

A300908


Number of ways to write n as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square , where w is a positive integer and x,y,z are nonnegative integers such that z or 2*z or 3*z is a square.


5



1, 3, 1, 2, 6, 1, 2, 3, 2, 7, 2, 2, 7, 1, 5, 2, 7, 9, 3, 6, 2, 3, 4, 1, 9, 7, 3, 4, 5, 7, 2, 3, 5, 6, 3, 4, 7, 3, 8, 6, 7, 6, 4, 3, 7, 3, 2, 2, 4, 13, 5, 8, 6, 5, 3, 1, 8, 8, 3, 6, 5, 1, 11, 2, 16, 5, 4, 8, 1, 8, 2, 7, 11, 7, 5, 4, 2, 13, 2, 6
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

Conjecture 1: a(n) > 0 for all n > 0. Moreover, any positive integer can be written as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square, where w is a positive integer and x,y,z are nonnegative integers for which one of z, z/2, z/3 is a square and z/2 (or z/3) is a power of 4 (including 1).
Conjecture 2: Each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with (3*x)^2 + (4*y)^2 + (12*z)^2 a square, where x,y,z,w are nonnegative integers for which x or z is a square or y = 2^k for some k = 0,1,2,....
Conjecture 3. Let a,b,c be positive integers with a <= b <= c and gcd(a,b,c) = 1. If each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that (a*x)^2 + (b*y)^2 + (c*z)^2 is a square, then (a,b,c) must be among the three triples (3,4,12), (12,15,20) and (12,21,28).
We have verified Conjectures 1 and 2 for n up to 5*10^5 and 10^6 respectively.
See also A300791 and A300844 for similar conjectures.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, Refining Lagrange's foursquare theorem, J. Number Theory 175(2017), 167190.
ZhiWei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 20172018.


EXAMPLE

a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 0 = 0^2 and (3*1)^2 + (4*1)^2 + (12*0)^2 = 5^2.
a(14) = 1 since 14 = 3^2 + 0^2 + 1^2 + 2^2 with 1 = 1^2 and (3*3)^2 + (4*0)^2 + (12*1)^2 = 15^2.
a(69) = 1 since 69 = 0^2 + 8^2 + 2^2 + 1^2 with 2*2 = 2^2 and (3*0)^2 + (4*8)^2 + (12*2)^2 = 40^2.
a(671) = 1 since 671 = 18^2 + 17^2 + 3^2 + 7^2 with 3*3 = 3^2 and (3*18)^2 + (4*17)^2 + (12*3)^2 = 94^2.
a(1175) = 1 since 1175 = 30^2 + 5^2 + 9^2 + 13^2 with 9 = 3^2 and (3*30)^2 + (4*5)^2 + (12*9)^2 = 142^2.
a(12151) = 1 since 12151 = 50^2 + 71^2 + 49^2 + 47^2 with 49 = 7^2 and (3*50)^2 + (4*71)^2 + (12*49)^2 = 670^2.
a(16204) = 1 since 16204 = 90^2 + 90^2 + 0^2 + 2^2 with 0 = 0^2 and (3*90)^2 + (4*90)^2 + (12*0)^2 = 450^2.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[z]SQ[2z]SQ[3z], Do[If[SQ[nx^2y^2z^2]&&SQ[(3x)^2+(4y)^2+(12z)^2], r=r+1], {x, 0, Sqrt[n1z^2]}, {y, 0, Sqrt[n1x^2z^2]}]], {z, 0, Sqrt[n1]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]


CROSSREFS

Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300791, A300792, A300844.
Sequence in context: A246432 A112571 A051277 * A080818 A334354 A052417
Adjacent sequences: A300905 A300906 A300907 * A300909 A300910 A300911


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Mar 15 2018


STATUS

approved



