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

1,2

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.

Links

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

Example

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.

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]

Prog

(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

Crossrefs

Cf. A000040, A000142, A386827, A386828.

Sequence in context: A290649 A118259 A060845 * A393188 A059478 A175329

Adjacent sequences: A386847 A386848 A386849 * A386851 A386852 A386853

Keyword

nonn

Author

Zhi-Wei Sun, Aug 05 2025

Status

approved