A386828 Least prime n < p <= 2*n^2 + 1 such that the polynomial Sum_{k=1..n} x^(n-k)/k^2 is irreducible modulo p, or 1 if such a prime p does not exist.

1, 3, 19, 13, 7, 17, 19, 13, 17, 13, 17, 31, 139, 151, 19, 181, 113, 157, 79, 89, 89, 71, 37, 31, 197, 31, 199, 149, 83, 37, 127, 59, 647, 89, 47, 47, 157, 197, 97, 79, 601, 59, 79, 67, 71, 487, 223, 577, 359, 83, 269, 269, 251, 461, 229, 67, 1777, 859, 1091, 701

Offset

1,2

Comments

Conjecture: a(n) > 1 for all n > 1.

We also have similar conjectures for Sum_{k=1..n} x^(n-k)/k^s with other values of s.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..300

Example

a(3) = 19 since 19 = 2*3^2 + 1 is the least prime p > 3 such that the polynomial x^2 + x/2 + 1/3 is irreducible modulo p.

Mathematica

P[n_, x_]:=P[n, x]=Sum[x^(n-k)/k^2, {k, 1, n}];
tab={}; Do[Do[If[IrreduciblePolynomialQ[P[n, x], Modulus->Prime[k]]==True, tab=Append[tab, Prime[k]]; Goto[aa]], {k, PrimePi[n]+1, PrimePi[2n^2+1]}]; tab=Append[tab, 1]; Label[aa]; Continue, {n, 1, 60}]; Print[tab]

Prog

(PARI) a(n) = forprime(p=n+1, 2*n^2+1, if (polisirreducible(Mod(sum(k=1, n, x^(n-k)/k^2), p)), return(p))); 1; \\ Michel Marcus, Aug 05 2025

Crossrefs

Cf. A000040, A386827.

Sequence in context: A358979 A178985 A357435 * A266704 A185446 A172032

Adjacent sequences: A386825 A386826 A386827 * A386829 A386830 A386831

Keyword

nonn

Author

Zhi-Wei Sun, Aug 05 2025

Status

approved