

A290491


Number of ways to write 12*n+1 as x^2 + 4*y^2 + 8*z^4, where x and y are positive integers and z is a nonnegative integer.


4



2, 2, 2, 1, 2, 3, 2, 2, 2, 1, 3, 4, 3, 2, 3, 5, 3, 1, 4, 3, 3, 4, 3, 2, 2, 5, 5, 1, 3, 2, 4, 3, 3, 3, 2, 6, 3, 2, 1, 3, 6, 4, 2, 2, 3, 5, 3, 3, 1, 2, 4, 3, 4, 2, 5, 4, 5, 5, 3, 2, 7, 4, 4, 3, 2, 6, 3, 4, 4, 3, 9, 3, 2, 3, 3, 6, 5, 4, 4, 3, 8, 5, 2, 2, 3, 6, 3, 3, 4, 4, 5, 7, 2, 3, 3, 8, 6, 1, 5, 4
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OFFSET

1,1


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4, 10, 18, 28, 39, 49, 98, 142, 163, 184, 208, 320, 382, 408, 814, 910, 1414, 2139, 2674, 3188, 3213, 4230, 6279, 25482.
(ii) All the numbers 16*n+5 (n = 0,1,2,...) can be written as x^4 + 4*y^2 + z^2, where x,y,z are integers with y > 0 and z > 0.
(iii) All the numbers 24*n+1 (n = 0,1,2,...) can be written as 12*x^4 + 4*y^2 + z^2 with x,y,z integers. Also, all the numbers 24*n+9 (n = 0,1,2,...) can be written as 2*x^4 + 6*y^2 + z^2 with x,y,z positive integers.
(iv) All the numbers 24*n+2 (n = 0,1,2,...) can be written as x^4 + 9*y^2 + z^2, where x,y,z are integers with z > 0. Also, all the numbers 24*n+17 (n = 0,1,2,...) can be written as x^4 + 16*y^2 + z^2, where x,y,z are integers with y > 0 and z > 0.
(v) All the numbers 30*n+3 (n = 1,2,3,...) can be written as 2*x^4 + 3*y^2 + z^2 with x,y,z positive integers. Also, all the numbers 30*n+21 (n = 0,1,2,...) can be written as 2*x^4 + 3*y^2 + z^2, where x,y,z are integers with z > 0.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..10000
ZhiWei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), 13671396.
ZhiWei Sun, On universal sums x(ax+b)/2+y(cy+d)/2+z(ez+f)/2, arXiv:1502.03056 [math.NT], 20152017.
ZhiWei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97120.


EXAMPLE

a(10) = 1 since 12*10+1 = 7^2 + 4*4^2 + 8*1^4.
a(28) = 1 since 12*28+1 = 9^2 + 4*8^2 + 8*0^4.
a(49) = 1 since 12*49+1 = 19^2 + 4*5^2 + 8*2^4.
a(3188) = 1 since 12*3188+1 = 103^2 + 4*80^2 + 8*4^4.
a(3213) = 1 since 12*3213+1 = 91^2 + 4*87^2 + 8*0^4.
a(4230) = 1 since 12*4230+1 = 223^2 + 4*16^2 + 8*1^4.
a(6279) = 1 since 12*6279+1 = 19^2 + 4*75^2 + 8*9^4.
a(25482) = 1 since 12*25482+1 = 531^2 + 4*58^2 + 8*6^4.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[12n+18x^44y^2], r=r+1], {x, 0, ((12n+1)/8)^(1/4)}, {y, 1, Sqrt[(12n+18x^4)/4]}]; Print[n, " ", r], {n, 1, 100}]


CROSSREFS

Cf. A000290, A000583, A270566, A286885, A286944, A287616, A290342, A290472.
Sequence in context: A185636 A333212 A182597 * A194314 A006371 A000177
Adjacent sequences: A290488 A290489 A290490 * A290492 A290493 A290494


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Aug 03 2017


STATUS

approved



