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.

A294891 Number of proper divisors d of n such that Stern polynomial B(d,x) is irreducible.

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 1, 1, 2, 0, 2, 0, 2, 2, 1, 0, 2, 0, 2, 2, 2, 0, 2, 1, 2, 1, 2, 0, 3, 0, 1, 2, 2, 2, 2, 0, 2, 2, 2, 0, 3, 0, 2, 2, 2, 0, 2, 1, 3, 2, 2, 0, 2, 2, 2, 2, 2, 0, 3, 0, 2, 2, 1, 2, 3, 0, 2, 2, 3, 0, 2, 0, 2, 3, 2, 2, 3, 0, 2, 1, 2, 0, 3, 2, 2, 2, 2, 0, 3, 2, 2, 2, 2, 2, 2, 0, 2, 2, 3, 0, 3, 0, 2, 3
Offset: 1

Views

Author

Antti Karttunen, Nov 10 2017

Keywords

Examples

			For n=50, with proper divisors [1, 2, 5, 10, 25], 2, 5, and 25 are larger than one and included in A186891, thus a(50) = 3.
		

Crossrefs

Cf. also A294881, A294901.
Differs from A087624 for the first time at n=50.

Programs

  • PARI
    ps(n) = if(n<2, n, if(n%2, ps(n\2)+ps(n\2+1), 'x*ps(n\2)));
    A283991(n) = polisirreducible(ps(n));
    A294891(n) = sumdiv(n,d,(dA283991(d));

Formula

a(n) = Sum_{d|n, dA283991(d).
a(n) + A294892(n) = A032741(n).
a(n) = A294893(n) - A283991(n).

A294892 Number of proper divisors d of n such that either d=1 or Stern polynomial B(d,x) is reducible.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 3, 1, 3, 1, 3, 1, 1, 1, 5, 1, 1, 2, 3, 1, 4, 1, 4, 1, 1, 1, 6, 1, 1, 1, 5, 1, 4, 1, 3, 3, 1, 1, 7, 1, 2, 1, 3, 1, 5, 1, 5, 1, 1, 1, 8, 1, 1, 3, 5, 1, 4, 1, 3, 1, 4, 1, 9, 1, 1, 2, 3, 1, 4, 1, 7, 3, 1, 1, 8, 1, 1, 1, 5, 1, 8, 1, 3, 1, 1, 1, 9, 1, 3, 3, 5, 1, 4, 1, 5, 4
Offset: 1

Views

Author

Antti Karttunen, Nov 10 2017

Keywords

Examples

			For n=48, its proper divisors are [1, 2, 3, 4, 6, 8, 12, 16, 24]. After 1, the divisors 4, 6, 8, 12, 16 and 24 are not found in A186891, thus a(48) = 1+6 = 7.
For n=50, its proper divisors are [1, 2, 5, 10, 25]. After 1, only 10 is not found in A186891, thus a(50) = 1+1 = 2.
		

Crossrefs

Programs

  • PARI
    ps(n) = if(n<2, n, if(n%2, ps(n\2)+ps(n\2+1), 'x*ps(n\2)));
    A283991(n) = polisirreducible(ps(n));
    A294892(n) = sumdiv(n,d,(dA283991(d)));

Formula

a(n) = Sum_{d|n, dA283991(d)).
a(n) + A294891(n) = A032741(n).
a(n) = A294894(n) + A283991(n) - 1.

A294894 Number of divisors d of n such that either d=1 or Stern polynomial B(d,x) is reducible.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 2, 4, 1, 4, 1, 4, 2, 2, 1, 6, 1, 2, 3, 4, 1, 5, 1, 5, 2, 2, 2, 7, 1, 2, 2, 6, 1, 5, 1, 4, 4, 2, 1, 8, 2, 3, 2, 4, 1, 6, 1, 6, 2, 2, 1, 9, 1, 2, 4, 6, 1, 5, 1, 4, 2, 5, 1, 10, 1, 2, 3, 4, 1, 5, 1, 8, 4, 2, 1, 9, 2, 2, 2, 6, 1, 9, 1, 4, 2, 2, 1, 10, 1, 4, 4, 6, 1, 5, 1, 6, 5
Offset: 1

Views

Author

Antti Karttunen, Nov 10 2017

Keywords

Examples

			For n=25, with divisors [1, 5, 25], both B(5,x) and B(25,x) are irreducible, so only 1 is counted and a(25)=1.
		

Crossrefs

Cf. also A294884, A294904.
Differs from A033273 for the first time at n=25.

Programs

  • PARI
    ps(n) = if(n<2, n, if(n%2, ps(n\2)+ps(n\2+1), 'x*ps(n\2)));
    A283991(n) = polisirreducible(ps(n));
    A294894(n) = sumdiv(n,d,(0==A283991(d)));

Formula

a(n) = Sum_{d|n} (1-A283991(d)).
a(n) + A294893(n) = A000005(n).
a(n) = 1 + A294892(n) - A283991(n).

A294883 Number of divisors of n that are irreducible when their binary expansion is interpreted as polynomial over GF(2).

Original entry on oeis.org

0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 0, 2, 1, 1, 2, 2, 0, 2, 1, 2, 1, 2, 0, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 3, 0, 2, 1, 1, 1, 2, 1, 2, 1, 2, 0, 2, 2, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 0, 2, 1, 2, 2, 2, 2, 3, 0, 1, 1, 2, 0, 3, 0, 1, 2, 2, 0, 2, 3, 1, 2, 2, 1, 2, 1, 2, 2, 2, 0, 2, 1, 2, 2
Offset: 1

Views

Author

Antti Karttunen, Nov 09 2017

Keywords

Comments

Number of terms of A014580 that divide n.

Crossrefs

Cf. A091209 (gives a subset of zeros).
Cf. also A234741, A234742, A294893.

Programs

  • PARI
    A294883(n) = sumdiv(n,d,polisirreducible(Mod(1, 2)*Pol(binary(d))));

Formula

a(n) = Sum_{d|n} A091225(d).
a(n) + A294884(n) = A000005(n).
a(n) = A294881(n) + A091225(n).

A294903 Number of divisors of n that are in A257691.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 3, 3, 2, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 2, 4, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 2, 4, 2, 3, 3, 3, 2, 3, 2, 4, 3, 3, 2, 3, 3, 3, 3, 3, 2, 4, 2, 3, 3, 2, 3, 4, 2, 3, 3, 4, 2, 3, 2, 3, 4, 3, 3, 4, 2, 3, 2, 3, 2, 4, 3, 3, 3, 3, 2, 4, 3, 3, 3, 3, 4, 3, 2, 3, 3, 4, 2, 4, 2, 3, 4, 3, 2, 3, 2, 4, 4, 3, 2, 4, 3, 3, 3, 3, 4, 4
Offset: 1

Views

Author

Antti Karttunen, Nov 10 2017

Keywords

Comments

Number of terms of A257691 that divide n.

Crossrefs

Programs

  • Mathematica
    q[n_] := DivisorSum[n, DigitCount[#, 2, 1] &] <= 2*DigitCount[n, 2, 1]; a[n_] := DivisorSum[n, 1 &, q[#] &]; Array[a, 100] (* Amiram Eldar, Jul 20 2023 *)
  • PARI
    A292257(n) = sumdiv(n,d,(dA294905(n) = (A292257(n) <= hammingweight(n));
    A294903(n) = sumdiv(n,d,A294905(d));

Formula

a(n) = Sum_{d|n} A294905(d).
a(n) = A294901(n) + A294905(n).
a(n) + A294904(n) = A000005(n).
Showing 1-5 of 5 results.