A220947 Least prime p such that sum_{k=0}^n F(k+1)*x^{n-k} is irreducible modulo p, where F(j) denotes the Fibonacci number A000045(j).

2, 3, 2, 11, 3, 2, 5, 3, 2, 11, 5, 41, 181, 31, 73, 89, 5, 7, 71, 11, 29, 5, 193, 41, 89, 61, 2, 43, 3, 31, 13, 191, 2, 61, 103, 97, 103, 47, 383, 367, 89, 17, 191, 1627, 193, 163, 5, 337, 349, 23, 149, 193, 199, 233, 173, 617, 593, 59, 113, 151

Offset: 1

Author

Zhi-Wei Sun, Apr 07 2013

Keywords

• nonn

Comments

Conjecture: a(n) <= n^2+12 for all n>0.

Such a phenomenon happens quite often. In fact, for many interesting integer sequences a(k) (k=1,2,3,...), each of the polynomials x^n + sum_{k=0}^n a(k)*x^{n-k} (n>0) is irreducible modulo some prime not exceeding a*n^2+b*n+c, where a, b, c are suitable nonnegative constants.

Examples

			a(2) = 3 since x^2+x+2 is irreducible modulo 3 but reducible modulo 2.
Note also that a(13) = 181 = 13^2+12.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..450

Crossrefs

Cf. A000040, A000045, A224416, A224417, A224418, A224480, A224210, A220072, A223934, A218465.

Programs

Mathematica

A[n_,x_]:=A[n,x]=Sum[Fibonacci[k+1]*x^(n-k),{k,0,n}]
Do[Do[If[IrreduciblePolynomialQ[A[n,x],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[n^2+12]}];
Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]