A080578 a(1)=1; for n > 1, a(n) = a(n-1) + 1 if n is already in the sequence, a(n) = a(n-1) + 3 otherwise.

1, 4, 7, 8, 11, 14, 15, 16, 19, 22, 23, 26, 29, 30, 31, 32, 35, 38, 39, 42, 45, 46, 47, 50, 53, 54, 57, 60, 61, 62, 63, 64, 67, 70, 71, 74, 77, 78, 79, 82, 85, 86, 89, 92, 93, 94, 95, 98, 101, 102, 105, 108, 109, 110, 113, 116, 117, 120, 123, 124, 125, 126

Offset: 1

Author

N. J. A. Sloane and Benoit Cloitre, Mar 23 2003

Keywords

• nonn

Comments

More generally for fixed r, there is a nice connection between the sequence a(1)=1, a(n) = a(n-1) + 1 if n is in the sequence, a(n) = a(n-1) + r + 1 otherwise and the so-called metafibonacci sequences. Indeed, (a(n)-n)/r is a generalized metafibonacci sequence of order r as defined in Ruskey's recent paper (reference given at A046699). - Benoit Cloitre, Feb 04 2007

In the Fokkink-Joshi paper, this sequence is the Cloitre (0,1,1,3)-hiccup sequence. - Michael De Vlieger, Jul 29 2025

Links

• Reinhard Zumkeller, 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 pp. 3, 5, 11.

Crossrefs

Cf. A080455, A080456, A080457, A080458, A080036, A080037, A080468.

Programs

Haskell

a080578 n = a080578_list !! (n-1)
a080578_list = 1 : f 2 [1] where
   f x zs@(z:_) = y : f (x + 1) (y : zs) where
     y = if x `elem` zs then z + 1 else z + 3
-- Reinhard Zumkeller, Sep 26 2014

Mathematica

l={1}; a=1; For[n=2, n<=100, If[MemberQ[l, n], a=a+1, a=a+3]; AppendTo[l, a]; n++]; l (* Indranil Ghosh, Apr 07 2017 *)

PARI

a(n)=if(n<2,1,a(n+1-2^floor(log(n)/log(2)))+2*2^floor(log(n)/log(2))-1) \\ Benoit Cloitre, Feb 04 2007

Python

l=[1]
a=1
for n in range(2, 101):
    a += 3 if n not in l else 1
    l.append(a)
print(l) # Indranil Ghosh, Apr 07 2017

Formula

a(n) = 2n + O(1); a(2^n) = 2^(n+1). - Benoit Cloitre, Oct 12 2003

a(1) = 1, for n >= 2 a(n) = a(n + 1 - 2^floor(log(n)/log(2))) + 2*2^floor(log(n)/log(2)) - 1; (a(n) - n)/2 = A046699(n) for n >= 2. - Benoit Cloitre, Feb 04 2007

a(n) = A055938(n-1) + 2 (conjectured). - Ralf Stephan, Dec 27 2013