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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.

%I A386850 #9 Aug 12 2025 18:46:10
%S A386850 1,3,7,7,13,41,13,31,29,31,37,23,97,331,53,101,47,89,43,199,53,43,47,
%T A386850 107,83,61,149,37,353,127,113,199,173,107,67,401,349,101,347,47,79,89,
%U A386850 83,241,139,641,673,103,491,179,383,293,61,439,397,547,79,1301,379,277
%N 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.
%C A386850 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.
%C A386850 Note that Sum_{k>0}x^k/k! = e^x - 1.
%H A386850 Zhi-Wei Sun, <a href="/A386850/b386850.txt">Table of n, a(n) for n = 1..400</a>
%e A386850 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.
%t A386850 P[n_, x_]:=P[n, x]=Sum[x^(n-k)/k!, {k, 1, n}];
%t A386850 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]
%o A386850 (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
%Y A386850 Cf. A000040, A000142, A386827, A386828.
%K A386850 nonn
%O A386850 1,2
%A A386850 _Zhi-Wei Sun_, Aug 05 2025