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A300441
Number of the integers 4^k*(4*u(m)^2+1) (k,m = 0,1,2,...) such that n^2 - 4^k*(4*u(m)^2+1) can be written as the sum of two squares, where u(0) = 0, u(1) = 1, and u(j+1) = 4*u(j) - u(j-1) for j = 1,2,3,....
17
1, 1, 2, 1, 3, 2, 1, 1, 4, 3, 3, 2, 3, 1, 3, 1, 3, 4, 4, 3, 5, 3, 3, 2, 4, 3, 3, 1, 5, 3, 3, 1, 6, 3, 4, 4, 5, 4, 4, 3, 6, 5, 4, 3, 5, 3, 4, 2, 5, 4, 5, 3, 4, 3, 5, 1, 5, 5, 3, 3, 3, 3, 5, 1, 5, 6, 3, 3, 6, 4, 4, 4, 6, 5, 5, 4, 6, 4, 5, 3
OFFSET
1,3
COMMENTS
Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k*m (k = 0,1,2,... and m = 1, 7).
This curious conjecture indicates that any positive square can be written as (2^k)^2 + (2^(k+1)*u(m))^2 + x^2 + y^2 with k,m,x,y nonnegative integers. In the 2017 JNT paper, the author proved that each n = 1,2,3,... can be written as 4^k*(1+4*x^2+y^2)+z^2 with k,x,y,z nonnegative integers.
We have verified that a(n) > 0 for all n = 1..10^7.
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(1) = 1 since 1^2 - 4^0*(4*u(0)^2+1) = 1 is 1^2 + 0^2.
a(5) = 3 since 5^2 - 4^0*(4*u(1)^2+1) = 20 = 4^2 + 2^2, 5^2 - 4^1*(4*u(1)^2+1) = 5 = 2^2 + 1^2, and 5^2 - 4^2*(4*u(0)^2+1) = 9 = 3^2 + 0^2.
a(7) = 1 since 7^2 - 4^1*(4*u(0)^2+1) = 45 = 6^2 + 3^2.
MATHEMATICA
u[0]=0;
u[1]=1;
u[n_]:=u[n]=4u[n-1]-u[n-2];
f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[If[Mod[Part[Part[f[n], i], 1]-3, 4]==0&&Mod[Part[Part[f[n], i], 2], 2]==1, 1, 0], {i, 1, Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
tab={}; Do[r=0; Do[m=0; Label[cc]; If[4u[m]^2+1>n^2/4^k, Goto[bb]]; If[QQ[n^2-4^k*(4u[m]^2+1)], r=r+1, m=m+1; Goto[cc]];
Label[bb], {k, 0, Log[2, n]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 05 2018
STATUS
approved