

A282161


Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x and (12*x)^2 + (5*y10*z)^2 both squares, where x,y,z are nonnegative integers and w is a positive integer.


1



1, 3, 2, 2, 5, 4, 2, 2, 4, 6, 4, 3, 4, 6, 3, 1, 9, 7, 5, 6, 7, 7, 1, 4, 8, 11, 7, 1, 11, 10, 2, 3, 8, 9, 6, 9, 8, 11, 5, 5, 15, 7, 4, 5, 13, 9, 2, 2, 8, 15, 10, 8, 10, 17, 3, 7, 12, 4, 10, 4, 11, 16, 3, 2, 18, 16, 6, 9, 15, 11, 4, 6, 8, 16, 12, 3, 13, 13, 1, 5
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OFFSET

1,2


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m (k = 0,1,2,... and m = 1, 23, 28, 79, 119, 191, 223, 263, 463, 703, 860, 1052).
(ii) Any positive integer n can be written as x^2 + y^2 + z^2 + w^2 with x and (35*x)^2 + (12*y24*z)^2 both squares, where x,y,z are nonnegative integers and w is a positive integer.
The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares.
See also A281976, A281977, A282013 and A282014 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], 2017.


EXAMPLE

a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 = 0^2 and (12*0)^2 + (5*010*0)^2 = 0^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with 1 = 1^2 and (12*1)^2 + (5*310*2)^2 = 13^2.
a(28) = 1 since 28 = 1^2 + 1^2 + 1^2 + 5^2 with 1 = 1^2 and (12*1)^2 + (5*110*1)^2 = 13^2.
a(79) = 1 since 79 = 1^2 + 5^2 + 2^2 + 7^2 with 1 = 1^2 and (12*1)^2 + (5*510*2)^2 = 13^2.
a(119) = 1 since 119 = 1^2 + 9^2 + 1^2 + 6^2 with 1 = 1^2 and (12*1)^2 + (5*910*1)^2 = 37^2.
a(191) = 1 since 191 = 9^2 + 5^2 + 7^2 + 6^2 with 9 = 3^2 and (12*9)^2 + (5*510*7)^2 = 117^2.
a(223) = 1 since 223 = 1^2 + 13^2 + 7^2 + 2^2 with 1 = 1^2 and (12*1)^2 + (5*1310*7)^2 = 13^2.
a(263) = 1 since 263 = 9^2 + 13^2 + 2^2 + 3^2 with 9 = 3^2 and (12*9)^2 + (5*1310*2)^2 = 117^2.
a(463) = 1 since 463 = 1^2 + 19^2 + 10^2 + 1^2 with 1 = 1^2 and (12*1)^2 + (5*1910*10)^2 = 13^2.
a(703) = 1 since 703 = 1^2 + 13^2 + 7^2 + 22^2 with 1 = 1^2 and (12*1)^2 + (5*1310*7)^2 = 13^2.
a(860) = 1 since 860 = 4^2 + 18^2 + 18^2 + 14^2 with 4 = 2^2 and (12*4)^2 + (5*1810*18)^2 = 102^2.
a(1052) = 1 since 1052 = 4^2 + 30^2 + 6^2 + 10^2 with 4 = 2^2 and (12*4)^2 + (5*3010*6)^2 = 102^2.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[nx^4y^2z^2]&&SQ[144x^4+(5y10z)^2], r=r+1], {x, 0, (n1)^(1/4)}, {y, 0, Sqrt[n1x^4]}, {z, 0, Sqrt[n1x^4y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]


CROSSREFS

Cf. A000118, A000290, A270969, A271714, A273108, A281939, A281941, A281975, A281976, A281977, A282013, A282014.
Sequence in context: A076118 A309045 A210956 * A205675 A342331 A058608
Adjacent sequences: A282158 A282159 A282160 * A282162 A282163 A282164


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Feb 07 2017


STATUS

approved



