A224416 Least prime p such that the polynomial sum_{k=0}^n C_k*x^{n-k} is irreducible modulo p, where C_k denotes the Catalan number binomial(2k,k)/(k+1).

2, 3, 2, 3, 17, 7, 47, 3, 53, 5, 137, 109, 79, 11, 37, 7, 59, 13, 53, 251, 251, 101, 467, 149, 79, 3, 83, 61, 239, 31, 79, 73, 73, 373, 199, 5, 337, 167, 17, 683, 523, 269, 37, 163, 431, 163, 163, 7, 487, 7, 167, 163, 197, 1549, 137, 503, 139, 263, 151, 283

Offset

1,1

Comments

Conjecture: (i) a(n) does not exceed n^2+n+5 for each n>0, and the Galois group of sum_{k=0}^n C_k*x^{n-k} over the rationals is isomorphic to the symmetric group S_n.

(ii) For any positive integer n, the polynomial sum_{k=0}^n binomial(2k,k)*x^{n-k} is irreducible modulo some prime if and only if n is not of the form 2k(k+1), where k is a positive integer.

(iii) For any positive integer n, the polynomial sum_{k=0}^n T_k*x^{n-k} is irreducible modulo some prime not exceeding n^2+n+5, where T_k referes to the central trinomial coefficient A002426(k) which is the coefficient of x^k in the expansion of (x^2+x+1)^k.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..350

Example

a(10) = 5 since sum_{k=0}^{10} C_k*x^{n-k} irreducible modulo 5 but reducible modulo any of 2 and 3.
Note also that a(11) = 137 coincides with 11^2+11+5.

Mathematica

A[n_, x_]:=A[n, x]=Sum[Binomial[2k, k]/(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+n+5]}];
Print[n, " ", counterexample]; Label[aa]; Continue, {n, 1, 100}]

Crossrefs

Cf. A000040, A000108, A224480, A224417, A224418, A220072, A223934, A224210, A217785, A217788, A224197, A002426.

Sequence in context: A077989 A109620 A138781 * A282049 A038700 A177799

Adjacent sequences: A224413 A224414 A224415 * A224417 A224418 A224419

Keyword

nonn

Author

Zhi-Wei Sun, Apr 06 2013

Status

approved