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.

A265640 Prime factorization palindromes (see comments for definition).

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 11, 12, 13, 16, 17, 18, 19, 20, 23, 25, 27, 28, 29, 31, 32, 36, 37, 41, 43, 44, 45, 47, 48, 49, 50, 52, 53, 59, 61, 63, 64, 67, 68, 71, 72, 73, 75, 76, 79, 80, 81, 83, 89, 92, 97, 98, 99, 100, 101, 103, 107, 108, 109, 112, 113, 116, 117, 121, 124, 125, 127, 128, 131, 137, 139, 144
Offset: 1

Views

Author

Vladimir Shevelev, Dec 11 2015

Keywords

Comments

a(66) is the first term at which this sequence differs from A119848.
A number N is called a prime factorization palindrome (PFP) if all its prime factors, taking into account their multiplicities, can be arranged in a row with central symmetry (see example). It is easy to see that every PFP-number is either a square or a product of a square and a prime. In particular, the sequence contains all primes.
Numbers which are both palindromes (A002113) and PFP are 1,2,3,4,5,7,9,11,44,99,101,... (see A265641).
If n is in the sequence, so is n^k for all k >= 0. - Altug Alkan, Dec 11 2015
The sequence contains all perfect numbers except 6 (cf. A000396). - Don Reble, Dec 12 2015
Equivalently, numbers that have at most one prime factor with odd multiplicity. - Robert Israel, Feb 03 2016
Numbers whose squarefree part is noncomposite. - Peter Munn, Jul 01 2020

Examples

			44 is a member, since 44=2*11*2.
52 is a member, since 52=2*13*2. [This illustrates the fact that the digits don't need to form a palindrome. This is not a base-dependent sequence. - _N. J. A. Sloane_, Oct 05 2024]
180 is a member, since 180=2*3*5*3*2.
		

Crossrefs

Cf. A000396, A000720, A002113, A265641, complement of A229153.
Disjoint union of A229125 and (A000290\{0}).
Cf. A013661 (zeta(2)).

Programs

  • Maple
    N:= 1000: # to get all terms <= N
    P:= [1,op(select(isprime, [2,seq(i,i=3..N,2)]))]:
    sort([seq(seq(p*x^2,x=1..floor(sqrt(N/p))),p=P)]); # Robert Israel, Feb 03 2016
  • Mathematica
    M = 200; P = Join[{1}, Select[Join[{2}, Range[3, M, 2]], PrimeQ]]; Sort[ Flatten[Table[Table[p x^2, {x, 1, Floor[Sqrt[M/p]]}], {p, P}]]] (* Jean-François Alcover, Apr 09 2019, after Robert Israel *)
  • PARI
    for(n=1, 200, if( ispseudoprime(core(n)) || issquare(n), print1(n, ", "))) \\ Altug Alkan, Dec 11 2015
    
  • Python
    from math import isqrt
    from sympy.ntheory.factor_ import core, isprime
    def ok(n): return isqrt(n)**2 == n or isprime(core(n))
    print([k for k in range(1, 145) if ok(k)]) # Michael S. Branicky, Oct 03 2024
    
  • Python
    from math import isqrt
    from sympy import primepi, mobius
    def A265640(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x):
            c = n-(a:=isqrt(x))
            for y in range(1,a+1):
                m = x//y**2
                c -= primepi(m)-sum(mobius(k)*(m//k**2) for k in range(1, isqrt(m)+1))
            return c
        return bisection(f,n,n) # Chai Wah Wu, Jan 30 2025

Formula

lim A(x)/pi(x) = zeta(2) where A(x) is the number of a(n) <= x and pi is A000720.

A266047 Smallest integers of each prime signature of prime factorization palindromes (A265640).

Original entry on oeis.org

1, 2, 4, 8, 12, 16, 32, 36, 48, 64, 72, 128, 144, 180, 192, 256, 288, 432, 512, 576, 720, 768, 900, 1024, 1152, 1296, 1728, 1800, 2048, 2304, 2592, 2880, 3072, 3600, 4096, 4608, 5184, 6300, 6480, 6912, 7200, 8192, 9216, 10368, 10800, 11520, 12288, 14400, 15552, 16384, 18432
Offset: 1

Views

Author

Vladimir Shevelev, Dec 20 2015

Keywords

Comments

A subsequence of A025487.
According to Hardy and Ramanujan, the number Q(x) of numbers
2^b_2*3^b_3*...*p^b_p <= x, (1)
where b_2>=b_3>=...>=b_p, is of order e^(2Pi/sqrt(3)(1+o(1))sqrt(log x/loglog x)).
If all b_i=2*c_i are even, then the number of such numbers is Q(sqrt(x)). Note that, if in (1) c_p>0, where p is n-th prime, then c_r>0, r=2 [Dusart], Eq(4.2),
p<=e*n*log(n)
Let K(x) be the number of a(n)<=x, q=nextprime(p). Then K(x)<=Q(sqrt(x))(1+Sum_{prime p}1/p)+1/3, where p satisfies (2) (+1/3, taking into account 1/q).
By [Rosser], Sum_{p<=x}1/p=loglog(x)+0.261497...+o(1). Hence K(x)<=Q(sqrt(x))*(loglog(e/2*log(x*loglogx))+1.594830...+o(1)).
Asymptotics of K(x) remain open.

Crossrefs

Showing 1-2 of 2 results.