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A224210
Least prime p such that sum_{k=0}^n (k+1)^2*x^{n-k} is irreducible modulo p.
8
2, 11, 7, 17, 11, 3, 7, 97, 3, 89, 31, 113, 43, 7, 23, 23, 17, 67, 23, 109, 17, 277, 103, 283, 59, 101, 157, 127, 29, 79, 23, 223, 73, 269, 433, 137, 5, 659, 109, 401, 419, 7, 373, 131, 89, 269, 149, 61, 829, 881
OFFSET
1,1
COMMENTS
Conjecture: a(n) does not exceed the (4n-3)-th prime for each n>0. Moreover, for any integers m>1 and n>0 the polynomial sum_{k=0}^n (k+1)^m*x^{n-k} is irreducible modulo some prime, and its Galois group over the rationals is isomorphic to the symmetric group S_n. Also, for m,n=2,3,... there are infinitely many integers b > n^m such that [n^m,...,2^m,1^m] in base b is prime.
We have a similar conjecture with the above (k+1)^m replaced by (2k+1)^m.
LINKS
EXAMPLE
a(3) = 7 since f(x) = x^3+4x^2+9x+16 is irreducible modulo 7 but reducible modulo any of 2, 3, 5. Note that
f(x)==x*(x-1)^2 (mod 2), f(x)==(x-1)*(x^2-x-1) (mod 3)
and
f(x)==(x+1)*(x-1)^2 (mod 5).
MATHEMATICA
A[n_, x_]:=Sum[(k+1)^2*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, Prime[4n-3]}];
Print[n, " ", counterexample]; Label[aa]; Continue, {n, 1, 100}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Apr 01 2013
STATUS
approved