login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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