OFFSET
0,3
COMMENTS
Conjecture: a(n) > 0 for all n = 0,1,2,....
The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares.
We have verified the conjecture for all n = 0..10^6.
Qing-Hu Hou at Tianjin University verified a(n) > 0 for n up to 10^8. - Zhi-Wei Sun, Jun 02 2019
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..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 = 0^2 + 0^2 + 0^2 + 1^2 with 0 = 0^2 and -7*0 - 8*0 + 8*0 + 16*1 = 4^2.
a(12) = 1 since 12 = 1^2 + 1^2 + 3^2 + 1^2 with 1 = 1^2 and -7*1 - 8*1 + 8*3 + 16*1 = 5^2.
a(17) = 1 since 17 = 1^2 + 0^2 + 4^2 + 0^2 with 1 = 1^2 and -7*1 - 8*0 + 8*4 + 16*0 = 5^2.
a(28) = 1 since 28 = 4^2 + 2^2 + 2^2 + 2^2 with 4 = 2^2 and -7*4 - 8*2 + 8*2 + 16*2 = 2^2.
a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 1 = 1^2 and -7*1 - 8*1 + 8*2 + 16*5 = 9^2.
a(40) = 1 since 40 = 4^2 + 2^2 + 2^2 + 4^2 with 4 = 2^2 and -7*4 -8*2 + 8*2 + 16*4 = 6^2.
a(41) = 1 since 41 = 1^2 + 2^2 + 6^2 + 0^2 with 1 = 1^2 and -7*1 - 8*2 + 8*6 + 16*0 = 5^2.
a(49) = 1 since 49 = 0^2 + 6^2 + 2^2 + 3^2 with 0 = 0^2 and -7*0 - 8*6 + 8*2 + 16*3 = 4^2.
a(241) = 1 since 241 = 9^2 + 4^2 + 12^2 + 0^2 with 9 = 3^2 and -7*9 - 8*4 + 8*12 + 16*0 = 1^2.
a(433) = 1 since 433 = 16^2 + 8^2 + 8^2 + 7^2 with 16 = 4^2 and -7*16 - 8*8 + 8*8 + 16*7 = 0^2.
a(1113) = 1 since 1113 = 1^2 + 30^2 + 4^2 + 14^2 with 1 = 1^2 and -7*1 - 8*30 + 8*4 + 16*14 = 3^2.
a(1521) = 1 since 1521 = 0^2 + 22^2 + 14^2 + 29^2 with 0 = 0^2 and -7*0 - 8*22 + 8*14 + 16*29 = 20^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[n-x^4-y^2-z^2]&&SQ[16*Sqrt[n-x^4-y^2-z^2]+8z-8y-7x^2], r=r+1], {x, 0, n^(1/4)}, {y, 0, Sqrt[n-x^4]}, {z, 0, Sqrt[n-x^4-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 04 2017
STATUS
approved