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 1-2 of 2 results.

A360013 Numbers whose exponent of 2 in their canonical prime factorization is larger than all the other exponents.

Original entry on oeis.org

2, 4, 8, 12, 16, 20, 24, 28, 32, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 104, 112, 116, 120, 124, 128, 132, 136, 140, 144, 148, 152, 156, 160, 164, 168, 172, 176, 184, 188, 192, 200, 204, 208, 212, 220, 224, 228, 232, 236, 240, 244, 248, 256
Offset: 1

Views

Author

Amiram Eldar, Jan 21 2023

Keywords

Comments

Numbers k such that A007814(k) > A051903(A000265(k)).
The powers of 2 (A000079), except for 1, are all terms.
The product of any two terms (not necessarily distinct) is also a term.
This sequence is a disjoint union of {2} and the subsequences of numbers m of the form 2^k*o where o = A000265(m), the odd part of m, is a k-free number, for k >= 2. These subsequences include, for k = 2, numbers of the form 4*o where o is an odd squarefree number (A056911); for k = 3, numbers of the form 8*o where o is an odd cubefree number; etc.
The asymptotic density of this sequence is Sum_{k>=2} 1/(zeta(k)*2*(2^k-1)) = 0.222707226888193809... .
The asymptotic mean of the exponent of 2 in the prime factorization of the terms of this sequence is Sum_{k>=2} k/(zeta(k)*2*(2^k-1)) / Sum_{k>=2} 1/(zeta(k)*2*(2^k-1)) = 3.10346728882748723133... . [corrected by Amiram Eldar, Jul 10 2025]
This sequence is a subsequence of A360015 and the asymptotic density of this sequence within A360015 is exactly 1/2.
Also even numbers whose multiset of prime factors has unique mode 2. - Gus Wiseman, Jul 10 2023

Examples

			From _Gus Wiseman_, Jul 09 2023: (Start)
108 = 2*2*3*3*3 is missing because its mode is not 2.
180 = 2*2*3*3*5 is missing because 2 is not the unique mode.
120 = 2*2*2*3*5 is present because its unique mode is 2.
The terms together with their prime factorizations begin:
   2 = 2
   4 = 2*2
   8 = 2*2*2
  12 = 2*2*3
  16 = 2*2*2*2
  20 = 2*2*5
  24 = 2*2*2*3
  28 = 2*2*7
  32 = 2*2*2*2*2
  40 = 2*2*2*5
  44 = 2*2*11
  48 = 2*2*2*2*3
  52 = 2*2*13
  56 = 2*2*2*7
  60 = 2*2*3*5
  64 = 2*2*2*2*2*2
(End)
		

Crossrefs

Equals A360015 \ A360014.
Partitions of this type are counted by A241131.
Allowing any unique mode gives A356862, complement A362605.
Allowing any unique co-mode gives A359178, complement A362606.
Not requiring the mode to be unique gives A360015.
The opposite version is A362616, counted by A362612.
For co-mode instead of mode we have A364061, counted by A364062.
With least prime factor instead of 2, we have A364160, counted by A364193.
With a different factorization, we have the subsequence A335738.
A124010 gives prime signature, ordered A118914.
A362611 counts modes in prime factorization, triangle A362614.
A362613 counts co-modes in prime factorization, triangle A362615.
A363486 gives least mode in prime indices, A363487 greatest.

Programs

  • Mathematica
    q[n_] := Module[{e = IntegerExponent[n, 2], m}, m = n/2^e; (m == 1 && e > 0) || AllTrue[FactorInteger[m][[;; , 2]], # < e &]]; Select[Range[256], q]
  • PARI
    is(n) = {my(e = valuation(n, 2), m = n >> e); (m == 1 && e > 0) || (m > 1 && vecmax(factor(m)[,2]) < e)};

Formula

a(n) = 2*A360015(n). - Gus Wiseman, Jul 10 2023

A364160 Numbers whose least prime factor has the greatest exponent.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 11, 12, 13, 16, 17, 19, 20, 23, 24, 25, 27, 28, 29, 31, 32, 37, 40, 41, 43, 44, 45, 47, 48, 49, 52, 53, 56, 59, 60, 61, 63, 64, 67, 68, 71, 72, 73, 76, 79, 80, 81, 83, 84, 88, 89, 92, 96, 97, 99, 101, 103, 104, 107, 109, 112, 113, 116
Offset: 1

Views

Author

Gus Wiseman, Jul 14 2023

Keywords

Comments

First differs from A334298 in having 600 and lacking 180.
Also numbers whose minimum part in prime factorization is a unique mode.
If k is a term, then so are all powers of k. - Robert Israel, Sep 17 2024

Examples

			The prime factorization of 600 is 2*2*2*3*5*5, and 3 > max(1,2), so 600 is in the sequence.
The prime factorization of 180 is 2*2*3*3*5, but 2 <= max(2,1), so 180 is not in the sequence.
The terms together with their prime indices begin:
     1: {}           29: {10}              67: {19}
     2: {1}          31: {11}              68: {1,1,7}
     3: {2}          32: {1,1,1,1,1}       71: {20}
     4: {1,1}        37: {12}              72: {1,1,1,2,2}
     5: {3}          40: {1,1,1,3}         73: {21}
     7: {4}          41: {13}              76: {1,1,8}
     8: {1,1,1}      43: {14}              79: {22}
     9: {2,2}        44: {1,1,5}           80: {1,1,1,1,3}
    11: {5}          45: {2,2,3}           81: {2,2,2,2}
    12: {1,1,2}      47: {15}              83: {23}
    13: {6}          48: {1,1,1,1,2}       84: {1,1,2,4}
    16: {1,1,1,1}    49: {4,4}             88: {1,1,1,5}
    17: {7}          52: {1,1,6}           89: {24}
    19: {8}          53: {16}              92: {1,1,9}
    20: {1,1,3}      56: {1,1,1,4}         96: {1,1,1,1,1,2}
    23: {9}          59: {17}              97: {25}
    24: {1,1,1,2}    60: {1,1,2,3}         99: {2,2,5}
    25: {3,3}        61: {18}             101: {26}
    27: {2,2,2}      63: {2,2,4}          103: {27}
    28: {1,1,4}      64: {1,1,1,1,1,1}    104: {1,1,1,6}
		

Crossrefs

Allowing any unique mode gives A356862, complement A362605.
Allowing any unique co-mode gives A359178, complement A362606.
The even case is A360013, counted by A241131.
For greatest instead of least we have A362616, counted by A362612.
These partitions are counted by A364193.
A027746 lists prime factors (with multiplicity).
A112798 lists prime indices, length A001222, sum A056239.
A362611 counts modes in prime factorization, triangle A362614.
A362613 counts co-modes in prime factorization, triangle A362615.
A363486 gives least mode in prime indices, A363487 greatest.

Programs

  • Maple
    filter:= proc(n) local F,i;
      F:= ifactors(n)[2];
      if nops(F) = 1 then return true fi;
      i:= min[index](F[..,1]);
      andmap(t -> F[t,2] < F[i,2], {$1..nops(F)} minus {i})
    end proc:
    filter(1):= true:
    select(filter, [$1..200]); # Robert Israel, Sep 17 2024
  • Mathematica
    Select[Range[100],First[Last/@FactorInteger[#]] > Max@@Rest[Last/@FactorInteger[#]]&]
Showing 1-2 of 2 results.