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

Author

Zhi-Wei Sun, Aug 05 2025

Keywords

• nonn

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A000040, A386827.

Programs

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]

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