A064437 -id:A064437 - cp's OEIS frontend

A079000 a(n) is taken to be the smallest positive integer greater than a(n-1) which is consistent with the condition "n is a member of the sequence if and only if a(n) is odd".

1, 4, 6, 7, 8, 9, 11, 13, 15, 16, 17, 18, 19, 20, 21, 23, 25, 27, 29, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 95, 97

Offset: 1

Matthew Vandermast, Feb 01 2003

a(a(n)) = 2n + 3 for n>1.

			a(2) cannot be 2 because 2 is even; it cannot be 3 because that would require 2 to be a member of the sequence. Hence a(2)=4 and the next odd member of the sequence is the fourth member.

Hsien-Kuei Hwang, S Janson, TH Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint, 2016; http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf. Also Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585
N. J. A. Sloane, Seven Staggering Sequences, in Homage to a Pied Puzzler, E. Pegg Jr., A. H. Schoen and T. Rodgers (editors), A. K. Peters, Wellesley, MA, 2009, pp. 93-110.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
N. J. A. Sloane, Seven Staggering Sequences.
Index entries for sequences of the a(a(n)) = 2n family

Cf. A079250-A079259, A079313, A079325, A064437, A003605, A079352, A079358.
Cf. also A080596, A080731, A080752, A121053, A080032.
Partial sums give A080566. Differences give A079948.

Digits := 50; A079000 := proc(n) local k,j; if n<=2 then n^2; else k := floor(evalf(log( (n+3)/6 )/log(2)) ); j := n-(9*2^k-3); 12*2^k-3+3*j/2 +abs(j)/2; fi; end;
A002264 := n->floor(n/3): A079944 := n->floor(log[2](4*(n+2)/3))-floor(log[2](n+2)): A000523 := n->floor(log[2](n)): f := n->A079944(A002264(n-4)): g := n->A000523(A002264(n+2)/2): A079000 := proc(n) if n>3 then RETURN(simplify(3*n+3-3*2^g(n)+(-1)^f(n)*(9*2^g(n)-n-3))/2) else if n>0 then RETURN([1,4,6][n]) else RETURN(0) fi fi: end;

a[1] = 1; a[n_] := (k = Floor[Log[2, (n+3)/6]]; j = n-(9*2^k - 3); 12*2^k-3 + 3*j/2 + Abs[j]/2); Table[a[n], {n, 1, 71}] (* Jean-François Alcover, May 21 2012, after Maple *)

a(1) = 1, a(2) = 4, then a(9*2^k-3+j) = 12*2^k-3+3*j/2+|j|/2 for k>=0, -3*2^k <= j <= 3*2^k. Also a(3n) = 3*b(n/3), a(3n+1) = 2*b(n)+b(n+1), a(3n+2) = b(n)+2*b(n+1) for n>=2, where b = A079905. - N. J. A. Sloane and Benoit Cloitre, Feb 20 2003
a(n+1) - 2*a(n) + a(n-1) = 1 for n = 9*2^k - 3, k>=0, = -1 for n = 2 and 3*2^k-3, k>=1 and = 0 otherwise.
a(n) = (3*n + 3 - 3*2^g(n) + (-1)^f(n)*(9*2^g(n) - n - 3))/2 for n>3, f(n) = A079944(A002264(n-4)) and g(n) = A000523(A002264(n+2)/2). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 23 2003
Also a(n) = n + 3*2^A000523(A002264(n+2)/2)*(1 - 3*A080584(n-4)) + A080584(n-4)*(n+3) for n>3, where A080584(n)=A079944(A002264(n)). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 24 2003

A007066 a(n) = 1 + ceiling((n-1)*phi^2), phi = (1+sqrt(5))/2.

Original entry on oeis.org

1, 4, 7, 9, 12, 15, 17, 20, 22, 25, 28, 30, 33, 36, 38, 41, 43, 46, 49, 51, 54, 56, 59, 62, 64, 67, 70, 72, 75, 77, 80, 83, 85, 88, 91, 93, 96, 98, 101, 104, 106, 109, 111, 114, 117, 119, 122, 125, 127, 130, 132, 135, 138, 140, 143, 145, 148, 151, 153, 156, 159, 161, 164, 166

Offset: 1

N. J. A. Sloane, Mira Bernstein

First column of dual Wythoff array, A126714.
Positions of 0's in A189479.
Skala (2016) asks if this sequence also gives the positions of the 0's in A283310. - N. J. A. Sloane, Mar 06 2017
Upper Wythoff sequence plus 2, when shifted by 1. - Michel Dekking, Aug 26 2019
In the Fokkink-Joshi paper, this sequence is the Cloitre (0,1,2,3)-hiccup sequence, i.e., a(1) = 1; for m < n, a(n) = a(n-1)+2 if a(m) = n, else a(n) = a(n-1)+3. - Michael De Vlieger, Jul 30 2025

Clark Kimberling, "Stolarsky interspersions," Ars Combinatoria 39 (1995) 129-138.
D. R. Morrison, "A Stolarsky array of Wythoff pairs," in A Collection of Manuscripts Related to the Fibonacci Sequence. Fibonacci Assoc., Santa Clara, CA, 1980, pp. 134-136.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 9.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See pp. 3-4, 7, 10.
Clark Kimberling, Interspersions
Matthew Skala, Graph Nimors, arXiv preprint arXiv:1604.04072 [math.CO], 2016.
N. J. A. Sloane, Classic Sequences

Cf. A064437.
Apart from initial terms, same as A026356 (Cloitre (0,2,2,3)-hiccup sequence).
First column of A126714.
Complement is (essentially) A026355.
Equals 1 + A004957, also n + A004956.
First differences give A076662.
Complement of A099267. [Gerald Hillier, Dec 19 2008]
Cf. A193214 (primes). Except for the first term equal to A001950 + 2.
Cf. A026352 (Cloitre (1,1,2,3)-hiccup sequence), A064437 (Cloitre (0,1,3,2)-hiccup sequence).

a007066 n = a007066_list !! (n-1)
a007066_list = 1 : f 2 [1] where
   f x zs@(z:_) = y : f (x + 1) (y : zs) where
     y = if x `elem` zs then z + 2 else z + 3
-- Reinhard Zumkeller, Sep 26 2014, Sep 18 2011

Digits := 100: t := (1+sqrt(5))/2; A007066 := proc(n) if n <= 1 then 1 else floor(1+t*floor(t*(n-1)+1)); fi; end;

t = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0, 1}}] &, {0}, 6] (*A189479*)
Flatten[Position[t, 0]] (*A007066*)
Flatten[Position[t, 1]] (*A099267*)
With[{grs=GoldenRatio^2},Table[1+Ceiling[grs(n-1)],{n,70}]] (* Harvey P. Dale, Jun 24 2011 *)

from math import isqrt
def A007066(n): return (n+1+isqrt(5*(n-1)**2)>>1)+n if n > 1 else 1 # Chai Wah Wu, Aug 25 2022

a(n) = floor(1+phi*floor(phi*(n-1)+1)), phi=(1+sqrt(5))/2, n >= 2.
a(1)=1; for n>1, a(n)=a(n-1)+2 if n is already in the sequence, a(n)=a(n-1)+3 otherwise. - Benoit Cloitre, Mar 06 2003
a(n+1) = floor(n*phi^2) + 2, n>=1. - Michel Dekking, Aug 26 2019

A080652 a(1)=2; for n>1, a(n)=a(n-1)+3 if n is already in the sequence, a(n)=a(n-1)+2 otherwise.

Original entry on oeis.org

2, 5, 7, 9, 12, 14, 17, 19, 22, 24, 26, 29, 31, 34, 36, 38, 41, 43, 46, 48, 50, 53, 55, 58, 60, 63, 65, 67, 70, 72, 75, 77, 79, 82, 84, 87, 89, 92, 94, 96, 99, 101, 104, 106, 108, 111, 113, 116, 118, 121, 123, 125, 128, 130, 133, 135, 137, 140, 142, 145

Offset: 1

N. J. A. Sloane, Mar 23 2003

nonn

In the Fokkink-Joshi paper, this sequence is the Cloitre (0,2,3,2)-hiccup sequence. - Michael De Vlieger, Jul 29 2025

Vincenzo Librandi, Table of n, a(n) for n = 1..5000
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 7.
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, 9.

Cf. A080455-A080458, A080036, A080037. Apart from start, equals A064437 - 1.

[Floor(n*(1+Sqrt(2)) + 1/(1+(1+Sqrt(2)))): n in [1..60]]; // Vincenzo Librandi, Oct 02 2018

a[1] = 2;
a[n_] := a[n] = If[MemberQ[Array[a, n-1], n], a[n-1] + 3, a[n-1] + 2];
Array[a, 60] (* Jean-François Alcover, Oct 01 2018 *)
Table[Floor[n (1 + Sqrt[2]) + 1 / (1 + (1 + Sqrt[2]))], {n, 60}] (* Vincenzo Librandi, Oct 02 2018 *)

a(n) = my(r=sqrt(2)+1); (r*(r+1)*n+1)\(r+1); \\ Altug Alkan, Oct 01 2018

a(n) = floor(n*r + 1/(1+r)) where r = 1+sqrt(2).

A080782 a(1)=1, a(n)=a(n-1)-1 if n is already in the sequence, a(n)=a(n-1)+2 otherwise.

Original entry on oeis.org

1, 3, 2, 4, 6, 5, 7, 9, 8, 10, 12, 11, 13, 15, 14, 16, 18, 17, 19, 21, 20, 22, 24, 23, 25, 27, 26, 28, 30, 29, 31, 33, 32, 34, 36, 35, 37, 39, 38, 40, 42, 41, 43, 45, 44, 46, 48, 47, 49, 51, 50, 52, 54, 53, 55, 57, 56, 58, 60, 59, 61, 63, 62, 64, 66, 65, 67, 69, 68

Offset: 1

Benoit Cloitre, Mar 07 2003

Permutation of the integers: exchange trisections starting with 2 and 3.

a(a(n)) = n. - Reinhard Zumkeller, Oct 29 2004

Guenther Schrack, Table of n, a(n) for n = 1..10000
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 (1,0,1,-1).
Index entries for sequences that are permutations of the natural numbers.

Cf. A064429, A079357, A079354, A080783, A064437.

Cf. A004442, A080412.

Array[#+Mod[#+1,3]&,70,0] (* or *) LinearRecurrence[{1,0,1,-1},{1,3,2,4},70] (* Harvey P. Dale, Mar 29 2013 *)
{#,#+1,#-1}[[Mod[#,3,1]]]&/@Range[99] (* Federico Provvedi, May 15 2021 *)

a(n) = A064429(n-1) + 1.
a(n) - n is periodic with period 3.
G.f.: x*(1+2*x-x^2+x^3)/(1-x-x^3+x^4). - Jaume Oliver Lafont, Mar 24 2009
a(0)=1, a(1)=3, a(2)=2, a(3)=4, a(n)=a(n-1)+0*a(n-2)+a(n-3)-a(n-4). - Harvey P. Dale, Mar 29 2013
a(n) = n + (2/sqrt(3))*sin(2*(n+2)*Pi/3). - Wesley Ivan Hurt, Sep 26 2017
From Guenther Schrack, Oct 23 2019: (Start)
a(n) = a(n-3) + 3 with a(1) = 1, a(2) = 3, a(3) = 2 for n > 3.
a(n) = n - (w^(2*n)*(2 + w) + w^n*(1 - w))/3 where w = (-1 + sqrt(-3))/2. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/(3*sqrt(3)) - log(2)/3. - Amiram Eldar, Jan 31 2023
From Charles L. Hohn, Sep 03 2024: (Start)
a(n) = n-1+n%3.
a(n) = A375336(n-2, 1) for n >= 6. (End)

A086377 a(1)=1; a(n)=a(n-1)+2 if n is in the sequence; a(n)=a(n-1)+2 if n and (n-1) are not in the sequence; a(n)=a(n-1)+3 if n is not in the sequence but (n-1) is in the sequence.

Original entry on oeis.org

1, 4, 6, 8, 11, 13, 16, 18, 21, 23, 25, 28, 30, 33, 35, 37, 40, 42, 45, 47, 49, 52, 54, 57, 59, 62, 64, 66, 69, 71, 74, 76, 78, 81, 83, 86, 88, 91, 93, 95, 98, 100, 103, 105, 107, 110, 112, 115, 117, 120, 122, 124, 127, 129, 132, 134, 136, 139, 141, 144, 146, 148, 151

Offset: 1

Benoit Cloitre, Sep 13 2003

nonn

From Joseph Biberstine (jrbibers(AT)indiana.edu), May 02 2006: (Start)
The continued fraction 4/Pi = 1 + 1/(3 + 4/(5 + 9/(7 + 16/(9 + 25/(11 + ...))))) (see A079037) suggests the recurrence b(n) = 2*n - 1 + n^2/b(n+1) with b(1) = 4/Pi. Solving the above recurrence in the other direction we would have b(n) = (n-1)^2/b(n-1 - 2*n + 3) with b(1) = 4/Pi.
Now consider this last defined sequence {b(n)}. It appears to grow linearly. (1) Does it? (2) What is the limit of b(n)/n as n->oo? (3) How does the limit depend on the initial term b(1)? (End)
From the recurrence relation, it follows that the limit L = lim_{b->oo} b(n)/n satisfies the following quadratic equation: L^2 - 2*L - 1 = 0 implying that L = 1+sqrt(2) or 1-sqrt(2). - Max Alekseyev, May 02 2006
Note that b(n)/n decreases, while b(n)/(n+1) increases. I speculate that 4/Pi is the only b(1) value such that b(n)/n converges to 1+sqrt(2) instead of 1-sqrt(2). - Don Reble, May 02 2006
It appears that round( b(n) ) = floor((1+sqrt(2))*n - 1/sqrt(2)) = A086377(n) = a(n). This is certainly true for the first 190 terms. Is there a formal proof? - Paul D. Hanna, May 02 2006
Is A086377 the sequence of positions of 0 in A189687? - Clark Kimberling, Apr 25 2011
The three conjectures by respectively Biberstein, Hanna, and Kimberling have all been proved, see the paper by Bosma et al. in the Links. - Michel Dekking, Oct 05 2017
In the Fokkink-Joshi paper, this sequence is the Cloitre (1,1,3,2)-hiccup sequence, - Michael De Vlieger, Jul 29 2025

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Wieb Bosma, Michel Dekking, and Wolfgang Steiner, A remarkable sequence related to Pi and sqrt(2), arXiv:1710.01498 [math.NT], 2017.
Wieb Bosma, Michel Dekking, and Wolfgang Steiner, A remarkable sequence related to Pi and sqrt(2), Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A4.
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 7.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See pp. 3, 9, 15.
Wolfgang Steiner, Continued fractions and S-adic sequences, Numeration systems: automata, combinatorics, dynamical systems, number theory, Univ. de Paris (France, 2021), see p. 33.

Cf. A189687, A081477.

[Floor((1+Sqrt(2))*n-1/Sqrt(2)): n in [1..70]]; // Vincenzo Librandi, Oct 05 2017

Block[{c, k}, c[] := False; k = 1; c[1] = True; {k}~Join~Reap[Do[If[c[n], k += 2, If[c[n - 1], k += 3, k += 2]]; c[k] = True; Sow[k], {n, 2, 120}] ][[-1, 1]] ] (* _Michael De Vlieger, Jul 02 2025 *)

a(n) = floor((1+sqrt(2))*n - 1/sqrt(2)).

n is in the sequence if A004641(n)=1 or A001030(n)=2. a(n) = A080652(n) - 1 = A064437(n+1) - 2 = A081841(n+2) - 3. - Ralf Stephan, Feb 23 2004

A081834 a(1)=1, a(n)=a(n-1)+4 if n is already in the sequence, a(n)=a(n-1)+3 otherwise.

Original entry on oeis.org

1, 4, 7, 11, 14, 17, 21, 24, 27, 30, 34, 37, 40, 44, 47, 50, 54, 57, 60, 63, 67, 70, 73, 77, 80, 83, 87, 90, 93, 97, 100, 103, 106, 110, 113, 116, 120, 123, 126, 130, 133, 136, 139, 143, 146, 149, 153, 156, 159, 163, 166, 169, 172, 176, 179, 182, 186, 189, 192, 196

Offset: 1

Benoit Cloitre, Apr 11 2003

nonn

In the Fokkink-Joshi paper, this sequence is the Cloitre (0,1,4,3)-hiccup sequence, - Michael De Vlieger, Jul 29 2025

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 7.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See p. 3.

Cf. A064437, A080590, A080600, A080667, A081842.

a081834 n = a081834_list !! (n-1)
a081834_list = 1 : f 2 [1] where
   f x zs@(z:_) = y : f (x + 1) (y : zs) where
     y = z + (if x `elem` zs then 4 else 3)
-- Reinhard Zumkeller, Sep 26 2014

Block[{c, k, nn}, nn = 10^4; c[] := False; k = 1; c[1] = True; {k}~Join~Reap[Do[If[c[n], k += 4, k += 3]; c[k] = True; Sow[k], {n, 2, nn}] ][[-1, 1]] ] (* _Michael De Vlieger, Jul 02 2025 *)

a(n) = floor(rn-(3r-1)/(r+1)) where r = (1/2) *(3+sqrt(13)).

A079357 a(1)=1; a(n)=a(n-1)-1 if n is already in the sequence, a(n)=a(n-1)+4 otherwise.

Original entry on oeis.org

1, 5, 9, 13, 12, 16, 20, 24, 23, 27, 31, 30, 29, 33, 37, 36, 40, 44, 48, 47, 51, 55, 54, 53, 57, 61, 60, 64, 63, 62, 61, 65, 64, 68, 72, 71, 70, 74, 78, 77, 81, 85, 89, 88, 92, 96, 95, 94, 98, 102, 101, 105, 104, 103, 102, 106, 105, 109, 113, 112, 111, 110, 109, 108, 107

Offset: 1

Benoit Cloitre, Feb 14 2003

If, in the defining recurrence, the rule a(n)=a(n-1)+4 when n is not already in the sequence is generalized to a(n)=a(n-1)+k, then the resulting sequence ultimately becomes periodic with period 1,3,10,35 for k=1,2,3,4, respectively. - John W. Layman, Apr 15 2003

Cf. A080782, A079354, A079355, A080783, A064437.

It appears that, for n >= 219, a(n)=n+b(n) where b(n) is the period-35 sequence (-1, 2, 5, 3, 6, 9, 7, 5, 8, 11, 9, 12, 10, 8, 6, 9, 7, 10, 13, 11, 9, 7, 5, 3, 1, -1, -3, -5, -7, -9, -11, -13, -10, -7, -4).

More terms from John W. Layman, Apr 15 2003

A080704 a(1)=2; for n>1, if n is in the sequence then a(n) is the smallest even integer > a(n-1), otherwise a(n) = a(n-1) + 3.

Original entry on oeis.org

2, 4, 7, 8, 11, 14, 16, 18, 21, 24, 26, 29, 32, 34, 37, 38, 41, 42, 45, 48, 50, 53, 56, 58, 61, 62, 65, 68, 70, 73, 76, 78, 81, 82, 85, 88, 90, 92, 95, 98, 100, 102, 105, 108, 110, 113, 116, 118, 121, 122, 125, 128, 130, 133, 136, 138, 141, 142, 145, 148

Offset: 1

Benoit Cloitre, Mar 01 2003

Amiram Eldar, Table of n, a(n) for n = 1..10000
Amiram Eldar, Plot of a(n)/n for n = 10000..100000

Cf. A080705, A064437, A080717.

a[1] = 2; a[n_] := a[n] = Module[{s = Array[a, n-1]}, a[n-1] + If[MemberQ[s, n], If[OddQ[a[n-1]], 1, 2], 3]]; Array[a, 100] (* Amiram Eldar, May 14 2022 *)

for(n=2,400,an[n]=if(setsearch(Set(vector(n-1,i,a(i))),n),if(a(n-1)%2,a(n-1)+1,a(n-1)+2),a(n-1)+3))

It seems that a(n)/n -> 2.4799.... = 62/25 ?

A081835 a(1)=1, a(n) = a(n-1) + 5 if n is already in the sequence, a(n) = a(n-1) + 4 otherwise.

Original entry on oeis.org

1, 5, 9, 13, 18, 22, 26, 30, 35, 39, 43, 47, 52, 56, 60, 64, 68, 73, 77, 81, 85, 90, 94, 98, 102, 107, 111, 115, 119, 124, 128, 132, 136, 140, 145, 149, 153, 157, 162, 166, 170, 174, 179, 183, 187, 191, 196, 200, 204, 208, 212, 217, 221, 225, 229, 234, 238, 242

Offset: 1

Benoit Cloitre, Apr 11 2003

nonn

In the Fokkink-Joshi paper, this sequence is the Cloitre (0,1,5,4)-hiccup sequence. - Michael De Vlieger, Jul 29 2025

			a(2) = a(1)+4 = 5 because 2 is not already in the sequence;
a(3) = a(2)+4 = 9 because 3 is not already in the sequence;
a(4) = a(3)+4 = 13 because 4 is not already in the sequence;
a(5) = a(4)+5 = 18 because 5 is already in the sequence.

Muniru A Asiru, Table of n, a(n) for n = 1..2000
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See p. 3.

Cf. A064437, A081843.

r:=2+sqrt(5): seq(floor(r*n-(4*r-1)/(r+1)),n=1..60); # Muniru A Asiru, Jun 06 2018

Module[{r=2+Sqrt[5],c},c=(4r-1)/(r+1);Table[Floor[r*n-c],{n,60}]] (* Harvey P. Dale, Feb 19 2013 *)

a(n) = floor(rn-(4r-1)/(r+1)) where r=2+sqrt(5).

A079352 a(1)=1, then a(n)=3a(n-1) if n is already in the sequence, a(n)=2a(n-1) otherwise.

Original entry on oeis.org

1, 2, 4, 12, 24, 48, 96, 192, 384, 768, 1536, 4608, 9216, 18432, 36864, 73728, 147456, 294912, 589824, 1179648, 2359296, 4718592, 9437184, 28311552, 56623104, 113246208, 226492416, 452984832, 905969664, 1811939328, 3623878656

Offset: 1

Benoit Cloitre, Feb 14 2003

nonn

Inspired by A079000. Cf. A064437.

a(n)=3*(3/2)^floor((log(n)-log(3))/log(2))*2^n

a(n+1)=3*a(n) for n=3 n of the form 3*2^k - 1, k>=2 . a(n+1)=2*a(n) otherwise. Hence a(n)=3*(3/2)^floor((log(n/3))/log(2))*2^n.

cp's OEIS Frontend

A079000 a(n) is taken to be the smallest positive integer greater than a(n-1) which is consistent with the condition "n is a member of the sequence if and only if a(n) is odd".

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A007066 a(n) = 1 + ceiling((n-1)*phi^2), phi = (1+sqrt(5))/2.

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Haskell

Maple

Mathematica

Python

Formula

A080652 a(1)=2; for n>1, a(n)=a(n-1)+3 if n is already in the sequence, a(n)=a(n-1)+2 otherwise.

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Magma

Mathematica

PARI

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A080782 a(1)=1, a(n)=a(n-1)-1 if n is already in the sequence, a(n)=a(n-1)+2 otherwise.

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Mathematica

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A086377 a(1)=1; a(n)=a(n-1)+2 if n is in the sequence; a(n)=a(n-1)+2 if n and (n-1) are not in the sequence; a(n)=a(n-1)+3 if n is not in the sequence but (n-1) is in the sequence.

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Magma

Mathematica

Formula

A081834 a(1)=1, a(n)=a(n-1)+4 if n is already in the sequence, a(n)=a(n-1)+3 otherwise.

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Haskell

Mathematica

Formula

A079357 a(1)=1; a(n)=a(n-1)-1 if n is already in the sequence, a(n)=a(n-1)+4 otherwise.

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Extensions

A079352 a(1)=1, then a(n)=3a(n-1) if n is already in the sequence, a(n)=2a(n-1) otherwise.