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
Keywords
Examples
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.
Links
- Zhi-Wei Sun, Table of n, a(n) for n = 1..350
Crossrefs
Programs
-
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}]
Comments