OFFSET
1,4
COMMENTS
It seems that a(n) = 1 only for n = 1, 2, 3, 14, 32, 33, 35, 110, 222, 689, 1009.
Conjecture 1: a(n) > 0 for all n > 0.
We also have the following similar conjecture.
Conjecture 2: Each natural number n can be written as 3*u^4 + v^2 + (p-1)^2/2 + (q-1)^2/4, where each of u and v is either zero or half of an even practical number, and each of p and q is either 1 or an odd prime.
Both conjectures have been verified for n <= 2*10^5.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
EXAMPLE
a(14) = 1 since 14 = 3*((3-1)/2)^4 + (3-1)^2/2 + (1-1)^2/4 + 3^2 with 3 prime and 2*3 practical.
a(33) = 1 since 33 = 3*((1-1)/2)^4 + (5-1)^2/2 + (7-1)^2/4 + 4^2 with 5 and 7 both prime, and 2*4 practical.
a(110) = 1 since 110 = 3*((3-1)/2)^4 + (7-1)^2/2 + (11-1)^2/4 + 8^2 with 3,7,11 all prime and 2*8 practical.
a(1009) = 1 since 1009 = 3*((1-1)/2)^4 + (1-1)^2/2 + (31-1)^2/4 + 28^2 with 31 prime and 2*28 practical.
MATHEMATICA
f[n_]:=f[n]=FactorInteger[n];
Pow[n_, i_]:=Pow[n, i]=Part[Part[f[n], i], 1]^(Part[Part[f[n], i], 2]);
Con[n_]:=Con[n]=Sum[If[Part[Part[f[n], s+1], 1]<=DivisorSigma[1, Product[Pow[n, i], {i, 1, s}]]+1, 0, 1], {s, 1, Length[f[n]]-1}];
pr[n_]:=pr[n]=n>0&&(n<3||Mod[n, 2]+Con[n]==0);
qq[n_]:=qq[n]=IntegerQ[n]&&pr[2n];
pq[n_]:=pq[n]=(n==0||PrimeQ[2n+1]);
tab={}; Do[m=0; Do[If[pq[x]&&pq[y]&&pq[z]&&qq[Sqrt[n-3x^4-2y^2-z^2]], m=m+1], {x, 0, ((n-1)/3)^(1/4)},
{y, 0, Sqrt[(n-1-3x^4)/2]}, {z, 0, Sqrt[n-1-3x^4-2y^2]}]; tab=Append[tab, m], {n, 1, 100}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Nov 20 2025
STATUS
approved
