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URL: https://oeis.org/A300396

⇱ A300396 - OEIS

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A300396
Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that 2*x or y is a power of 4 (including 4^0 = 1) and x + 63*y = 2^(2k+1) for some k = 0,1,2,....
29
0, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 3, 1, 3, 4, 2, 2, 4, 1, 2, 1, 1, 2, 4, 3, 1, 1, 2, 1, 6, 2, 2, 2, 5, 1, 4, 1, 2, 6, 3, 3, 3, 1, 2, 3, 4, 3, 3, 2, 4, 2, 2, 1, 7, 3, 1, 4, 1, 2, 8, 1, 3, 7, 3, 4, 6, 3, 4, 4, 6, 4, 3, 2, 4, 3, 1, 2
OFFSET
1,3
COMMENTS
Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 5, 13, 25, 29, 59, 61, 79, 91, 95, 101, 103, 1315, 2^k (k = 1,2,3,...), 2^(2k+1)*m (k = 0,1,2,... and m = 3, 5, 7, 11, 15, 19, 23, 887).
This is stronger than the conjecture that A300360(n) > 0 for all n > 1. Note that a(387) = 3 < A300360(387) = 4 and a(1774) = 1 < A300360(1774) = 2.
We have verified that a(n) > 0 for all n = 2..10^7.
See also A299537, A299794 and A300219 for similar conjectures.
LINKS
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], 2017-2018.
EXAMPLE
a(29) = 1 since 29^2 = 2^2 + 2^2 + 7^2 + 28^2 with 2*2 = 4^1 and 2 + 63*2 = 2^7.
a(86) = 2 since 65^2 + 1^2 + 19^2 + 53^2 = 65^2 + 1^2 + 31^2 + 47^2 with 1 = 4^0 and 65 + 63*1 = 2^7.
a(1774) = 1 since 1774^2 = 8^2 + 520^2 + 14^2 + 1696^2 with 2*8 = 4^2 and 8 + 63*520 = 2^15.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Pow[n_]:=Pow[n]=IntegerQ[Log[4, n]];
tab={}; Do[r=0; Do[If[Pow[y]||Pow[(2*4^k-63y)/2], Do[If[SQ[n^2-y^2-(2*4^k-63y)^2-z^2], r=r+1], {z, 0, Sqrt[Max[0, (n^2-y^2-(2*4^k-63y)^2)/2]]}]], {k, 0, Log[4, Sqrt[63^2+1]*n/2]}, {y, 0, Min[n, 2*4^k/63]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 05 2018
STATUS
approved