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.

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A052130 a(n) is the number of numbers between 1 and 2^m with m-n prime factors (counted with multiplicity), for m sufficiently large.

Original entry on oeis.org

1, 2, 7, 15, 37, 84, 187, 421, 914, 2001, 4283, 9184, 19611, 41604, 87993, 185387, 389954, 817053, 1709640, 3567978, 7433670, 15460810, 32103728, 66567488, 137840687, 285076323, 588891185, 1215204568, 2505088087, 5159284087
Offset: 0

Views

Author

Bernd-Rainer Lauber (br.lauber(AT)surf1.de), Jan 21 2000

Keywords

Comments

a(n) = number of products of half-odd-primes <= 2^n. E.g., a(2) = 7 since 1, 3/2, (3/2)^2, (3/2)^3, (3/2)*(5/2), 5/2, 7/2 are all <= 2^2. - David W. Wilson, Feb 01 2000
m is sufficiently large precisely when 2^m > 3^(m-n), i.e., when m >= floor(n*log(3)/log(1.5)) = A117630(n+1) = A126281(n) for n > 1. (Robert G. Wilson v asks if this conjecture holds in a comment to A126281.) - David A. Corneth, Apr 09 2015
From Robert G. Wilson v, Apr 13 2020: (Start)
This sequence shows a sufficiently large row of A126279 read backwards or a sufficiently large column of A126279 read vertically.
log(y) ~ a + b*x + c*x^2, where a=1.1422, b=0.7419, and c=-0.00035, with an r^2 of 1.0. (End)
[But what is y? - Editors, Jun 15 2021]

Examples

			Between 1 and 2^m there is just one number with m prime factors, namely 2^m, so a(0) = 1.
For m >= 3, up to 2^m there are 2 numbers with m-1 prime factors, 2^(m-1) and 3*2^(m-2), so a(1) = 2.
		

Crossrefs

Programs

  • Mathematica
    AlmostPrimePi[k_Integer, n_] := Module[{a, i}, a[0] = 1; If[k == 1, PrimePi[n], Sum[PrimePi[n/Times @@ Prime[ Array[a, k - 1]]] - a[k - 1] + 1, Evaluate[ Sequence @@ Table[{a[i], a[i - 1], PrimePi[(n/Times @@ Prime[Array[a, i - 1]])^(1/(k - i + 1))]}, {i, k - 1}]] ]]]; (* Eric W. Weisstein, Feb 07 2006 *)
    Table[ AlmostPrimePi[Floor[n(1 + 1/Sqrt@2)] + 2, 2^(n + Floor[n(1 + 1/Sqrt@2)]) + 2], {n, 2, 30}] (* Robert G. Wilson v, Feb 21 2006 *)
  • Python
    from math import prod, isqrt
    from sympy import primepi, primerange, integer_nthroot
    def A052130(n):
        if n<=1: return n+1
        def g(x,a,b,c,m): yield from (((d,) for d in enumerate(primerange(b,isqrt(x//c)+1),a)) if m==2 else (((a2,b2),)+d for a2,b2 in enumerate(primerange(b,integer_nthroot(x//c,m)[0]+1),a) for d in g(x,a2,b2,c*b2,m-1)))
        k = 1
        while 3**k<(r:=1<Chai Wah Wu, Dec 03 2024

Extensions

More terms from David W. Wilson, Feb 01 2000
a(24)-a(29) from Robert G. Wilson v, Feb 21 2006

A182769 Beatty sequence for (4 + sqrt(2))/2.

Original entry on oeis.org

2, 5, 8, 10, 13, 16, 18, 21, 24, 27, 29, 32, 35, 37, 40, 43, 46, 48, 51, 54, 56, 59, 62, 64, 67, 70, 73, 75, 78, 81, 83, 86, 89, 92, 94, 97, 100, 102, 105, 108, 110, 113, 116, 119, 121, 124, 127, 129, 132, 135, 138, 140, 143, 146, 148, 151, 154, 157, 159, 162, 165, 167, 170, 173, 175
Offset: 1

Views

Author

Clark Kimberling, Nov 30 2010

Keywords

Comments

Let u=1+sqrt(2) and v=sqrt(2). Jointly rank {j*u} and {k*v} as in the first comment at A182760; a(n) is the position of n*u.
Is this a shifted version of A126281? - R. J. Mathar, Jan 24 2011
The answer to R. J. Mathar's question is no: A126281 contains 65 while this sequence does not. - L. Edson Jeffery, Sep 02 2014

Crossrefs

Programs

  • Magma
    [Floor(n*(4 + Sqrt(2))/2): n in [1..50]]; // G. C. Greubel, Jan 27 2018
  • Mathematica
    Table[Floor[n*(4 + Sqrt[2])/2], {n, 1, 100}] (* G. C. Greubel, Jan 27 2018 *)
  • PARI
    a(n) = floor(n*(4+sqrt(2))/2); \\ Michel Marcus, Sep 02 2014
    

Formula

a(n) = floor(n*(4 + sqrt(2))/2).
Showing 1-2 of 2 results.