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 1-3 of 3 results.

A102393 A wicked evil sequence.

Original entry on oeis.org

1, 0, 0, 4, 0, 6, 7, 0, 0, 10, 11, 0, 13, 0, 0, 16, 0, 18, 19, 0, 21, 0, 0, 24, 25, 0, 0, 28, 0, 30, 31, 0, 0, 34, 35, 0, 37, 0, 0, 40, 41, 0, 0, 44, 0, 46, 47, 0, 49, 0, 0, 52, 0, 54, 55, 0, 0, 58, 59, 0, 61, 0, 0, 64, 0, 66, 67, 0, 69, 0, 0, 72, 73, 0, 0, 76, 0, 78, 79, 0, 81, 0, 0, 84, 0, 86
Offset: 0

Views

Author

Paul Barry, Jan 06 2005

Keywords

Comments

Elements of A026147 (evil numbers plus one) appear at positions indexed by the evil numbers A001969, 0 otherwise. A000027(n) = A102393(n) + A102394(n).
The following sequences all appear to have the same parity: A003071, A029886, A061297, A092524, A093431, A102393, A104258, A122248, A128975. - Jeremy Gardiner, Dec 28 2008

Crossrefs

Programs

  • Mathematica
    a[n_] := If[EvenQ @ DigitCount[n, 2, 1], n + 1, 0]; Array[a, 100, 0] (* Amiram Eldar, Aug 02 2020 *)

Formula

a(n) = (n+1)(1+(-1)^A000120(n))/2.
a(n) = (n+1)(1+(-1)^A010060(n))/2.

A102392 Odious numbers in odious places.

Original entry on oeis.org

0, 1, 2, 0, 4, 0, 0, 7, 8, 0, 0, 11, 0, 13, 14, 0, 16, 0, 0, 19, 0, 21, 22, 0, 0, 25, 26, 0, 28, 0, 0, 31, 32, 0, 0, 35, 0, 37, 38, 0, 0, 41, 42, 0, 44, 0, 0, 47, 0, 49, 50, 0, 52, 0, 0, 55, 56, 0, 0, 59, 0, 61, 62, 0, 64, 0, 0, 67, 0, 69, 70, 0, 0, 73, 74, 0, 76, 0, 0, 79, 0, 81, 82, 0, 84, 0
Offset: 0

Views

Author

Paul Barry, Jan 06 2005

Keywords

Comments

Odious numbers A000069(n) appear at positions indexed by the odious numbers, 0 otherwise.

Crossrefs

Programs

  • Mathematica
    odnQ[n_]:=OddQ[DigitCount[n,2,1]]; Table[If[odnQ[n],n,0],{n,0,100}] (* Harvey P. Dale, Oct 15 2019 *)
  • Python
    def A102392(n): return n*(n.bit_count()&1) # Chai Wah Wu, Nov 23 2023

Formula

A001477(n) = A102391(n)+a(n).
a(n) = if((1+floor(n/2))(1+(-1)^A000120(n))/2=0, n, 0).
a(n) = if((1+floor(n/2))(1+(-1)^A010060(n))/2=0, n, 0).
a(n) = n*A010060(n). - Ridouane Oudra, Apr 19 2025

A367513 The exponentially evil part of n: the largest unitary divisor of n that is an exponentially evil number (A262675).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 27, 1, 1, 1, 1, 32, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 27, 1, 8, 1, 1, 1, 1, 1, 1, 1, 64, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Nov 21 2023

Keywords

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := p^(e * (1 - ThueMorse[e])); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(hammingweight(f[i, 2])%2, 1, f[i, 1]^f[i, 2]));}
    
  • Python
    from math import prod
    from sympy import factorint
    def A367513(n): return prod(p**e for p, e in factorint(n).items() if e.bit_count()&1^1) # Chai Wah Wu, Nov 23 2023

Formula

Multiplicative with a(p^e) = p^(e*A010059(e)) = p^A102391(e).
a(n) = n/A367514(n).
A001221(a(n)) = A367512(n).
A034444(a(n)) = A367516(n).
a(n) >= 1, with equality if and only if n is an exponentially odious number (A270428).
a(n) <= n, with equality if and only if n is an exponentially evil number (A262675).
Showing 1-3 of 3 results.