A278085
1/4 of the number of primitive integral quadruples with sum = 3*n and sum of squares = 3*n^2.
Original entry on oeis.org
1, 1, 3, 0, 6, 3, 6, 0, 9, 6, 12, 0, 12, 6, 18, 0, 18, 9, 18, 0, 18, 12, 24, 0, 30, 12, 27, 0, 30, 18, 30, 0, 36, 18, 36, 0, 36, 18, 36, 0, 42, 18, 42, 0, 54, 24, 48, 0, 42, 30, 54, 0, 54, 27, 72, 0, 54, 30, 60, 0, 60, 30, 54, 0, 72, 36, 66, 0, 72, 36, 72, 0, 72, 36, 90, 0, 72, 36, 78, 0, 81, 42, 84, 0, 108, 42, 90, 0, 90, 54, 72, 0, 90, 48, 108, 0, 96, 42, 108, 0
Offset: 1
For the case r = s = 3, we have 4*a(3) = 12 because of (1,1,3,4) (12 permutations). Indeed, 1 + 1 + 3 + 4 = 9 = 3*3 and 1^2 + 1^2 + 3^2 + 4^2 = 27 = 3*3^2.
For the case r = s = 1, we have again 4*a(3) = 12 because of (3,3,3,3) - (1,1,3,4) = (2,2,0,-1) (12 permutations). Indeed, 2 + 2 + 0 + (-1) = 3 = 1*3 and 2^2 + 2^2 + 0^2 + (-1)^2 = 9 = 1*3^2.
- Andrew Howroyd, Table of n, a(n) for n = 1..500
- Petros Hadjicostas, Slight modification of Mallows' R program. [To get the total counts for n = 1 to 120, type gc(1:120, 3, 3), where r = 3 and s = 3. To get the 1/4 of these counts, type gc(1:120, 3, 3)[,3]/4. As stated in the comments, we get the same sequence with r = 1 and s = 1, i.e., we may type gc(1:120, 1, 1)[,3]/4.]
- Colin Mallows, R programs for A278081-A278086.
-
sqrtint = Floor[Sqrt[#]]&;
q[r_, s_, g_] := Module[{d = 2 s - 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[{s}, s = 3 n^2; Sum[q[3 n - i - j, s - i^2 - j^2, GCD[i, j]] , {i, -sqrtint[s], sqrtint[s]}, {j, -sqrtint[s - i^2], sqrtint[s - i^2]}]/4];
Table[an = a[n]; Print[n, " ", an]; an, {n, 1, 100}] (* Jean-François Alcover, Sep 20 2020, after Andrew Howroyd *)
-
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(s=3*n^2); sum(i=-sqrtint(s), sqrtint(s), sum(j=-sqrtint(s-i^2), sqrtint(s-i^2), q(3*n-i-j, s-i^2-j^2, gcd(i,j)) ))/4} \\ Andrew Howroyd, Aug 02 2018
A354778
Number of integer quadruples (u,v,w,x) such that u^2+v^2+w^2+x^2 = n^2 and u+v+w+x = n.
Original entry on oeis.org
1, 4, 8, 16, 8, 28, 32, 28, 8, 52, 56, 52, 32, 52, 56, 112, 8, 76, 104, 76, 56, 112, 104, 100, 32, 148, 104, 160, 56, 124, 224, 124, 8, 208, 152, 196, 104, 148, 152, 208, 56, 172, 224, 172, 104, 364, 200, 196, 32, 196, 296, 304, 104, 220, 320, 364, 56, 304, 248, 244, 224, 244, 248, 364, 8, 364, 416, 268, 152, 400, 392, 292, 104, 292, 296, 592, 152, 364, 416, 316, 56, 484, 344, 340, 224
Offset: 0
A354777
Irregular triangle read by rows: T(n,k) is the number of integer quadruples (u,v,w,x) such that u^2+v^2+w^2+x^2 = n and u+v+w+x = k (n>=0, 0 <= k <= A307531(n)).
Original entry on oeis.org
1, 0, 4, 12, 0, 6, 0, 12, 0, 4, 6, 0, 8, 0, 1, 0, 12, 0, 12, 24, 0, 24, 0, 12, 0, 16, 0, 12, 0, 4, 12, 0, 0, 0, 6, 0, 24, 0, 16, 0, 12, 24, 0, 30, 0, 24, 0, 6, 0, 12, 0, 24, 0, 12, 8, 0, 24, 0, 12, 0, 8, 0, 24, 0, 12, 0, 16, 0, 4, 48, 0, 24, 0, 24, 0, 24, 0, 36, 0, 24, 0, 24, 0, 12, 6, 0, 0, 0, 8, 0, 0, 0, 1, 0, 12, 0, 36, 0, 12, 0, 12
Offset: 0
The triangle begins:
[1],
[0, 4],
[12, 0, 6],
[0, 12, 0, 4],
[6, 0, 8, 0, 1],
[0, 12, 0, 12],
[24, 0, 24, 0, 12],
[0, 16, 0, 12, 0, 4],
[12, 0, 0, 0, 6],
[0, 24, 0, 16, 0, 12],
[24, 0, 30, 0, 24, 0, 6],
[0, 12, 0, 24, 0, 12],
[8, 0, 24, 0, 12, 0, 8],
[0, 24, 0, 12, 0, 16, 0, 4],
[48, 0, 24, 0, 24, 0, 24],
[0, 36, 0, 24, 0, 24, 0, 12],
[6, 0, 0, 0, 8, 0, 0, 0, 1],
[0, 12, 0, 36, 0, 12, 0, 12],
[36, 0, 48, 0, 48, 0, 30, 0, 12],
...
T(4,2) = 8 from the solutions (u,v,w,x) = (2,0,0,0) (4 such) and (1,1,1,-1) (4 such).
A353589
Number of nondecreasing nonnegative integer quadruples (m,p,q,r) such that m^2 + p^2 + q^2 + r^2 = n^2 and m +- p +- q +- r = +- n.
Original entry on oeis.org
1, 1, 2, 2, 2, 2, 4, 2, 2, 4, 4, 4, 4, 3, 4, 6, 2, 5, 8, 5, 4, 6, 8, 5, 4, 7, 6, 9, 4, 6, 12, 6, 2, 12, 10, 9, 8, 7, 10, 10, 4, 9, 12, 9, 8, 17, 10, 9, 4, 9, 14, 16, 6, 10, 18, 17, 4, 16, 12, 12, 12, 11, 12, 17, 2, 16, 24, 13, 10, 18, 18, 13, 8, 14, 14, 26, 10, 17, 20, 14, 4, 23
Offset: 0
For n = 1, (0, 0, 0, 1) is the only solution.
For n = 2, (0, 0, 0, 2) and (1, 1, 1, 1) are solutions, with 1 + 1 + 1 - 1 = 2.
-
apply( {A353589(n, show=0, cnt=0, n2=n^2, e=[1,-1]~)=
for(a=0,sqrtint(n2\4), for(b=a,sqrtint((n2-a^2)\3),
my(s=[a+b, b-a, a-b, -a-b]); foreach(sum2sqr(n2-a^2-b^2), cd, cd[1] >= b &&
vecsum(cd)+s[1] >= n && foreach(s, d, (vecsum(cd)+d==n || abs(cd*e+d)==n)&&
cnt++&& !(show && print1(concat([a, b], cd)))&& break)))); cnt}, [0..99]) \\ See A133388 for sum2sqr().
Showing 1-4 of 4 results.
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