A087561 Number of solutions to x^2 + 2y^2 == 0 (mod n).
1, 2, 5, 4, 1, 10, 1, 8, 21, 2, 21, 20, 1, 2, 5, 16, 33, 42, 37, 4, 5, 42, 1, 40, 25, 2, 81, 4, 1, 10, 1, 32, 105, 66, 1, 84, 1, 74, 5, 8, 81, 10, 85, 84, 21, 2, 1, 80, 49, 50, 165, 4, 1, 162, 21, 8, 185, 2, 117, 20, 1, 2, 21, 64, 1, 210, 133, 132, 5, 2, 1, 168, 145, 2, 125, 148, 21
Offset: 1
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..10000
- László Tóth, Counting Solutions of Quadratic Congruences in Several Variables Revisited, J. Int. Seq. 17 (2014), Article 14.11.6.
Programs
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Mathematica
a[n_] := If[n == 1, 1, Product[{p, e} = pe; Which[p == 2, 2^e, Abs[Mod[p, 8] - 2] != 1, (p^2)^Quotient[e, 2], True, (p + e (p-1)) p^(e-1)], {pe, FactorInteger[n]}]]; a /@ Range[1, 100] (* Jean-François Alcover, Sep 20 2019, from PARI *)
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PARI
a(n)={my(v=vector(n)); for(i=0, n-1, v[i^2%n + 1]++); sum(i=0, n-1, v[i+1]*v[(-2*i)%n + 1])} \\ Andrew Howroyd, Jul 16 2018
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PARI
a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); if(p==2, 2^e, if(abs(p%8-2)<>1, (p^2)^(e\2), (p+e*(p-1))*p^(e-1))))} \\ Andrew Howroyd, Jul 16 2018
Formula
Multiplicative with a(2^e) = 2^e, a(p^e) = p^(2*floor(e/2)) for p - 2 == +-3 (mod 8), a(p^e) = ((p-1)*e+p)*p^(e-1) for p - 2 == +-1 (mod 8). - Andrew Howroyd, Jul 16 2018
Sum_{k=1..n} a(k) ~ c * n^2, where c = Pi/(4*sqrt(2)*A309710) = 0.521595326207... . - Amiram Eldar, Nov 21 2023
Extensions
More terms from David Wasserman, Jun 07 2005