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.

A342131 a(n) = n/2 + floor(n/4) if n is even, otherwise (3*n+1)/2.

Original entry on oeis.org

0, 2, 1, 5, 3, 8, 4, 11, 6, 14, 7, 17, 9, 20, 10, 23, 12, 26, 13, 29, 15, 32, 16, 35, 18, 38, 19, 41, 21, 44, 22, 47, 24, 50, 25, 53, 27, 56, 28, 59, 30, 62, 31, 65, 33, 68, 34, 71, 36, 74, 37, 77, 39, 80, 40, 83, 42, 86, 43, 89, 45, 92, 46, 95, 48, 98, 49, 101, 51, 104
Offset: 0

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Author

Thomas Scheuerle, Mar 01 2021

Keywords

Comments

A permutation of the nonnegative integers related to the Collatz function (A014682).
Interspersion of A032766 and A016789. - Michel Marcus, Mar 04 2021

Crossrefs

Programs

  • MATLAB
    function [a] = A342131(max_n)
        for n = 1:max_n
            m = n-1;
            if floor(m/2) == m/2
                a(n) = (m/2)+floor(m/4);
            else
                a(n) = (m*3+1)/2;
            end
        end
    end
    
  • Magma
    &cat [[3*k,6*k+2,3*k+1,6*k+5]: k in [0..20]] // Bruno Berselli, Mar 05 2021
    
  • Mathematica
    a[n_] := If[EvenQ[n], n/2 + Floor[n/4], (3*n + 1)/2]; Array[a, 100, 0] (* Amiram Eldar, Mar 03 2021 *)
    Table[(12 n + 4 - (3 n + 3 - (-1)^(n/2)) (1 + (-1)^n))/8, {n, 0, 70}] (* Bruno Berselli, Mar 05 2021 *)
  • PARI
    a(n) = if (n%2, (3*n+1)/2, n/2 + n\4); \\ Michel Marcus, Mar 04 2021
    
  • Python
    def A342131(n): return (3*n+1)//2 if n % 2 else n//2+n//4 # Chai Wah Wu, Mar 05 2021

Formula

a(n) = 9*n - 2*a(n-1) - 2*a(n-2) - 2*a(n-3) - a(n-4) - 17 for n >= 4.
a(n) = a(n-2) + a(n-4) - a(n-6).
a(n) = A006368(n+1) - 1.
G.f.: (x^4+2*x^3+3*x^2+x+2)*x/((x^2+1)*(x-1)^2*(x+1)^2). - Alois P. Heinz, Mar 01 2021
E.g.f.: (cos(x) + (6*x - 1)*cosh(x) + (2 + 3*x)*sinh(x))/4. - Stefano Spezia, Mar 02 2021
From Bruno Berselli, Mar 05 2021: (Start)
a(n) = (12*n + 4 - (3*n + 3 - (-1)^(n/2))*(1 + (-1)^n))/8. Therefore:
a(4*k) = 3*k;
a(4*k+1) = 6*k + 2;
a(4*k+2) = 3*k + 1;
a(4*k+3) = 6*k + 5. (End)