A080458 a(1)=4; for n>1, a(n)=a(n-1) if n is already in the sequence, a(n)=a(n-1)+4 otherwise.
4, 8, 12, 12, 16, 20, 24, 24, 28, 32, 36, 36, 40, 44, 48, 48, 52, 56, 60, 60, 64, 68, 72, 72, 76, 80, 84, 84, 88, 92, 96, 96, 100, 104, 108, 108, 112, 116, 120, 120, 124, 128, 132, 132, 136, 140, 144, 144, 148, 152, 156, 156, 160, 164, 168, 168, 172, 176
Offset: 1
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
- Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
- Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
- Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See p. 2.
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).
Programs
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Mathematica
LinearRecurrence[{1, 0, 0, 1, -1}, {4, 8, 12, 12, 16}, 60] (* Jean-François Alcover, Sep 20 2018 *) -
PARI
a(n) = 4 + 4*(n-2-(n-4)\4); \\ Michel Marcus, May 06 2016
Formula
a(n) = 4 + 4*(n-2-floor((n-4)/4)).
From Chai Wah Wu, Jul 17 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.
G.f.: 4*x*(x^2 + x + 1)/(x^5 - x^4 - x + 1). (End)
From Ilya Gutkovskiy, Jul 17 2016: (Start)
E.g.f.: (3*x + 1)*cosh(x) + (3*x + 2)*sinh(x) - cos(x) - sin(x).
a(n) = (6*n - (-1)^n - 2*sqrt(2)*sin(Pi*n/2+Pi/4) + 3)/2. (End)