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.

A321643 a(n) = 5*2^n - (-1)^n.

Original entry on oeis.org

4, 11, 19, 41, 79, 161, 319, 641, 1279, 2561, 5119, 10241, 20479, 40961, 81919, 163841, 327679, 655361, 1310719, 2621441, 5242879, 10485761, 20971519, 41943041, 83886079, 167772161, 335544319, 671088641, 1342177279, 2684354561, 5368709119, 10737418241, 21474836479
Offset: 0

Views

Author

Paul Curtz, Dec 03 2018

Keywords

Crossrefs

Programs

  • GAP
    List([0..30],n->5*2^n-(-1)^n); # Muniru A Asiru, Dec 05 2018
    
  • Maple
    [5*2^n-(-1)^n$n=0..30]; # Muniru A Asiru, Dec 05 2018
  • Mathematica
    a[n_] := 5*2^n - (-1)^n; Array[a, 30, 0] (* Amiram Eldar, Dec 03 2018 *)
  • PARI
    Vec((4 + 7*x) / ((1 + x)*(1 - 2*x)) + O(x^40)) \\ Colin Barker, Dec 04 2018
    
  • Python
    for n in range(0,30): print(5*2**n - (-1)**n) # Stefano Spezia, Dec 05 2018

Formula

a(n+2) - a(n) = a(n+1) + a(n) = 15*2^n, n >= 0.
a(n) - 2*a(n-1) = period 2: repeat [3, -3], n > 0, a(0)=4, a(1)=11.
a(n+1) = 10*A051049(n) + period 2: repeat [1, 9].
a(n) = 12*2^n - A321483(n), n >= 0.
a(n) = 2^(n+2) + 3*A001045(n), n >= 0.
a(n) == A070366(n+4) (mod 9).
From Colin Barker, Dec 04 2018: (Start)
G.f.: (4 + 7*x) / ((1 + x)*(1 - 2*x)).
a(n) = a(n-1) + 2*a(n-2) for n > 1. (End)
E.g.f.: exp(-x)*(5*exp(3*x) - 1). - Elmo R. Oliveira, Aug 17 2024

A322417 a(n) - 2*a(n-1) = period 2: repeat [3, 0] for n > 0, a(0)=5, a(1)=13.

Original entry on oeis.org

5, 13, 26, 55, 110, 223, 446, 895, 1790, 3583, 7166, 14335, 28670, 57343, 114686, 229375, 458750, 917503, 1835006, 3670015, 7340030, 14680063, 29360126, 58720255, 117440510, 234881023, 469762046, 939524095, 1879048190, 3758096383, 7516192766
Offset: 0

Views

Author

Paul Curtz, Dec 07 2018

Keywords

Comments

a(n) mod 9 = period 6: repeat [5, 4, 8, 1, 2, 7]. See A177883(n+2).
a(n+1) mod 10 = period 4: repeat [3, 6, 5, 0].

Crossrefs

Programs

  • GAP
    a:=[13,26];; for n in [3..30] do a[n]:=a[n-2]+21*2^(n-2); od; Concatenation([5],a); # Muniru A Asiru, Dec 07 2018
    
  • Mathematica
    a[0] = 5; a[1] = 13; a[n_] := a[n] = a[n - 2] + 21*2^(n - 2); Array[a, 30, 0] (* Amiram Eldar, Dec 07 2018 *)
    LinearRecurrence[{2, 1, -2}, {5, 13, 26}, 31] (* Jean-François Alcover, Jan 28 2019 *)
  • PARI
    Vec((5 + 3*x - 5*x^2) / ((1 - x)*(1 + x)*(1 - 2*x)) + O(x^40)) \\ Colin Barker, Dec 07 2018

Formula

a(n) = A166920(n) + A166920(n+1) + A166920(n+2) for n >= 2.
a(n) = a(n-2) + 21*2^(n-2) for n >= 2.
a(n) = a(n-1) + A321483(n) for n > 0.
From Colin Barker, Dec 07 2018: (Start)
G.f.: (5 + 3*x - 5*x^2) / ((1 - x)*(1 + x)*(1 - 2*x)).
a(n) = 7*2^n - 2 for n even.
a(n) = 7*2^n - 1 for n odd.
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n > 2.
(End)
a(2*n+1) = A206372(n). a(2*n+2) = 2*A206372(n) for n > 0.

Extensions

First formula corrected by Jean-François Alcover, Feb 01 2019
Showing 1-2 of 2 results.