OFFSET
1,2
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Any nonnegative integer can be written as x^4 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that a*y^2 + b*y*z + c*z^2 is a square, whenever (a,b,c) is among the ordered triples (1,484,44), (1,666,9), (16,1336,169), (25,900,36).
(iii) For each c = 1, 49, any nonnegative integer can be written as x^4 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that 120*(x^2+y)*z + c*z^2 is a square.
By the linked JNT paper, each n = 0,1,2,... can be expressed as the sum of a fourth power and three squares.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.
EXAMPLE
a(1) = 1 since 1 = 0^4 + 1^2 + 0^2 + 0^2 with 1^2 + 64*0^2 + 1024*1*0 = 1^2.
a(31) = 1 since 31 = 1^4 + 2^2 + 1^2 + 5^2 with 2^2 + 64*1^2 + 1024*2*1 = 46^2.
a(47) = 1 since 47 = 1^4 + 6^2 + 3^2 + 1^2 with 6^2 + 64*3^2 + 1024*6*3 = 138^2.
a(79) = 1 since 79 = 1^4 + 7^2 + 2^2 + 5^2 with 7^2 + 64*2^2 + 1024*7*2 = 121^2.
a(156) = 1 since 156 = 3^4 + 5^2 + 5^2 + 5^2 with 5^2 + 64*5^2 + 1024*5*5 = 165^2.
a(184) = 1 since 184 = 0^4 + 12^2 + 6^2 + 2^2 with 12^2 + 64*6^2 + 1024*12*6 = 276^2.
a(316) = 1 since 316 = 2^4 + 10^2 + 10^2 + 10^2 with 10^2 + 64*10^2 + 1024*10*10 = 330^2.
a(380) = 1 since 380 = 1^4 + 3^2 + 3^2 + 19^2 with 3^2 + 64*3^2 + 1024*3*3 = 99^2.
a(2383) = 1 since 2383 = 3^4 + 22^2 + 33^2 + 27^2 with 22^2 + 64*33^2 + 1024*22*33 = 902^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^4-y^2-z^2]&&SQ[y^2+1024y*z+64z^2], r=r+1], {x, 0, (n-1)^(1/4)}, {y, 1, Sqrt[n-x^4]}, {z, 0, Sqrt[n-x^4-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 18 2017
STATUS
approved