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.

A245481 Numbers k such that the k-th cyclotomic polynomial has a root mod 13.

Original entry on oeis.org

1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156, 169, 338, 507, 676, 1014, 2028, 2197, 4394, 6591, 8788, 13182, 26364, 28561, 57122, 85683, 114244, 171366, 342732, 371293, 742586, 1113879, 1485172, 2227758, 4455516, 4826809, 9653618, 14480427, 19307236, 28960854
Offset: 1

Views

Author

Eric M. Schmidt, Jul 23 2014

Keywords

Comments

Numbers of the form d*13^j for d a divisor of 12.

Examples

			The 4th cyclotomic polynomial x^2 + 1 considered modulo 13 has a root x = 5, so 4 is in the sequence.

References

  • Trygve Nagell, Introduction to Number Theory. New York: Wiley, 1951, pp. 164-168.

Links

Crossrefs

Cf. A018309, A038754, A245478, A245479, A245480.

Programs

  • Mathematica
    LinearRecurrence[{0,0,0,0,0,13},{1,2,3,4,6,12},50] (* Harvey P. Dale, Aug 19 2021 *)
  • PARI
    for(n=1,10^6,if(#polrootsmod(polcyclo(n),13),print1(n,", "))) /* by definition; rather inefficient. - Joerg Arndt, Jul 28 2014 */
  • PARI
    Vec(-x*(12*x^5+6*x^4+4*x^3+3*x^2+2*x+1)/(13*x^6-1) + O(x^100)) \\ Colin Barker, Jul 30 2014
  • PARI
    a(n)=[12,1,2,3,4,6][n%6+1]*13^((n-1)\6) \\ Charles R Greathouse IV, Jan 12 2017
  • Sage
    def A245481(n) : return [12,1,2,3,4,6][n%6]*13^((n-1)//6)

Formula

a(n) = 13*a(n-6). G.f.: -x*(12*x^5+6*x^4+4*x^3+3*x^2+2*x+1) / (13*x^6-1). - Colin Barker, Jul 30 2014

A253235 Numbers n such that the n-th cyclotomic polynomial has no root mod p for all primes p <= n.

Original entry on oeis.org

1, 12, 15, 24, 28, 30, 33, 35, 36, 40, 44, 45, 48, 51, 56, 60, 63, 65, 66, 69, 70, 72, 75, 76, 77, 80, 84, 85, 87, 88, 90, 91, 92, 95, 96, 99, 102, 104, 105, 108, 112, 115, 117, 119, 120, 123, 124, 126, 130, 132, 133, 135, 138, 140, 141, 143, 144, 145, 150, 152, 153, 154
Offset: 1

Views

Author

Eric Chen, Apr 19 2015

Keywords

Comments

Numbers n such that A253236(n) = 0.
Numbers n such that all divisors of Phi_n(b) are congruent to 1 (mod n) for every natural number b.
If p is prime, k, r are natural numbers, then:
Every n = p^r is not in this sequence.
Every n = 2p^r is not in this sequence.
n = 3p^r (p>3) is in this sequence iff p != 1 (mod 3).
n = 4p^r (p>4) is in this sequence iff p != 1 (mod 4).
n = 5p^r (p>5) is in this sequence iff p != 1 (mod 5).
...
n = kp^r (p>k) is in this sequence iff p != 1 (mod k).

Links

Crossrefs

For A253236(n) = 2, 3, 5, 7, 11, 13, see A000079, A038754, A245478, A245479, A245480, A245481.

Programs

  • PARI
    is(n)=my(P=polcyclo(n), f=factor(n)[, 1]); for(i=1, #f, if(#polrootsmod(P, f[i]), return(0))); 1 \\ Charles R Greathouse IV, Apr 20 2015

A245478 Numbers k such that the k-th cyclotomic polynomial has a root mod 5.

Original entry on oeis.org

1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500, 625, 1250, 2500, 3125, 6250, 12500, 15625, 31250, 62500, 78125, 156250, 312500, 390625, 781250, 1562500, 1953125, 3906250, 7812500, 9765625, 19531250, 39062500, 48828125, 97656250, 195312500, 244140625, 488281250
Offset: 1

Views

Author

Eric M. Schmidt, Jul 23 2014

Keywords

Comments

Numbers of the form 2^i * 5^j for 0 <= i <= 2 and j >= 0.

Examples

			The 4th cyclotomic polynomial x^2 + 1 considered modulo 5 has a root x = 2, so 4 is in the sequence.

References

  • Trygve Nagell, Introduction to Number Theory. New York: Wiley, 1951, pp. 164-168.

Links

Crossrefs

Cf. A000079, A038754, A245479, A245480, A245481.

Programs

  • PARI
    for(n=1,10^6,if(#polrootsmod(polcyclo(n),5),print1(n,", "))) /* by definition; rather inefficient. - Joerg Arndt, Jul 28 2014 */
  • PARI
    is(n)=n%8 && 2^valuation(n,2)*5^valuation(n,5)==n \\ Charles R Greathouse IV, Jul 29 2014
  • PARI
    Vec(-x*(4*x^2+2*x+1)/(5*x^3-1) + O(x^100)) \\ Colin Barker, Aug 01 2014
  • Sage
    def A245478(n) : return 2^((n-1)%3)*5^((n-1)//3)

Formula

a(3j + i) = 2^(i-1)*5^j for i = 1,2,3 and j >= 0.
a(n) = 5*a(n-3). G.f.: -x*(4*x^2+2*x+1) / (5*x^3-1). - Colin Barker, Aug 01 2014

A245480 Numbers n such that the n-th cyclotomic polynomial has a root mod 11.

Original entry on oeis.org

1, 2, 5, 10, 11, 22, 55, 110, 121, 242, 605, 1210, 1331, 2662, 6655, 13310, 14641, 29282, 73205, 146410, 161051, 322102, 805255, 1610510, 1771561, 3543122, 8857805, 17715610, 19487171, 38974342, 97435855, 194871710, 214358881, 428717762, 1071794405, 2143588810
Offset: 1

Views

Author

Eric M. Schmidt, Jul 23 2014

Keywords

Comments

Numbers of the form d*11^j for d=1,2,5,10.

Examples

			The 5th cyclotomic polynomial x^4 + x^3 + x^2 + x + 1 considered modulo 11 has a root x = 3, so 5 is in the sequence.

References

  • Trygve Nagell, Introduction to Number Theory. New York: Wiley, 1951, pp. 164-168.

Links

Crossrefs

Cf. A000079, A038754, A245478, A245479, A245481.

Programs

  • Mathematica
    CoefficientList[Series[x(2x+1)(5x^2+1)/(1-11x^4), {x, 0, 20}], x] (* Benedict W. J. Irwin, Jul 24 2016 *)
LinearRecurrence[{0,0,0,11},{1,2,5,10},40] (* Harvey P. Dale, Aug 04 2021 *)
  • PARI
    for(n=1,10^6,if(#polrootsmod(polcyclo(n),11),print1(n,", "))) /* by definition; rather inefficient. - Joerg Arndt, Jul 28 2014 */
  • PARI
    a(n)=11^((n-1)\4)*[10,1,2,5][n%4+1] \\ Charles R Greathouse IV, Jun 11 2015
  • Sage
    def A245480(n) : return [10,1,2,5][n%4]*11^((n-1)//4)

Formula

From Benedict W. J. Irwin, Jul 29 2016: (Start)
a(n) = 11*a(n-4).
G.f.: x*(1 + 2*x)*(1 + 5*x^2)/(1 - 11*x^4).
a(n) appears to satisfy x*Prod_{n>=0} (1 + a(2^n+1)x^(2^n)) = Sum_{n>=1} a(n)*x^n.
Then a(n+1) = a(2^x+1)*a(2^y+1)*a(2^z+1)..., where n=2^x+2^y+2^z+... .
For example, n=31=2^0+2^1+2^2+2^3+2^4, then a(31+1)=a(2)*a(3)*a(5)*a(9)*a(17) i.e. 194871710=2*5*11*121*14641.
(End)

A253236 The unique prime p <= n such that n-th cyclotomic polynomial has a root mod p, or 0 if no such p exists.

Original entry on oeis.org

0, 2, 3, 2, 5, 3, 7, 2, 3, 5, 11, 0, 13, 7, 0, 2, 17, 3, 19, 5, 7, 11, 23, 0, 5, 13, 3, 0, 29, 0, 31, 2, 0, 17, 0, 0, 37, 19, 13, 0, 41, 7, 43, 0, 0, 23, 47, 0, 7, 5, 0, 13, 53, 3, 11, 0, 19, 29, 59, 0, 61, 31, 0, 2, 0, 0, 67, 17, 0, 0, 71, 0
Offset: 1

Views

Author

Eric Chen, Apr 07 2015

Keywords

Comments

There is at most one prime p <= n such that n-th cyclotomic polynomial has a root mod p.
For prime n and every natural number k, a(n^k) = n.
If a(n) != 0, then a(n)|n.
a(n) is either 0 or A006530(n). See Corollary 23 of the Shevelev et al. link. - Robert Israel, Sep 07 2016

Links

Crossrefs

For a(n) = 0, see A253235.
For a(n) = 2, see A000079.
For a(n) = 3, see A038754.
For a(n) = 5, see A245478.
For a(n) = 7, see A245479.
For a(n) = 11, see A245480.
For a(n) = 13, see A245481.
Cf. A006530.

Programs

  • Maple
    N:= 1000: # to get a(1) to a(N)
f:= proc(n) local p,x,C,v;
    C:= numtheory:-cyclotomic(n,x);
    p:= max(numtheory:-factorset(n));
    for v from 0 to p-1 do
      if eval(C,x=v) mod p = 0 then return p fi
    od:
    0
end proc:
f(1):= 0:
seq(f(n),n=1..N); # Robert Israel, Sep 07 2016
  • Mathematica
    a[n_] := Module[{p, x, c, v}, c[x_] = Cyclotomic[n, x]; p = FactorInteger[ n][[-1, 1]]; For[v=0, vJean-François Alcover, Jul 27 2020, after Maple *)
  • PARI
    a(n) = forprime(p=2, n, if(#polrootsmod(polcyclo(n), p), return(p)))
  • PARI
    a(n)=my(P=polcyclo(n),f=factor(n)[,1]); for(i=1,#f, if(#polrootsmod(P, f[i]), return(f[i]))); 0 \\ Charles R Greathouse IV, Apr 07 2015
Showing 1-5 of 5 results.

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