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.

A224417 Least prime p such that sum_{k=0}^n B_k*x^{n-k} is irreducible modulo p, where B_k refers to the Bell number A000110(k).

Original entry on oeis.org

2, 3, 2, 11, 3, 2, 193, 113, 2, 29, 71, 167, 19, 3, 7, 13, 199, 5, 101, 59, 13, 41, 3, 359, 7, 11, 2, 31, 197, 139, 3, 59, 2, 139, 83, 37, 23, 193, 587, 199, 67, 47, 401, 41, 571, 73, 1063, 229, 1163, 47, 53, 239, 347, 223, 577, 499, 271, 269, 11, 179
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 06 2013

Keywords

Comments

Conjecture: a(n) < 4n^2-1 for all n>0.

Examples

			a(5) = 3 since the polynomial sum_{k=0}^5 B_5*x^{5-k} = x^5+x^4+2*x^3+5*x^2+15*x+52 is irreducible modulo 3 but reducible modulo 2.
Note also that a(7) = 193 < 4*7^2-1 = 195.

Links

Crossrefs

Cf. A000040, A000110, A220072, A223934, A224210, A224416, A218465, A217785, A217788, A224197.

Programs

  • Mathematica
    A[n_,x_]:=A[n,x]=Sum[BellB[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[4n^2-2]}];
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.

Links

Crossrefs

Cf. A000040, A000041, A224417, A224416, A220072, A223934, A224210, A217788, A224197.

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.

Links

Crossrefs

Cf. A000040, A220947, A220072, A223934, A224210, A224416, A224417, A224418, A217788, A224197, A218465, A217785.

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.

Links

Crossrefs

Cf. A000040, A000045, A224416, A224417, A224418, A224480, A224210, A220072, A223934, A218465.

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.

Links

Crossrefs

Cf. A000040, A218465, A224210, A224416, A224417, A224418, A224480, A220072, A223934.

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.

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