OFFSET
1,2
COMMENTS
Conjecture: a(n) > 1 for all n > 1. In other words, for each n = 2,3,... the polynomial x^(n-1) + tau(2)*x^(n-2) + ... + tau(n) is irreducible modulo some prime p < 2n*(n+1).
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..500
EXAMPLE
a(5) = 59 since the prime 59 is smaller than 2*5*(5+1) = 60, and 59 is the least prime p such that the polynomial tau(1)*x^4 + tau(2)*x^3 + tau(3)*x^2 + tau(4)*x + tau(5) is irreducible modulo p.
MATHEMATICA
Tau[n_]:=Tau[n]=RamanujanTau[n];
P[n_, x_]:=P[n, x]=Sum[Tau[k]x^(n-k), {k, 1, n}];
tab={}; Do[Do[If[IrreduciblePolynomialQ[P[n, x], Modulus->Prime[k]]==True, tab=Append[tab, Prime[k]]; Goto[aa]], {k, 1, PrimePi[2n(n+1)-1]}]; tab=Append[tab, 1]; Label[aa]; Continue, {n, 1, 63}]; Print[tab]
PROG
(PARI) a(n) = forprime(p=2, 2*n*(n+1)-1, if (polisirreducible(Mod(sum(k=1, n, ramanujantau(k)*x^(n-k)), p)), return(p))); 1; \\ Michel Marcus, Aug 04 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 03 2025
STATUS
approved
