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A389525
Number of ways to write 2*n + 1 as prime(i^2) + prime(j^2) + prime(k^2) + 2*prime(m^2) with i >= j >= k >= m >= 1.
2
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 2, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1
OFFSET
0,58
COMMENTS
Conjecture: a(n) > 0 for n > 3200.
This has been verified for n <= 250000.
It seems that 5930 is the largest value of n with a(n) = 1.
See also A389517 for a similar conjecture.
From Chai Wah Wu, Oct 15 2025: (Start)
Conjecture: for all k, there exists n_k such that a(m)>k for all m >= n_k.
a(n) > 0 for 3201 <= n <= 7.7*10^6.
k conjectured largest value of n for which a(n) = k
----------------
2 8360
3 10844
4 12371
5 11579
6 16229
7 18225
8 18686
9 20327
10 20838
(End)
EXAMPLE
a(20) = 1 with prime(3^2) + prime(2^2) + prime(2^2) + 2*prime(1^2) = 23 + 7 + 7 + 2*2 = 2*20 + 1.
a(5104) = 1 with prime(32^2) + prime(14^2) + prime(11^2) + 2*prime(5^2) = 8161 + 1193 + 661 + 2*97 = 2*5104 + 1.
a(5930) = 1 with prime(25^2) + prime(22^2) + prime(22^2) + 2*prime(6^2) = 4637 + 3461 + 3461 + 2*151 = 2*5930 + 1.
MATHEMATICA
p[n_]:=p[n]=Prime[n^2];
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PQ[n_]:=PQ[n]=PrimeQ[n]&&SQ[PrimePi[n]];
tab={}; Do[r=0; Do[If[PQ[2n+1-2p[m]-p[k]-p[j]], r=r+1],
{m, 1, Sqrt[PrimePi[(2n+1)/5]]}, {k, m, Sqrt[PrimePi[(2n+1-2p[m])/3]]}, {j, k, Sqrt[PrimePi[(2n+1-2p[m]-p[k])/2]]}];
tab=Append[tab, r], {n, 0, 100}]; Print[tab]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 07 2025
STATUS
approved