|
| |
|
|
A278081
|
|
a(n) is 1/12 of the number of primitive quadruples with sum = 0 and sum of squares = 2*m^2, where m = 2*n - 1.
|
|
7
|
|
|
|
1, 2, 6, 8, 6, 10, 14, 12, 16, 18, 16, 24, 30, 18, 30, 32, 20, 48, 38, 28, 40, 42, 36, 48, 56, 32, 54, 60, 36, 58, 62, 48, 84, 66, 48, 72, 72, 60, 80, 80, 54, 82, 96, 60, 88, 112, 64, 108, 96, 60, 102, 104, 96, 106, 110, 76, 112, 144, 84, 128, 110, 80, 150, 128
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Set b(m) = a(n) for m = 2*n-1, and b(m) = 0 for m even.
Conjecture: b(m) is multiplicative: for k >= 1, b(2^k) = 0; for p an odd prime, b(p*k) = p^(k-1)*b(p); b(p)= p + 1 for p == (5, 7, 13, 23) (mod 24); b(p) = p-1 for p == (1, 11, 17, 19) (mod 24); and b(3) = 3. It would be nice to have a proof of this.
This sequence applies also to the case sum = 4*m and ssq = 6*m^2. Generally, there is a 1-to-1 correspondence between a quadruple (h,i,j,k) with sum = r*m and ssq = s*m^2 and another with r'*m and s'*m^2, resp., if r + r'= 4, s - r = s' - r', namely (h',i',j',k') = (m,m,m,m) - (h,i,j,k). [Edited by Petros Hadjicostas, Apr 21 2020]
|
|
|
LINKS
|
Petros Hadjicostas, Slight modification of Mallows' R program. [To get the total counts for n = 1 to 120, with the zeros, i.e., the sequence (b(n): n >= 1) shown in the comments above, type gc(1:120, 0, 2), where r = 0 and s = 2. To get the 1/12 of these counts with no zeros, type gc(seq(1,59,2), 0, 2)[,3]/12. As stated in the comments, we get the same sequence with r = 4 and s = 6, i.e., we may type gc(seq(1,59,2), 4, 6)[,3]/12.]
|
|
|
EXAMPLE
|
For the case r = 0 and s = 2, we have a(2) = 2 = b(3) because of (-3,-1,2,2) and (-2,-2,1,3) (12 permutations each). For example, (-3) + (-1) + 2 + 2 = 0 but (-3)^2 + (-1)^2 + 2^2 + 2^2 = 18 = 2*3^2 = 2*(2*2-1)^2 (with n = 2 and m = 3).
For the case r = 4 and s = 6, we again have a(2) = 2 = b(3) because of (3,3,3,3) - (-3,-1,2,2) = (6,4,1,1) and (3,3,3,3) - (-2,-2,1,3) = (5,5,2,0) (12 permutations each). For example, 5 + 5 + 2 + 0 = 12 = 4*3 and 5^2 + 5^2 + 2^2 + 0^2 = 54 = 6*3^2 (with n = 2 and m = 3).
|
|
|
MATHEMATICA
|
sqrtint = Floor[Sqrt[#]]&;
q[r_, s_, g_] := Module[{d = 2s - r^2, h}, If[d <= 0, d==0 && Mod[r, 2]==0 && GCD[g, r/2]==1, h = Sqrt[d]; If[IntegerQ[h] && Mod[r+h, 2]==0 && GCD[g, GCD[(r+h)/2, (r-h)/2]]==1, 2, 0]]] /. {True -> 1, False -> 0};
a[n_] := Module[{m = 2n - 1, s}, s = 2m^2; Sum[q[i + j, s - i^2 - j^2, GCD[i, j]], {i, -sqrtint[s], sqrtint[s]}, {j, -sqrtint[s - i^2], sqrtint[s - i^2]}]/12];
|
|
|
PROG
|
(PARI)
q(r, s, g)={my(d=2*s - r^2); if(d<=0, d==0 && r%2==0 && gcd(g, r/2)==1, my(h); if(issquare(d, &h) && (r+h)%2==0 && gcd(g, gcd((r+h)/2, (r-h)/2))==1, 2, 0))}
a(n)={my(m=2*n-1, s=2*m^2); sum(i=-sqrtint(s), sqrtint(s), sum(j=-sqrtint(s-i^2), sqrtint(s-i^2), q(i+j, s-i^2-j^2, gcd(i, j)) ))/12} \\ Andrew Howroyd, Aug 02 2018
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
