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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A279796 Erroneous duplicate of A285022.

Original entry on oeis.org

820, 1276, 1422, 1926, 2080, 2640, 3160, 3186, 3250, 4446, 4720, 4930, 5370, 6006
Offset: 1

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Keywords

A280877 Occurrences of decrease of the probability density P(a(n)) of coprime numbers k,m, satisfying 1 <= k <= a(n) and 1 <= m <= a(n); i.e., P(a(n)) < P(a(n)-1).

Original entry on oeis.org

2, 4, 6, 8, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 28, 30, 32, 33, 34, 36, 38, 40, 42, 44, 45, 46, 48, 50, 52, 54, 56, 58, 60, 62, 63, 64, 66, 68, 70, 72, 74, 75, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 99, 100, 102, 104, 105, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 128
Offset: 1

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Author

A.H.M. Smeets, Jan 09 2017

Keywords

Comments

Probability densities satisfying P(a(n)) < P(a(n)-1).
A285022 is a subset.
Related to Euler phi function by P(a(n)) = ((2*Sum_{1 <= m <= a(n)} phi(m))-1)/a(n)^2.
The sequence is very regular in the sense that all {0 < i} 2i appear in this sequence, as well as all {0 < i} 30i - 15 appear in this sequence.
Presuming P(n) > 0.6: phi(n)/n < 1/2 for n congruent to 0 mod 2, P(n) < P(n-1).
Presuming P(n) > 0.6: phi(n)/n < 8/15 for n congruent to 15 mod 30, P(n) < P(n-1).
A280877 = {i > 0 | 2i} union {i > 0 | 30i - 15} union A280878 union A280879.
The irregular appearances are given in the two disjoint sequences A280878 and A280879.
See also A285022.
Experimental observation: n/a(n) < Euler constant (A001620).
Probability density P(a(n)) = A018805(a(n))/a(n)^2.
There seems, with good reason, to be a high correlation between the odd numbers in this sequence and A079814. - Peter Munn, Apr 11 2021

Crossrefs

Programs

  • Mathematica
    P[n_] := P[n] = (2 Sum[CoprimeQ[i, j] // Boole, {i, n}, {j, i-1}] + 1)/n^2;
    Select[Range[2, 200], P[#] < P[#-1]&] (* Jean-François Alcover, Nov 15 2019 *)
  • PARI
    P(n) = sum(i=1, n, sum(j=1, n, gcd(i,j)==1))/n^2;
    isok(n) = P(n) < P(n-1); \\ Michel Marcus, Jan 28 2017
  • Python
    from math import gcd
    t = 1
    to = 1
    i = 1
    x = 1
    while x < 10000:
        x = x + 1
        y = 0
        while y < x:
            y = y + 1
            if gcd(x,y) == 1:
                t = t + 2
        e = t*(x-1)*(x-1) - to*x*x
        if e < 0:
            print(i,x)
            i = i + 1
        to = t
    
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