A390795 Number of ways to write n as ((p-1)/2)^4 + x^2 + 2*y^2 + 3*z^2, where p is 1 or an odd prime, and x,y,z are in the set {m >= 0: 2*m is practical}.

1, 2, 2, 3, 4, 3, 3, 3, 2, 4, 3, 2, 6, 5, 3, 4, 5, 4, 5, 7, 6, 7, 8, 4, 4, 6, 1, 5, 9, 5, 8, 8, 5, 4, 7, 6, 10, 11, 5, 7, 6, 3, 2, 4, 7, 9, 9, 5, 6, 8, 6, 7, 10, 3, 6, 7, 2, 7, 3, 3, 9, 7, 3, 3, 8, 6, 8, 9, 5, 4, 5, 3, 7, 7, 1, 6, 10, 4, 1, 4, 5, 9, 11, 6, 11, 11, 6, 6, 6, 6, 5, 6, 9, 9, 6, 6, 8, 8, 3, 8, 15

Offset

0,2

Comments

Conjecture: a(n) > 0 for every natural number n. This has been verified for n <= 2*10^5.

It seems that a(n) = 1 only for n = 0, 26, 74, 123, 251, 2714, 3707, 4047.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Example

a(74) = 1 since 74 = ((3-1)/2)^4 + 1^2 + 2*6^2 + 3*0^2 with 3 prime, and 2*1 and 2*6 both practical.
a(251) = 1 since 251 = ((1-1)/2)^4 + 0^2 + 2*2^2 + 3*9^2 with 2*2 and 2*9 both practical.
a(3707) = 1 since 3707 = ((13-1)/2)^4 + 24^2 + 2*16^2 + 3*21^2 with 13 prime, and 2*24, 2*16, 2*21 all practical.
a(4047) = 1 since 4047 = ((13-1)/2)^4 + 50^2 + 2*2^2 + 3*9^2 with 13 prime, and 2*50, 2*2, 2*9 all 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]&&(n==0||pr[2n]);
pq[n_]:=pq[n]=(n==0||PrimeQ[2n+1]);
tab={}; Do[m=0; Do[If[pq[x]&&qq[y]&&qq[z]&&qq[Sqrt[n-x^4-2y^2-3z^2]], m=m+1], {x, 0, n^(1/4)},
{y, 0, Sqrt[(n-x^4)/2]}, {z, 0, Sqrt[(n-x^4-2y^2)/3]}]; tab=Append[tab, m], {n, 0, 100}]; Print[tab]

Crossrefs

Cf. A000040, A000290, A000583, A005097, A005153, A389523, A390762, A390793, A390799.

Sequence in context: A262824 A242899 A347865 * A391037 A306606 A229989

Adjacent sequences: A390792 A390793 A390794 * A390796 A390797 A390798

Keyword

nonn

Author

Zhi-Wei Sun, Nov 19 2025

Status

approved