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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).
3
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
1,1
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.
LINKS
EXAMPLE
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.
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}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 07 2013
STATUS
approved