A386850 Least prime n < p <= (n-1)*(2n-1) such that Sum_{k=1..n} x^(n-k)/k! is irreducible modulo p, or 1 if such a prime p does not exist.

1, 3, 7, 7, 13, 41, 13, 31, 29, 31, 37, 23, 97, 331, 53, 101, 47, 89, 43, 199, 53, 43, 47, 107, 83, 61, 149, 37, 353, 127, 113, 199, 173, 107, 67, 401, 349, 101, 347, 47, 79, 89, 83, 241, 139, 641, 673, 103, 491, 179, 383, 293, 61, 439, 397, 547, 79, 1301, 379, 277

Offset: 1

Author

Zhi-Wei Sun, Aug 05 2025

Keywords

• nonn

Comments

Conjecture: a(n) > 1 for all n > 1. In other words, for any integer n > 1, there is a prime p with n < p <= (n-1)*(2n-1) such that the polynomial Sum_{k=1..n}x^(n-k)/k! is irreducible modulo p.

Note that Sum_{k>0}x^k/k! = e^x - 1.

Examples

			a(2) = 3 since 3 is the only prime in the interval (2, (2-1)*(2*2-1)] and x + 1/2 is irreducible modulo 3.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..400

Crossrefs

Cf. A000040, A000142, A386827, A386828.

Programs

Mathematica

P[n_, x_]:=P[n, x]=Sum[x^(n-k)/k!, {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[(n-1)(2n-1)]}];tab=Append[tab,1]; Label[aa]; Continue, {n, 1, 60}];Print[tab]

PARI

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