OFFSET
0,2
COMMENTS
Conjecture 1: a(n) > 0 for any nonnegative integer n.
Conjecture 2: Each n = 0,1,2,... can be written as f(u,v,x,y,z) with u,v,x,y,z integers, where f is any of the following polynomials: u^4 + (v*(v+1)/2)^2 + (x*(3x+1)/2)^2 + (y*(5y+1)/2)^2 + (z*(5z+3)/2)^2, u^4 + (v*(v+1)/2)^2 + (x*(3x+1)/2)^2 + (y*(5y+1)/2)^2 + (z*(3z+2))^2, (u*(u+1)/2)^2 + (v*(3v+1)/2)^2 + (x*(5x+1)/2)^2 + (y*(5y+3)/2)^2 + (z*(3z+2))^2, (u*(u+1)/2)^2 + (v*(3v+1)/2)^2 + (x*(5x+1)/2)^2 + (y*(5y+3)/2)^2 + (z*(4z+3))^2, (u*(u+1)/2)^2 + (v*(3v+1)/2)^2 + (x*(5x+1)/2)^2 + (y*(5y+3)/2)^2 + (z*(9z+7)/2)^2.
We have verified Conjectures 1 and 2 for n up to 2*10^6 and 10^6 respectively.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
EXAMPLE
a(8) = 1 with 8 = 0^4 + (0*(0+1)/2)^2 + (1*(3*1+1)/2)^2 + ((-1)*(5*(-1)+1)/2)^2 + (0*(9*0+1)/2)^2.
a(2953) = 1 with 2953 = 6^4 + (8*(8+1)/2)^2 + (0*(3*0+1)/2)^2 + (0*(5*0+1)/2)^2 + (2*(9*2+1)/2)^2.
a(8953) = 1 with 8953 = 2^4 + (7*(7+1)/2)^2 + (6*(3*6+1)/2)^2 + ((-1)*(5*(-1)+1)/2)^2 + ((-4)*(9*(-4)+1)/2)^2.
MATHEMATICA
t[x_]:=t[x]=(x(x+1)/2)^2; f[x_]:=f[x]=(x(5x+1)/2)^2; g[x_]:=g[x]=(x(9x+1)/2)^2; SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]; PQ[n_]:=PQ[n]=SQ[n]&&SQ[24*Sqrt[n]+1];
tab={}; Do[r=0; Do[If[PQ[n-k^4-t[x]-f[y]-g[z]], r=r+1], {k, 0, n^(1/4)}, {x, 0, (Sqrt[8*Sqrt[n-k^4]+1]-1)/2}, {y, -Floor[(Sqrt[40*Sqrt[n-k^4-t[x]]+1]+1)/10], (Sqrt[40*Sqrt[n-k^4-t[x]]+1]-1)/10}, {z, -Floor[(Sqrt[72*Sqrt[n-k^4-t[x]-f[y]]+1]+1)/18], (Sqrt[72*Sqrt[n-k^4-t[x]-f[y]]+1]-1)/18}]; tab=Append[tab, r], {n, 0, 100}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Mar 05 2019
STATUS
approved