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.

A025003 a(1) = 2; a(n+1) = a(n)-th nonprime, where nonprimes begin at 1.

Original entry on oeis.org

2, 4, 8, 14, 22, 33, 48, 66, 90, 120, 156, 202, 256, 322, 400, 494, 604, 734, 888, 1067, 1272, 1512, 1790, 2107, 2472, 2890, 3364, 3903, 4515, 5207, 5990, 6875, 7868, 8984, 10238, 11637, 13207, 14959, 16909, 19075, 21483, 24173, 27149, 30436, 34080, 38103
Offset: 1

Views

Author

Keywords

Comments

Index of first occurrence of n in A090532.
Let b(n) (n >= 0) be the smallest integer k >= 1 that takes n steps to reach 1 iterating the map f: k -> k - pi(k). The sequence {b(n), n >= 0} begins 1, 2, 4, 8, 14, 22, 33, 48, 66, 90, 120, 156, ... and agrees with the present sequence except for b(0). - Ya-Ping Lu, Sep 07 2020

Examples

			From _Ya-Ping Lu_, Sep 07 2020: (Start)
a(1) = 2 because f(2) = 2 - pi(2) = 1 and m(2) = 1;
For the integer 3, since f(3) = 1. m(3) = 1, which is not bigger than m(1) or m(2). So, 3 is not a term in the sequence;
a(2) = 4 because f^2(4) = f(2) = 1 and m(4) = 2;
a(3) = 8 because f^3(8) = f^2(4) = 1 and m(8) = 3. (End)
		

Crossrefs

Programs

  • Maple
    N:= 50: # to get a(0)..a(N)
    V:= Array(0..N):
    V[0]:= 1: V[1]:= 2:
    m:= 2: p:= 3: g:= 1: n:= 1:
    do
      if g+p-m-1 >= V[n] then
        m:= V[n]+m-g;
        n:= n+1;
        V[n]:= m;
        if n = N then break fi;
        g:= V[n-1];
      else
        g:= g+p-m;
        m:= p+1;
        p:= nextprime(m);
      fi;
    od;
    convert(V, list); # Robert Israel, Sep 08 2020
  • Python
    from sympy import prime, primepi
    n_last = 0
    pi_last = 0
    ct_max = -1
    for n in range(1, 100001):
        ct = 0
        pi = pi_last + primepi(n) - primepi(n_last)
        n_c = n
        pi_c = pi
        while n_c > 1:
            nc -= pi_c
            ct += 1
            pi_c -= primepi(n_c + pi_c) - primepi(n_c)
        if ct > ct_max:
            print(n)
            ct_max = ct
        n_last = n
        pi_last = pi # Ya-Ping Lu, Sep 07 2020

Formula

a(n) = min(k: f^n(k) = 1), where f = A062298 and n-fold iteration of f is denoted by f^n. - Ya-Ping Lu, Sep 07 2020

A334614 a(n) = pi(prime(n) - n) + n, where pi is the prime counting function.

Original entry on oeis.org

1, 2, 4, 6, 8, 10, 11, 13, 15, 18, 19, 21, 22, 24, 26, 28, 30, 32, 34, 35, 36, 38, 40, 42, 45, 47, 48, 50, 51, 53, 55, 57, 60, 61, 65, 66, 67, 68, 70, 72, 74, 76, 77, 79, 81, 82, 85, 88, 89, 91, 93, 94, 95, 99, 101, 102, 104, 105, 106, 107, 108, 112, 116, 117
Offset: 1

Views

Author

Ya-Ping Lu, Sep 08 2020

Keywords

Comments

It can be shown that a(n) > a(n-1) >= 1 and a(n) <= 2*n - 1 < 2*n (see proofs in the Links section).

Crossrefs

Programs

  • Mathematica
    Table[PrimePi[Prime[n] - n] + n, {n, 1, 64}] (* Amiram Eldar, Sep 09 2020 *)
  • PARI
    a(n) = n + primepi(prime(n) - n); \\ Michel Marcus, Sep 09 2020
  • Python
    from sympy import prime, primepi
    for n in range(1, 100001):
        a_n = primepi(prime(n) - n) + n
        print(a_n)
    

Formula

a(n) = A000720(A014689(n)) + n.
a(n) = A065328(n) + n. - Michel Marcus, Sep 12 2020
Showing 1-2 of 2 results.