 4, 1, 2, 1, 8, 9, 2, 1, 4, 1, 2, 1, 16, 1, 18, 1, 4, 1, 2, 1, 8, 25, 2, 9, 4, 1, 2, 1, 32, 1, 2, 1, 36, 1, 2, 1, 8, 1, 2, 1, 4, 9, 2, 1, 16, 49, 50, 1, 4, 1, 18, 1, 8, 1, 2, 1, 4, 1, 2, 9, 64, 1, 2, 1, 4, 1, 2, 1, 72, 1, 2, 25, 4, 1, 2, 1, 16, 81, 2
Offset: 1
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
invoG[n_] := invoG[n] = Sum[If[Mod[(x + I y)^2, n] == 0, 1, 0], {x, 0, n - 1}, {y, 0, n - 1}]; Table[invoG[n], {n, 1, 104}] f[p_, e_] := p^(2*Floor[e/2]); f[2, e_] := 2^e; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 13 2022 *)
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PARI
a(n)={sum(i=0, n-1, sum(j=0, n-1, (i^2 - j^2)%n == 0 && 2*i*j%n == 0))} \\ Andrew Howroyd, Aug 06 2018
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PARI
a(n)={my(f=factor(n)); prod(i=1, #f~, my([p,e]=f[i,]); p^if(p==2, e, e - e%2))} \\ Andrew Howroyd, Aug 06 2018
Formula
Multiplicative with a(2^e) = 2^e, a(p^e) = p^(2*floor(e/2)). - Andrew Howroyd, Aug 06 2018
Sum_{k=1..n} a(k) ~ c * n^(3/2), where c = (2/21)*(3+sqrt(2))*zeta(3/2)/zeta(3) = 0.91363892007.... - Amiram Eldar, Nov 13 2022
Comments