

A282013


Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 49*x + 48*(yz) are squares.


9



1, 2, 3, 3, 2, 1, 3, 2, 1, 4, 4, 2, 2, 2, 1, 2, 2, 4, 8, 4, 3, 2, 3, 2, 3, 5, 5, 7, 3, 2, 5, 1, 3, 7, 6, 5, 5, 3, 5, 3, 2, 3, 9, 5, 2, 6, 3, 1, 3, 5, 5, 10, 6, 2, 8, 4, 3, 5, 6, 3, 3, 3, 4, 4, 2, 5, 9, 8, 5, 4, 6, 1, 5, 6, 5, 9, 2, 3, 7, 1, 1
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OFFSET

0,2


COMMENTS

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 5, 8, 14, 31, 47, 71, 79, 143, 248, 463, 1039).
The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares.
We have verified a(n) > 0 for all n = 0..10^7.
See also A281976, A281977 and A282014 for similar conjectures.
QingHu Hou at Tianjin University verified a(n) > 0 for n up to 10^9.  ZhiWei Sun, Jun 02 2019


LINKS

ZhiWei Sun, Table of n, a(n) for n = 0..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(5) = 1 since 5 = 1^2 + 0^2 + 0^2 + 2^2 with 1 = 1^2 and 49*1 + 48*(00) = 7^2.
a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^2 with 0 = 0^2 and 49*0 + 48*(22) = 0^2.
a(14) = 1 since 14 = 1^2 + 2^2 + 3^2 + 0^2 with 1 = 1^2 and 49*1 + 48*(23) = 1^2.
a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(12) = 1^2.
a(47) = 1 since 47 = 1^2 + 6^2 + 1^2 + 3^2 with 1 = 1^2 and 49*1 + 48*(61) = 17^2.
a(71) = 1 since 71 = 1^2 + 5^2 + 6^2 + 3^2 with 1 = 1^2 and 49*1 + 48*(56) = 1^2.
a(79) = 1 since 79 = 1^2 + 7^2 + 2^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(72) = 17^2.
a(143) = 1 since 143 = 1^2 + 5^2 + 6^2 + 9^2 with 1 = 1^2 and 49*1 + 48*(56) = 1^2.
a(248) = 1 since 248 = 4^2 + 6^2 + 0^2 + 14^2 with 4 = 2^2 and 49*4 + 48*(60) = 22^2.
a(463) = 1 since 463 = 9^2 + 6^2 + 15^2 + 11^2 with 9 = 3^2 and 49*9 + 48*(615) = 3^2.
a(1039) = 1 since 1039 = 1^2 + 22^2 + 23^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(2223) = 1^2.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[nx^4y^2z^2]&&SQ[49x^2+48(yz)], r=r+1], {x, 0, n^(1/4)}, {y, 0, Sqrt[nx^4]}, {z, 0, Sqrt[nx^4y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]


CROSSREFS

Cf. A000118, A000290, A270969, A271518, A281939, A281941, A281975, A281976, A281977, A282014.
Sequence in context: A132312 A090431 A107336 * A275299 A205100 A156613
Adjacent sequences: A282010 A282011 A282012 * A282014 A282015 A282016


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Feb 04 2017


STATUS

approved



