cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 06 2013

Keywords

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.

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.
		

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}]

A224418 Least prime q such that sum_{k=0}^n p(k)*x^{n-k} is irreducible modulo q, where p(k) refers to the partition number A000041(k).

Original entry on oeis.org

2, 3, 2, 11, 2, 13, 19, 19, 13, 29, 73, 47, 19, 43, 7, 59, 13, 29, 3, 13, 179, 29, 173, 19, 3, 163, 23, 3, 101, 71, 131, 977, 5, 157, 43, 13, 73, 2, 89, 197, 151, 151, 313, 3, 13, 31, 23, 97, 173, 241, 181, 109, 487, 157, 17, 29, 89, 109, 257, 317
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 06 2013

Keywords

Comments

Conjecture: a(n) < n^2 for all n > 1.

Examples

			a(2) = 3 since sum_{k=0}^2 p(k)*x^{n-k} = x^2 + x + 2  is irreducible modulo 3 but reducible modulo 2.
		

Crossrefs

Programs

  • Mathematica
    A[n_,x_]:=A[n,x]=Sum[PartitionsP[k]*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[Max[1,n^2-1]]}];
    Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]

A224480 Least prime q such that x^n + sum_{k=1}^n p_k*x^{n-k} is irreducible modulo q, where p_k denotes the k-th prime.

Original entry on oeis.org

2, 11, 2, 2, 2, 2, 2, 53, 13, 3, 5, 2, 2, 2, 2, 421, 29, 19, 7, 2, 29, 37, 2, 743, 41, 23, 13, 47, 5, 2, 269, 139, 211, 31, 73, 307, 2, 2, 5, 89, 23, 839, 181, 379, 173, 89, 2, 353, 101, 307, 3, 29, 389, 2, 863, 71, 503, 619, 193, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 07 2013

Keywords

Comments

Conjecture: a(n) <= (n+4)*(n+5)+1 for all n>0.

Examples

			a(10) = 3 since P(x) = x^{10} + 2*x^9 + 3*x^8 + 5*x^7 + 7*x^6
+ 11*x^5 + 13*x^4 + 17*x^3 + 19*x^2 + 23*x + 29 is irreducible modulo 3, but reducible modulo 2, for,
    P(x)==(x+1)^2*(x^3+x+1)*(x^5+x^3+1) (mod 2).
Note also that a(16) = 421 = (16+4)*(16+5)+1.
		

Crossrefs

Programs

  • Mathematica
    A[n_,x_]:=A[n,x]=Sum[x^n+Prime[k]*x^(n-k),{k,1,n}]
    Do[Do[If[IrreduciblePolynomialQ[A[n,x],Modulus->Prime[k]]==True,Print[n," ",Prime[k]];Goto[aa]],{k,1,PrimePi[n^2+9n+21]}];
    Print[n," ",counterexample];Label[aa];Continue,{n,1,100}]

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).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 07 2013

Keywords

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.

Examples

			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.
		

Crossrefs

Programs

  • 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}]

A220949 Least prime p such that sum_{k=0}^n (2k+1)*x^(n-k) is irreducible modulo p.

Original entry on oeis.org

2, 2, 3, 2, 5, 3, 71, 23, 11, 2, 5, 2, 13, 23, 47, 47, 269, 2, 7, 19, 53, 101, 7, 53, 113, 11, 23, 2, 43, 347, 53, 283, 191, 17, 41, 2, 239, 677, 3, 281, 37, 641, 613, 41, 17, 269, 181, 137, 383, 41, 127, 2, 71, 739, 71, 353, 59, 2, 83, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 07 2013

Keywords

Comments

Conjecture: a(n) <= n^2+22 for all n>0.
We have similar conjectures with 2k+1 in the definition replaced by (2k+1)^m (m=2,3,...).

Examples

			a(5) = 5 since f(x) = x^5+3*x^4+5*x^3+7*x^2+9*x+11 is irreducible modulo 5, but f(x)==(x+1)*(x^2+x+1)^2 (mod 2) and f(x)==(x+1)^4*(x-1) (mod 3).
Note also that a(7) = 71 = 7^2+22.
		

Crossrefs

Programs

  • Mathematica
    A[n_,x_] := A[n,x] = Sum[(2k+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+22]}]; Print[n," ",counterexample]; Label[aa]; Continue,{n,1,100}]
Showing 1-5 of 5 results.