1,2

a(n) is multiplicative and, for a prime p, a(p) = p^2. Hence a(n) = n^2 if n is square-free.

See newsgroup sci.math.research; subject: Re: Pythagorean triples mod n / Solution enhanced; author: Gottfried Helms; Message-ID: brv5i2$28v$1(AT)news.ks.uiuc.edu; Date: Fri, Dec 19 2003 15: 30: 10 +0000 (UTC)

T. D. Noe, Table of n, a(n) for n=1..1000

a(n) is multiplicative. For the powers of primes p, there are four cases. For p=2, there are cases for even and odd powers: a(2^(2k-1)) = 2^(3k-1) (2^k-1) and a(2^(2k)) = 2^(3k) (2^(k+1)-1). Similarly, for odd primes p, a(p^(2k-1)) = p^(3k-2) (p^k+p^(k-1)-1) and a(p^(2k)) = p^(3k-1) (p^(k+1)+p^k-1). - T. D. Noe (noe(AT)sspectra.com), Dec 22 2003

Thanks to T. D. Noe (noe(AT)sspectra.com) and Gottfried Helms (helms(AT)uni-kassel.de) for additional comments, Dec 23, 2003.

If the canonical form of n is n = 2^i*3^j*5^k*...p^q then it appears that a(n) = n*f(2, i)*f(3, j)*f(5, k)*...*f(p, q) where f(p, 1) = p for any prime p; f(2, i) = 2^i + 2^i - 2^ceil(i/2); f(p, i) = p^i + p^(i-1) - p^floor((i-1)/2) for any odd prime p. For example a(7) = 49 because a(7) = 7*f(7, 1) = 7*7; a(16) = 448 because a(16) = a(2^4)= 16 * f(2, 4) = 16 * (16+16-4) = 16*28 = 448; a(12) = 216 because a(12) = a(3*2^2)= 12*f(2, 2)*f(3, 1) = 12*(4+4-2)*3 = 216. - Gottfried Helms (helms(AT)uni-kassel.de), May 13 2004

Table[cnt=0; Do[If[Mod[x^2+y^2-z^2, n]==0, cnt++ ], {x, 0, n-1}, {y, 0, n-1}, {z, 0, n-1}]; cnt, {n, 50}]

Cf. A091143 (number of solutions to x^2 + y^2 = z^2 mod 2^n).

Cf. A060968, A063454.

Sequence in context: A131607 A027119 A067801 this_sequence A046422 A056575 A056032

Adjacent sequences: A062772 A062773 A062774 this_sequence A062776 A062777 A062778

nonn,nice,mult

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 18 2001

More terms from Sascha Kurz (sascha.kurz(AT)uni-bayreuth.de), Mar 25 2002