OFFSET
0,4
COMMENTS
Conjecture: a(n) > 0 for any nonnegative integer n.
I'd like to call this the 1-2-3-4-5 conjecture. I have verified it for all n = 0..4*10^5.
It seems that a(n) = 1 only for n = 0,1,2.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
EXAMPLE
Let f(x) = x*(3x-1)/2. Then
a(2) = 1 with 2 = f(0)^2 + 2*f(1)^2 + 3*f(0)^2 + 4*f(0)^2 + 5*f(0)^2,
a(415) = 2 with 415 = f(-1)^2 + 2*f(3)^2 + 3*f(1)^2 + 4*f(2)^2 + 5*f(-1)^2 = f(3)^2 + 2*f(0)^2 + 3*f(2)^2 + 4*f(-2)^2 + 5*f(0)^2,
a(427) = 2 with 427 = f(-3)^2 + 2*f(2)^2 + 3*f(-2)^2 + 4*f(0)^2 + 5*f(1)^2 = f(-3)^2 + 2*f(1)^2 + 3*f(2)^2 + 4*f(0)^2 + 5*f(2)^2.
MATHEMATICA
f[x_]:=f[x]=(x(3x-1)/2)^2;
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PQ[n_]:=PQ[n]=SQ[n]&&SQ[24Sqrt[n]+1];
tab={}; Do[r=0; Do[If[PQ[n-5f[x]-4f[y]-3f[z]-2f[w]], r=r+1], {x, -Floor[(Sqrt[24Sqrt[n/5]+1]-1)/6], (Sqrt[24Sqrt[n/5]+1]+1)/6}, {y, -Floor[(Sqrt[24Sqrt[(n-5f[x])/4]+1]-1)/6], (Sqrt[24Sqrt[(n-5f[x])/4]+1]+1)/6}, {z, -Floor[(Sqrt[24Sqrt[(n-5f[x]-4f[y])/3]+1]-1)/6], (Sqrt[24Sqrt[(n-5f[x]-4f[y])/3]+1]+1)/6}, {w, -Floor[(Sqrt[24Sqrt[(n-5f[x]-4f[y]-3f[z])/2]+1]-1)/6], (Sqrt[24Sqrt[(n-5f[x]-4f[y]-3f[z])/2]+1]+1)/6}]; tab=Append[tab, r], {n, 0, 100}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 26 2019
STATUS
approved