A391037 Number of ways to write n as ((p-1)/2)^4 + (q-1)^2/2 + (r-1)^2/4 + (s+1)^2/4, where each of p,q,r,s is either 1 or an odd prime.

1, 2, 2, 3, 4, 3, 3, 3, 3, 6, 5, 4, 7, 4, 2, 4, 5, 7, 7, 7, 7, 5, 6, 4, 5, 9, 6, 9, 9, 1, 5, 5, 3, 8, 6, 7, 8, 7, 6, 5, 5, 7, 7, 8, 7, 4, 6, 5, 5, 6, 4, 9, 10, 8, 10, 7, 2, 9, 9, 8, 10, 2, 8, 7, 3, 4, 7, 10, 7, 7, 9, 2, 5, 11, 7, 11, 9, 2, 7, 5, 7, 12, 12, 9, 11, 12, 9, 10, 8, 15, 16, 12, 12, 7, 4, 7, 7, 14, 13, 13

Offset

1,2

Comments

Conjecture: For any c = 1, 5, 7, every positive integer n can be written as c*((p-1)/2)^4 + (q-1)^2/2 + (r-1)^2/4 + (s+1)^2/4, where each of p,q,r,s is either 1 or an odd prime.

This has been verified for n <= 5*10^5 in the case c = 1, and for n <= 3*10^5 when c is 5 or 7.

See also A391038 for a smilar conjecture.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Example

a(1) = 1 since 1 = ((1-1)/2)^4 + (1-1)^2/2 + (1-1)^2/4 + (1+1)^2/4.
a(30) = 1 since 30 = ((3-1)/2)^4 + (1-1)^2/2 + (11-1)^2/4 + (3+1)^2/4 with 3 and 11 prime.

Mathematica

pp[n_]:=pp[n]=IntegerQ[n]&&(n==1||PrimeQ[2n-1]); pq[n_]:=pq[n]=(n==0||PrimeQ[2n+1]);
tab={}; Do[m=0; Do[If[pq[x]&&pq[y]&&pq[z]&&pp[Sqrt[n-x^4-2y^2-z^2]], m=m+1], {x, 0, (n-1)^(1/4)},
{y, 0, Sqrt[(n-1-x^4)/2]}, {z, 0, Sqrt[n-1-x^4-2y^2]}]; tab=Append[tab, m], {n, 1, 100}]; Print[tab]

Crossrefs

Cf. A000040, A000290, A000583, A005097, A006254, A390762, A390793, A390795, A390799, A390818, A391038.

Sequence in context: A242899 A347865 A390795 * A306606 A229989 A126688

Adjacent sequences: A391034 A391035 A391036 * A391038 A391039 A391040

Keyword

nonn

Author

Zhi-Wei Sun, Nov 26 2025

Status

approved