A278086 -id:A278086 - cp's OEIS frontend

A278085 1/4 of the number of primitive integral quadruples with sum = 3n and sum of squares = 3n^2.

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

Colin Mallows, Nov 14 2016

nonn

Conjecture: a(n) is multiplicative, with a(2) = 1, a(2^k) = 0 for k>=2, and for k >= 1 and p an odd prime, a(p^k) = p^(k-1)*a(p), with a(p) = p+1 for p == 5 (mod 6), a(p) = p-1 for p=1 (mod 6), and a(3) = 3. It would be nice to have a proof of this. [See A354766 for additional conjectures. - N. J. A. Sloane, Jun 19 2022]

This is also 1/4 of the number of primitive integral quadruples with sum = n and sum of squares = n^2. See A354766, A354777, A354778 for the total number of solutions. - N. J. A. Sloane, Jun 27 2022

			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.

Cf. A046897, A278081, A278082, A278083, A278084, A278086, A354766, A353589, A354777, A354778.

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

Example edited by Petros Hadjicostas, Apr 21 2020

A278081 a(n) is 1/12 of the number of primitive quadruples with sum = 0 and sum of squares = 2m^2, where m = 2n - 1.

Original entry on oeis.org

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

Offset: 1

Colin Mallows, Nov 14 2016

nonn

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]

			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).

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, 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.]
Colin Mallows, R programs for A278081-A278086.

Cf. A005886, A046897, A278082, A278083, A278084, A278085, A278086.

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];
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(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

Terms a(51) and beyond from Andrew Howroyd, Aug 02 2018

Name and example section edited by Petros Hadjicostas, Apr 21 2020

A005886 Theta series of f.c.c. lattice with respect to tetrahedral hole.

Original entry on oeis.org

4, 12, 12, 16, 24, 12, 24, 36, 12, 28, 36, 24, 36, 36, 24, 24, 60, 36, 28, 48, 12, 60, 60, 24, 48, 48, 36, 48, 60, 24, 52, 84, 48, 24, 60, 36, 48, 96, 36, 72, 48, 36, 72, 60, 48, 52, 96, 36, 60, 96, 24, 72, 108, 24, 48, 60, 72, 96, 84, 60, 48, 108, 36, 52, 72, 60, 108, 108, 36, 48, 108

Offset: 0

N. J. A. Sloane

Empirically, the number of integral quadruples with sum = 1, sum-of-squares = 2n-1. - Colin Mallows, Dec 31 2016

			4 + 12*x + 12*x^2 + 16*x^3 + 24*x^4 + 12*x^5 + 24*x^6 + 36*x^7 + 12*x^8 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 0..1000
G. Nebe and N. J. A. Sloane, Home page for this lattice
N. J. A. Sloane and B. K. Teo, Theta series and magic numbers for close-packed spherical clusters, J. Chem. Phys. 83 (1985) 6520-6534.
Index entries for sequences related to f.c.c. lattice

Cf. A005869, A005878, A008443. Partial sums is A121054. Cf also A278081-A278086.

QP = QPochhammer; CoefficientList[4(QP[q^2]^2/QP[q])^3 + O[q]^50, q] (* Jean-François Alcover, Jul 04 2017, after Michael Somos *)

a(n) = 1/2 * A005878(n) = 2 * A005869(n) = 4 * A008443(n). - Michael Somos, May 31 2012

Terms a(50) onward added by G. C. Greubel, Feb 20 2018

A278082 1/12 of the number of primitive quadruples with sum = n and sum of squares = 3*n^2.

Original entry on oeis.org

1, 1, 2, 0, 4, 2, 8, 0, 6, 4, 11, 0, 14, 8, 8, 0, 18, 6, 20, 0, 16, 11, 22, 0, 20, 14, 18, 0, 30, 8, 30, 0, 22, 18, 32, 0, 36, 20, 28, 0, 42, 16, 44, 0, 24, 22, 46, 0, 56, 20, 36, 0, 52, 18, 44, 0, 40, 30, 58, 0, 62, 30, 48, 0, 56, 22, 66, 0, 44, 32, 70, 0, 74, 36, 40, 0, 88, 28, 80, 0, 54, 42, 84, 0, 72, 44, 60, 0, 88, 24, 112, 0, 60, 46, 80, 0, 96, 56, 66, 0

Offset: 1

Colin Mallows, Nov 14 2016

nonn

Conjecture: a(n) is multiplicative, with a(2) = 1, a(2^k) = 0 (k >= 2); a(p^k) = p^(k-1)*a(p); a(p) = p + 1 for p == (2, 6, 7, 8, 10)(mod 11), a(p) = p - 1 for p == (1, 3, 4, 5, 9)(mod 11); and p(11) = 11. It would be nice to have a proof of this.

This sequence applies also to the case sum = 3*n and ssq = 5*n^2. - Colin Mallows, Nov 30 2016 [Edited by Petros Hadjicostas, Apr 20 2020]

			For the case r = 1 and r = 3, we have 12*a(3) = 24 because of (-3,1,1,4) and (-1,-1,0,5) (12 permutations each). For example, (-3) + 1 + 1 + 4 = 3 = 1*3 and (-3)^2 + 1^2 + 1^2 + 4^2 = 27 = 3*3^2.
For the case r = 3 and m = 5, we again have 12*a(3) = 24 because of (3,3,3,3) - (-3,1,1,4) = (6,2,2,-1) and (3,3,3,3) - (-1,-1,0,5) = (4,4,3,-2) (12 permutations each). For example, 6 + 2 + 2 + (-1) = 9 = 3*3 and 6^2 + 2^2 + 2^2 + (-1)^2  = 45 = 5*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, 1, 3), where r = 1 and s = 3. To get the 1/12 of these counts, type gc(1:120, 1, 3)[,3]/12. As stated in the comments, we get the same sequence with r = 3 and s = 5, i.e., we may type gc(1:120, 3, 5)[,3]/12.]
Colin Mallows, R programs for A278081-A278086.

Cf. A046897, A278081, A278083, A278084, A278085, A278086.

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[{s = 3n^2}, Sum[q[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]}]/12];
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(n-i-j, s-i^2-j^2, gcd(i,j)) ))/12} \\ Andrew Howroyd, Aug 02 2018

Example section edited by Petros Hadjicostas, Apr 21 2020

A278083 a(n) is 1/6 of the number of primitive integral quadruples with sum = 2m and sum of squares = 2m^2, where m = 2*n-1.

Original entry on oeis.org

1, 4, 4, 8, 12, 12, 12, 16, 16, 20, 32, 24, 20, 36, 28, 32, 48, 32, 36, 48, 40, 44, 48, 48, 56, 64, 52, 48, 80, 60, 60, 96, 48, 68, 96, 72, 72, 80, 96, 80, 108, 84, 64, 112, 88, 96, 128, 80, 96, 144, 100, 104, 128, 108, 108, 144, 112, 96, 144, 128, 132, 160

Offset: 1

Colin Mallows, Nov 14 2016

nonn

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; b(p^k) = p^(k-1)*b(p) for p an odd prime; b(p) = p+1 for p == 3 (mod 4); b(p) = p-1 for p == 1 (mod 4). It would be nice to have a proof of this.

			6*a(2) = 24 = 6*b(3) because of (-1,2,2,3) and (0,1,1,4) (12 permutations each). For example, (-1) + 2 + 2 + 3 = 6 = 2*3 and (-1)^2 + 2^2 + 2^2 + 3^2 = 18 = 2*3^2 (with n = 2 and m = 3 = 2*n - 1).

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, with the zeros, i.e., the sequence (b(n): n >= 1) shown in the comments above, type gc(1:120, 2, 2), where r = 2 and s = 2. To get the 1/6 of these counts with no zeros, type gc(seq(1,59,2), 2, 2)[,3]/6.]
Colin Mallows, R programs for A278081-A278086.

Cf. A046897, A278081, A278082, A278084, A278085, 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[{m = 2n - 1, s}, s = 2m^2; Sum[q[2m - i - j, s - i^2 - j^2, GCD[i, j]] , {i, -sqrtint[s], sqrtint[s]}, {j, -sqrtint[s - i^2], sqrtint[s - i^2]}]/6];
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(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(2*m-i-j, s-i^2-j^2, gcd(i,j)) ))/6} \\ Andrew Howroyd, Aug 02 2018

Terms a(51) and beyond from Andrew Howroyd, Aug 02 2018

Name and example section edited by Petros Hadjicostas, Apr 21 2020

A278084 a(n) is 1/24 of the number of primitive integral quadruples with sum = 2m and sum of squares = 6m^2, where m = 2*n-1.

Original entry on oeis.org

1, 2, 5, 6, 6, 12, 14, 10, 18, 20, 12, 22, 25, 18, 28, 32, 24, 30, 38, 28, 40, 42, 30, 46, 42, 36, 54, 60, 40, 60, 60, 36, 70, 66, 44, 72, 74, 50, 72, 80, 54, 82, 90, 56, 88, 84, 64, 100, 98, 72, 100, 102, 60, 106, 108, 76, 114, 110, 84, 108, 132, 80, 125, 126

Offset: 1

Colin Mallows, Nov 14 2016

nonn

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, and for p an odd prime, b(p^k) = p^(k-1)*b(p), with b(p) = p + 1 for p == (11, 13, 17, 19) (mod 20), b(p) = p - 1 for p == (1, 3, 7, 9) (mod 20), b(5) = 5. It would be nice to have a proof of this.

			24*a(2) = 48 = 24*b(3) because of (-4,2,3,5) and (-2,0,1,7) (24 permutations each). For example, (-2) + 0 + 1 + 7 = 6 = 2*3 and (-2)^2 + 0^2 + 1^2 + 7^2 = 54 = 6*3^2 (with n = 2 and m = 3 = 2*2 - 1).

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, with the zeros, i.e., the sequence (b(n): n >= 1) shown in the comments above, type gc(1:120, 2, 6), where r = 2 and s = 6. To get the 1/24 of these counts with no zeros, type gc(seq(1,59,2), 2, 6)[,3]/24.]
Colin Mallows, R programs for A278081-A278086.

Cf. A046897, A278081, A278082, A278083, A278085, A278086.

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 = 6m^2; Sum[q[2m - i - j, s - i^2 - j^2, GCD[i, j]] , {i, -sqrtint[s], sqrtint[s]}, {j, -sqrtint[s - i^2], sqrtint[s - i^2]}]/24];
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(m=2*n-1, s=6*m^2); sum(i=-sqrtint(s), sqrtint(s), sum(j=-sqrtint(s-i^2), sqrtint(s-i^2), q(2*m-i-j, s-i^2-j^2, gcd(i,j)) ))/24} \\ Andrew Howroyd, Aug 02 2018

Terms a(51) and beyond from Andrew Howroyd, Aug 02 2018

Name and example section edited by Petros Hadjicostas, Apr 21 2020

cp's OEIS Frontend

A278085 1/4 of the number of primitive integral quadruples with sum = 3*n and sum of squares = 3*n^2.

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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.

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A005886 Theta series of f.c.c. lattice with respect to tetrahedral hole.

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A278082 1/12 of the number of primitive quadruples with sum = n and sum of squares = 3*n^2.

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A278083 a(n) is 1/6 of the number of primitive integral quadruples with sum = 2*m and sum of squares = 2*m^2, where m = 2*n-1.

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A278084 a(n) is 1/24 of the number of primitive integral quadruples with sum = 2*m and sum of squares = 6*m^2, where m = 2*n-1.

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A278085 1/4 of the number of primitive integral quadruples with sum = 3n and sum of squares = 3n^2.

A278081 a(n) is 1/12 of the number of primitive quadruples with sum = 0 and sum of squares = 2m^2, where m = 2n - 1.

A278083 a(n) is 1/6 of the number of primitive integral quadruples with sum = 2m and sum of squares = 2m^2, where m = 2*n-1.

A278084 a(n) is 1/24 of the number of primitive integral quadruples with sum = 2m and sum of squares = 6m^2, where m = 2*n-1.