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 1-4 of 4 results.

A061142 Replace each prime factor of n with 2: a(n) = 2^bigomega(n), where bigomega = A001222, number of prime factors counted with multiplicity.

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 2, 8, 4, 4, 2, 8, 2, 4, 4, 16, 2, 8, 2, 8, 4, 4, 2, 16, 4, 4, 8, 8, 2, 8, 2, 32, 4, 4, 4, 16, 2, 4, 4, 16, 2, 8, 2, 8, 8, 4, 2, 32, 4, 8, 4, 8, 2, 16, 4, 16, 4, 4, 2, 16, 2, 4, 8, 64, 4, 8, 2, 8, 4, 8, 2, 32, 2, 4, 8, 8, 4, 8, 2, 32, 16, 4, 2, 16, 4, 4, 4, 16, 2, 16, 4, 8, 4, 4, 4
Offset: 1

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Author

Henry Bottomley, May 29 2001

Keywords

Comments

The inverse Möbius transform of A162510. - R. J. Mathar, Feb 09 2011

Examples

			a(100)=16 since 100=2*2*5*5 and so a(100)=2*2*2*2.
		

Crossrefs

Programs

  • Maple
    with(numtheory): seq(2^bigomega(n),n=1..95);
  • Mathematica
    Table[2^PrimeOmega[n], {n, 1, 95}] (* Jean-François Alcover, Jun 08 2013 *)
  • PARI
    a(n)=direuler(p=1,n,1/(1-2*X))[n] /* Ralf Stephan, Mar 28 2015 */
    
  • PARI
    a(n) = 2^bigomega(n); \\ Michel Marcus, Aug 08 2017

Formula

a(n) = Sum_{d divides n} 2^(bigomega(d)-omega(d)) = Sum_{d divides n} 2^(A001222(d) - A001221(d)). - Benoit Cloitre, Apr 30 2002
a(n) = A000079(A001222(n)), i.e., a(n)=2^bigomega(n). - Emeric Deutsch, Feb 13 2005
Totally multiplicative with a(p) = 2. - Franklin T. Adams-Watters, Oct 04 2006
Dirichlet g.f.: Product_{p prime} 1/(1-2*p^(-s)). - Ralf Stephan, Mar 28 2015
a(n) = A001316(A156552(n)). - Antti Karttunen, May 29 2017
Dirichlet g.f.: zeta(s)^2 * Product_{p prime} 1/(1 - 1/(p^s - 1)^2). - Vaclav Kotesovec, Mar 14 2023

A347046 Greatest divisor of n with exactly half as many prime factors (counting multiplicity) as n, or 1 if there are none.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 1, 3, 5, 1, 1, 1, 7, 5, 4, 1, 1, 1, 1, 7, 11, 1, 6, 5, 13, 1, 1, 1, 1, 1, 1, 11, 17, 7, 9, 1, 19, 13, 10, 1, 1, 1, 1, 1, 23, 1, 1, 7, 1, 17, 1, 1, 9, 11, 14, 19, 29, 1, 15, 1, 31, 1, 8, 13, 1, 1, 1, 23, 1, 1, 1, 1, 37, 1, 1, 11, 1, 1, 1, 9
Offset: 1

Views

Author

Gus Wiseman, Aug 16 2021

Keywords

Comments

Problem: What are the positions of last appearances > 1?

Examples

			The divisors of 90 with half bigomega are: 6, 9, 10, 15, so a(90) = 15.
		

Crossrefs

The greatest divisor without the condition is A006530 (smallest: A020639).
Positions of 1's are A026424.
The case of powers of 2 is A072345.
Positions of first appearances are A123667 (sorted: A123666).
Divisors of this type are counted by A345957 (rounded: A096825).
The rounded version is A347044.
The smallest divisor of this is A347045 (rounded: A347043).
A000005 counts divisors.
A001221 counts distinct prime factors.
A001222 counts all prime factors (also called bigomega).
A056239 adds up prime indices, row sums of A112798.
A207375 lists central divisors (min: A033676, max: A033677).
A340387 lists numbers whose sum of prime indices is twice bigomega.
A340609 lists numbers whose maximum prime index divides bigomega.
A340610 lists numbers whose maximum prime index is divisible by bigomega.
A347042 counts divisors d|n such that bigomega(d) divides bigomega(n).

Programs

  • Mathematica
    Table[If[#=={},1,Max[#]]&@Select[Divisors[n], PrimeOmega[#]==PrimeOmega[n]/2&],{n,100}]
    a[n_] := Module[{p = Flatten[Table[#[[1]], {#[[2]]}] & /@ FactorInteger[n]], np}, np = Length[p]; If[OddQ[np], 1, Times @@ p[[np/2+1 ;; np]]]]; Array[a, 100] (* Amiram Eldar, Nov 02 2024 *)
  • Python
    from sympy import divisors, factorint
    def a(n):
        npf = len(factorint(n, multiple=True))
        for d in divisors(n)[-1:0:-1]:
            if 2*len(factorint(d, multiple=True)) == npf: return d
        return 1
    print([a(n) for n in range(1, 82)]) # Michael S. Branicky, Aug 18 2021
    
  • Python
    from math import prod
    from sympy import factorint
    def A347046(n):
        fs = factorint(n,multiple=True)
        q, r = divmod(len(fs),2)
        return 1 if r else prod(fs[q:]) # Chai Wah Wu, Aug 20 2021

Formula

a(n) = Product_{k=A001222(n)/2+1..A001222(n)} A027746(n,k) if A001222(n) is even, and 1 otherwise. - Amiram Eldar, Nov 02 2024

A123666 Numbers with an even number of prime factors, at least half of which are 2.

Original entry on oeis.org

1, 4, 6, 10, 14, 16, 22, 24, 26, 34, 36, 38, 40, 46, 56, 58, 60, 62, 64, 74, 82, 84, 86, 88, 94, 96, 100, 104, 106, 118, 122, 132, 134, 136, 140, 142, 144, 146, 152, 156, 158, 160, 166, 178, 184, 194, 196, 202, 204, 206, 214, 216, 218, 220, 224, 226, 228, 232, 240
Offset: 1

Views

Author

Keywords

Comments

This is the lexicographically earliest maximal set of nonprime integers that has unique factorization. Products of any number of terms from A100484, with repetition allowed.

Crossrefs

Programs

  • Mathematica
    fn[b_]:=Total[Last/@FactorInteger[b]];Join[{1},Select[Range[240],EvenQ[#]&&EvenQ[fn[#]]&&First[FactorInteger[#]][[2]]>=fn[#]/2&]] (* James C. McMahon, Nov 23 2024 *)

A328854 Dirichlet g.f.: Product_{p prime} 1 / (1 - 2 * p^(-s))^2.

Original entry on oeis.org

1, 4, 4, 12, 4, 16, 4, 32, 12, 16, 4, 48, 4, 16, 16, 80, 4, 48, 4, 48, 16, 16, 4, 128, 12, 16, 32, 48, 4, 64, 4, 192, 16, 16, 16, 144, 4, 16, 16, 128, 4, 64, 4, 48, 48, 16, 4, 320, 12, 48, 16, 48, 4, 128, 16, 128, 16, 16, 4, 192, 4, 16, 48, 448, 16, 64, 4, 48, 16, 64, 4, 384, 4, 16, 48
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 28 2019

Keywords

Comments

Dirichlet convolution of A061142 with itself.

Crossrefs

Programs

  • Mathematica
    Table[2^PrimeOmega[n] DivisorSigma[0, n], {n, 1, 75}]
    f[p_, e_] := (e + 1)*2^e; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Dec 02 2020 *)
  • PARI
    a(n) = numdiv(n)*2^bigomega(n); \\ Michel Marcus, Dec 02 2020
    
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, 1/(1 - 2*X)^2)[n], ", ")) \\ Vaclav Kotesovec, Aug 22 2021

Formula

If n = Product (p_j^k_j) then a(n) = Product (2^k_j * (k_j + 1)).
a(n) = 2^bigomega(n) * tau(n), where bigomega = A001222 and tau = A000005.
Showing 1-4 of 4 results.