OFFSET
1,5
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 3, 15, 31, 39, 71, 79, 195, 311, 319, 403, 559, 591, 683, 719, 1031, 1439, 1643, 2519, 6879, 2^k, 2^(2k+1)*39 (k = 0,1,2,...). Also, any positive integer can be written as w^2 + x^2 + y^2 + z^2 with x a positive integer and w,y,z nonnegative integer such that 6*w^2*x^2 + 12*x^2*y^2 + 52*y^2*z^2 + 27*z^2*w^2 is a square.
(ii) For each triple (a,b,c) = (1,3,2), (1,11,9), (1,14,4),(1,20,25), (1,27,18), (1,36,9), (1,56,4), (4,32,25), (9,15,25), (9,35,25), (25,8,64), (25,15,54), (25,32,28), (25,35,49), (28,32,49), any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z integers such that a*w^2*x^2 + b*x^2*y^2 + c*y^2*z^2 is a square.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723, 2016.
EXAMPLE
a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 1 > 0 and 6*0^2*0^2 + 12*0^2*0^2 + 52*0^2*1^2 + 27*1^2*0^2 = 0^2.
a(2) = 1 since 2 = 0^2 + 1^2 + 0^2 + 1^2 with 1 > 0 and 6*0^2*1^2 + 12*1^2*0^2 + 52*0^2*1^2 + 27*1^2*0^2 = 0^2.
a(3) = 1 since 3 = 0^2 + 1^2 + 1^2 + 1^2 with 1 > 0 and 6*0^2*1^2 + 12*1^2*1^2 + 52*1^2*1^2 + 27*1^2*0^2 = 8^2.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^2 + 1^2 with 1 > 0 and 6*2^2*3^2 + 12*3^2*1^2 + 52*1^2*1^2 = 22^2.
a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 5 > 0 and 6*1^2*1^2 + 12*1^2*2^2 + 52*2^2*5^2 + 27*5^2*1^2 = 77^2.
a(39) = 1 since 39 = 2^2 + 1^2 + 5^2 + 3^2 with 3 > 0 and 6*2^2*1^2 + 12*1^2*5^2 + 52*5^2*3^2 + 27*3^2*2^2 = 114^2.
a(71) = 1 since 71 = 3^2 + 1^2 + 6^2 + 5^2 with 5 > 0 and 6*3^2*1^2 + 12*1^2*6^2 + 52*6^2*5^2 + 27*5^2*3^2 = 231^2.
a(78) = 1 since 78 = 2^2 + 7^2 + 4^2 + 3^2 with 3 > 0 and 6*2^2*7^2 + 12*7^2*4^2 + 52*7^2*4^2 + 27*3^2*2^2 = 138^2.
a(79) = 1 since 79 = 2^2 + 5^2 + 7^2 + 1^2 with 1 > 0 and 6*2^2*5^2 + 12*5^2*7^2 + 52*7^2*1^2 + 27*1^2*2^2 = 134^2.
a(195) = 1 since 195 = 3^2 + 7^2 + 4^2 + 11^2 with 11 > 0 and 6*3^2*7^2 + 12*7^2*4^2 + 52*4^2*11^2 + 27*11^2*3^2 = 377^2.
a(311) = 1 since 311 = 14^2 + 9^2 + 3^2 + 5^2 with 5 > 0 and 6*14^2*9^2 + 12*9^2*3^2 + 52*3^2*5^2 + 27*5^2*14^2 = 498^2.
a(319) = 1 since 319 = 6^2 + 3^2 + 7^2 + 15^2 with 15 > 0 and 6*6^2*3^2 + 12*3^2*7^2 + 52*7^2*15^2 + 27*15^2*6^2 = 894^2.
a(403) = 1 since 403 = 3^2 + 13^2 + 12^2 + 9^2 with 9 > 0 and 6*3^2*13^2 + 12*13^2*12^2 + 52*12^2*9^2 + 27*9^2*3^2 = 963^2.
a(559) = 1 since 559 = 5^2 + 23^2 + 2^2 + 1^2 with 1 > 0 and 6*5^2*23^2 + 12*23^2*2^2 + 52*2^2*1^2 + 27*1^2*5^2 = 325^2.
a(591) = 1 since 591 = 21^2 + 11^2 + 2^2 + 5^2 with 5 > 0 and 6*21^2*11^2 + 12*11^2*2^2 + 52*2^2*5^2 + 27*5^2*21^2 = 793^2.
a(683) = 1 since 683 = 0^2 + 11^2 + 21^2 + 11^2 with 11 > 0 and 6*0^2*11^2 + 12*11^2*21^2 + 52*21^2*11^2 + 27*11^2*0^2 = 1848^2.
a(719) = 1 since 719 = 10^2 + 3^2 + 21^2 + 13^2 with 13 > 0 and 6*10^2*3^2 + 12*3^2*21^2 + 52*21^2*13^2 + 27*13^2*10^2 = 2094^2.
a(1031) = 1 since 1031 = 26^2 + 15^2 + 9^2 + 7^2 with 7 > 0 and 6*26^2*15^2 + 12*15^2*9^2 + 52*9^2*7^2 + 27*7^2*26^2 = 1494^2.
a(1439) = 1 since 1439 = 13^2 + 27^2 + 10^2 + 21^2 with 21 > 0 and 6*13^2*27^2 + 12*27^2*10^2 + 52*10^2*21^2 + 27*21^2*13^2 = 2433^2.
a(1643) = 1 since 1643 = 36^2 + 17^2 + 3^2 + 7^2 with 7 > 0 and 6*36^2*17^2 + 12*17^2*3^2 + 52*3^2*7^2 + 27*7^2*36^2 = 2004^2.
a(2519) = 1 since 2519 = 27^2 + 7^2 + 30^2 + 29^2 with 29 > 0 and 6*27^2*7^2 + 12*7^2*30^2 + 52*30^2*29^2 + 27*29^2*27^2 = 7527^2.
a(6879) = 1 since 6879 = 38^2 + 53^2 + 49^2 + 15^2 with 15 > 0 and 6*38^2*53^2 + 12*53^2*49^2 + 52*49^2*15^2 + 27*15^2*38^2 = 11922^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[6*(n-x^2-y^2-z^2)*x^2+12*x^2*y^2+52*y^2*z^2+27*z^2*(n-x^2-y^2-z^2)], r=r+1], {x, 0, Sqrt[n-1]}, {y, 0, Sqrt[n-1-x^2]}, {z, 1, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 16 2016
STATUS
approved