A354778 -id:A354778 - 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

A354766 1/4 of the total number of integral quadruples with sum = n and sum of squares = n^2.

Original entry on oeis.org

1, 2, 4, 2, 7, 8, 7, 2, 13, 14, 13, 8, 13, 14, 28, 2, 19, 26, 19, 14, 28, 26, 25, 8, 37, 26, 40, 14, 31, 56, 31, 2, 52, 38, 49, 26, 37, 38, 52, 14, 43, 56, 43, 26, 91, 50, 49, 8, 49, 74, 76, 26, 55, 80, 91, 14, 76, 62, 61, 56, 61, 62, 91, 2, 91, 104, 67, 38, 100, 98, 73, 26, 73, 74, 148, 38, 91, 104, 79, 14, 121, 86, 85, 56

Offset: 1

N. J. A. Sloane, Jun 19 2022, based on an email from Colin Mallows, Jun 12 2022

If instead we count only primitive quadruples (meaning quadruples (h,i,j,k) with gcd(h,i,j,k) = 1) we get A278085(n).
Conjectures from Colin Mallows, Jun 12 2022: (Start)
Given a natural number n, a "quad" for n is a quadruple q = (h,i,j,k) of integers with sum(q) = h+i+j+k = n and sum(q^2) = h^2+i^2+j^2+k^2 = n^2.
A quad q is "primitive" if gcd(h,i,j,k) = 1. Define pq(n) = A278085(n) to be the number of distinct primitive quads for n, and tq(n) (the present sequence) to be the total number of quads for n.
Conjecture 1: (Based on the data for n <= 5000) pq/4 and tq/4 are multiplicative sequences.
Conjecture 2: When n = p^k, p prime and k >= 1:
  if p = 2, k =  1 then pq(q)/4 = 1 and tq(n)/4 = 2;
  if p = 2, k >= 2 then pq(q)/4 = 0 and tq(n)/4 = 2;
  if p = 3, k >= 1 then pq(q)/4 = n and tq(n)/4 = (3*n-1)/2;
  if p == 5 (mod 6), k >= 1 then pq(q)/4 = (p+1)*n/p and tq(n)/4 = n + 2*(n-1)/(p-1);
  if p == 1 (mod 6), k >= 1 then pq(q)/4 = (p-1)*n/p and tq(n)/4 = n.
(End)
Conjecture: the numbers n for which a(n) = n have a positive asymptotic density.

			Solutions for n = 1: (1,0,0,0) and all permutations thereof.
n=2: (2,0,0,0) and (1,1,1,-1).
n=3: (3,0,0,0) and (2,2,-1,0).
n=4: (4,0,0,0) and (2,2,2,-2). Eight solutions, so a(4) = 8/4 = 2. None are primitive, so A278085(4) = 0.
n=5: (5,0,0,0) and (4,2,-2,1). 4+24 solutions, so a(5) = 28/4 = 7. 24 are primitive, so A278085(5) = 24/4 = 6.

Robert Israel, Table of n, a(n) for n = 1..650

Cf. A278085, A354777, A354778.

See also A353589 (counts nondecreasing nonnegative (h,i,j,k) such that (+-h, +-i, +-j, +-k) is a solution).

f:= proc(n) local d; add(g3(n-d, n^2 - d^2), d=-n .. n)/4 end proc:
g3:= proc(x,y) option remember; local m,c;
   if x^2 > 3*y then return 0 fi;
   m:= floor(sqrt(y));
   add(g2(x-c,y - c^2), c=- m.. m)
end proc:
g2:= proc(x,y) option remember;
   local v;
   v:= 2*y - x^2;
   if not issqr(v) then 0
   elif v = 0 then 1
   else 2
   fi
end proc:
map(f, [$1..100]); # Robert Israel, Feb 16 2023

f[n_] := Sum[g3[n - d, n^2 - d^2], {d, -n, n}]/4 ;
g3[x_, y_] := g3[x, y] = Module[{m}, If[x^2 > 3*y, 0, m = Floor[Sqrt[y]]; Sum[g2[x - c, y - c^2], {c, -m, m}]]];
g2[x_, y_] := g2[x, y] = Module[{v}, v = 2*y - x^2; Which[!IntegerQ@Sqrt[v], 0, v == 0, 1, True, 2]];
f /@ Range[100] (* Jean-François Alcover, Mar 09 2023, after Robert Israel *)

A354780 a(n) is the bitwise OR of (the binary expansions of) b(n+1) to b(2*n), where b is A354169.

Original entry on oeis.org

2, 12, 27, 115, 252, 1004, 2013, 4031, 16307, 32631, 65279, 261375, 524270, 2096110, 4194253, 8386527, 16773119, 67096575, 134217659, 536854459, 1073741623, 2147450751, 4294901759, 17179672575, 34359737599, 137438690559, 274877382143, 549754765311, 2199022205950, 4398044412927, 8796093022189, 35184367894509, 70368744175567

Offset: 1

N. J. A. Sloane, Jul 05 2022

If the binary expansion of a(n) has a 1 in the 2^i's bit (for any i >= 0) then A354169(2*n+1) must have a 0 in that bit.
A354169(2*n+1) is the smallest number not yet in A354169 which satisfies that condition (this follows at once from the definition of A354169).
This sequence bears the same relation to A354169 as A355057 does to A090252.

			Consider n=6. Then b(7) to b(12) are 32, 64, 12, 128, 256, 512. The bitwise OR of those 6 numbers is 1111101100_2 = 1004_10 = a(6). The bitwise complement of 1004_10 is 10011_2 = 19_10 = A354781(6), and A354169(6) = 17_10 = 10001_2.
On the other hand, for n=5, b(6) to b(10) are 16, 32, 64, 12, 128, whose bitwise OR is 11111100_2 = 252_10 = a(5). The bitwise complement of 252_10 is 3_10 = 11_2 = A354781(5). However, 3 has already appeared in A354169, and the smallest available number whose binary expansion is disjoint from 252_10 = 11111100_2 is 2^8 = 100000000_2 = 256_10 = 2^8 = A354169(5).

N. J. A. Sloane, Table of n, a(n) for n = 1..256

A bisection of A354757.
Cf. A354169, A354680, A354757, A354758, A354767, A354773, A354774, A354778, A355150, A354781.
See also A090252, A355057.

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

N. J. A. Sloane, Jun 27 2022

Row n has width A307531(n).

			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).

Cf. A000118, A307531.

T(n^2,n) = A354778(n). See also A278085 and A354766.

A354781 If the binary expansion of A354780(n) is 1 d_1 d_2 ... d_k, then the binary expansion of a(n) is c_1 c_2 ... c_k, where c_i = 1 - d_i.

Original entry on oeis.org

1, 3, 4, 12, 3, 19, 34, 64, 76, 136, 256, 768, 17, 1041, 50, 2080, 4096, 12288, 68, 16452, 200, 32896, 65536, 196608, 768, 262912, 524800, 1048576, 1049601, 2098176, 18, 4194322, 2096, 8390656, 16777216, 50331648, 12288, 67121152, 134225920, 268435456, 268451844, 536887296, 72, 1073741896, 32960, 2147516416, 4294967296, 12884901888

Offset: 1

N. J. A. Sloane, Jul 05 2022

			See A354780.

N. J. A. Sloane, Table of n, a(n) for n = 1..2400

Cf. A354169, A354680, A354757, A354758, A354767, A354773, A354774, A354778, A355150, A354780.

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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A354766 1/4 of the total number of integral quadruples with sum = n and sum of squares = n^2.

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A354780 a(n) is the bitwise OR of (the binary expansions of) b(n+1) to b(2*n), where b is A354169.

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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)).

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A354781 If the binary expansion of A354780(n) is 1 d_1 d_2 ... d_k, then the binary expansion of a(n) is c_1 c_2 ... c_k, where c_i = 1 - d_i.

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