0,2

In more detail, the sequence is constructed as follows: Start with a(0) = 0. The missing numbers are 1 2 3 4 5 6 ... Add the first two, and we get 3, which is therefore a(1). Cross 1, 2, and 1+2=3 off the missing list. The first two missing numbers are now 4 and 5, so a(2) = 4+5 = 9. Cross off 4,5,9 from the missing list. Repeat.

In other words, this is the sum of consecutive pairs in the sequence 1, 2, 4, 5, 6, 7, 8, 10, 11, 12, 14, 15, ..., (A249031) the complement to the present one in the natural numbers. For example, a(1)=1+2=3, a(2)=4+5=9, a(3)=6+7=13, ... - Philippe Lallouet (philip.lallouet(AT)orange.fr), May 08 2008

The new definition is due to Philippe Lalloue (philip.lallouet(AT)orange.fr), May 08 2008, while the name "anti-Fibonacci numbers" is due to D. R. Hofstadter, Oct 23 2014.

Original definition: second members of pairs in A075325.

If instead we take the sum of the last used non-term and the most recent (i.e., 1+2, 2+4, 4+5, 5+7, etc.), we get A008585. - Jon Perry, Nov 01 2014

The sequences a = A075325, b = A047215, and c = A075326 are the solutions of the system of complementary equations defined recursively as follows:

a(n) = least new,

b(n) = least new,

c(n) = a(n) + b(n),

where "least new k" means the least positive integer not yet placed. For anti-tribonacci numbers, see A265389; for anti-tetranacci, see A299405. - Clark Kimberling, May 01 2018

We see the Fibonacci numbers 3, 13, 89 and 233 occur in this sequence of anti-Fibonacci numbers. Are there infinitely many Fibonacci numbers occurring in (a(n))? The answer is yes: at least 13% of the Fibonacci numbers occur in (a(n)). This follows from Thomas Zaslavsky's formula, which implies that the sequence A017305 = (10n+3) is a subsequence of (a(n)). The Fibonacci sequence A000045 modulo 10 equals A003893, and has period 60. In this period, the number 3 occurs 8 times. - Michel Dekking, Feb 14 2019

From Augusto Santi, Aug 16 2025: (Start)

If we apply the anti-Fibonacci algorithm to the set of natural numbers minus the multiples of 3, we get 5, 10, 20, 25, 35, 40, 50, ...; that is, all the multiples of 5 present in the restricted set used. It is quite curious that in this particular case the algorithm can be applied recursively to its own output, generating, at the generic step s, the subset of multiples of 5^s (see Mathematics StackExchange link).

Conjectures:

After the first 0, the residues (mod 5) all fall in the classes 3 and 4. More generally, for k-nacci sequences the residue classes (mod k^2+1) all fall in k consecutive ones, the first being ceiling((k^2+1)/2).

It is known that the sequence contains the arithmetic progression 10k+3, 20k+9 and 40k+18. These three progressions cover, experimentally, the 87.5% = 7/8 of the entire sequence. The remaining terms all belong to two forms: 40k+38 and 40k+39.

The anti-Fibonacci sequence contains all the squares of the numbers of the form 10k+3 and 10k+7, and all the cubes of the numbers of the form 10k+7, for k>=0. (End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Wieb Bosma, Rene Bruin, Robbert Fokkink, Jonathan Grube, Anniek Reuijl, and Thian Tromp, Using Walnut to solve problems from the OEIS, arXiv:2503.04122 [math.NT], 2025. See pp. 7, 14.

Robbert Fokkink and Gandhar Joshi, Anti-recurrence sequences, arXiv:2506.13337 [math.NT], 2025. See pp. 2, 18.

D. R. Hofstadter, Anti-Fibonacci numbers, Oct 23 2014.

Augusto Santi, A conjecture on Anti-k-nacci numbers, Mathematics StackExchange, 2025.

Thomas Zaslavsky, Anti-Fibonacci Numbers: A Formula, Sep 26 2016

See Zaslavsky (2016) link.

# Maple code for M+1 terms of sequence, from N. J. A. Sloane, Oct 26 2014

c:=0; a:=[c]; t:=0; M:=100;

for n from 1 to M do

s:=t+1; if s in a then s:=s+1; fi;

t:=s+1; if t in a then t:=t+1; fi;

c:=s+t;

a:=[op(a), c];

od:

[seq(a[n], n=1..nops(a))];

(* Three sequences a, b, c as in Comments *)

z = 200;

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);

a = {}; b = {}; c = {};

Do[AppendTo[a,

mex[Flatten[{a, b, c}], If[Length[a] == 0, 1, Last[a]]]];

AppendTo[b, mex[Flatten[{a, b, c}], Last[a]]];

AppendTo[c, Last[a] + Last[b]], {z}];

Take[a, 100] (* A075425 *)

Take[b, 100] (* A047215 *)

Take[c, 100] (* A075326 *)

Grid[{Join[{"n"}, Range[0, 20]], Join[{"a(n)"}, Take[a, 21]],

Join[{"b(n)"}, Take[b, 21]], Join[{"c(n)"}, Take[c, 21]]},

Alignment -> ".",

Dividers -> {{2 -> Red, -1 -> Blue}, {2 -> Red, -1 -> Blue}}]

(* Peter J. C. Moses, Apr 26 2018 *)

********

(* Sequence "a" via A035263 substitutions *)

Accumulate[Prepend[Flatten[Nest[Flatten[# /. {0 -> {1, 1}, 1 -> {1, 0}}] &, {0}, 7] /. Thread[{0, 1} -> {{5, 5}, {6, 4}}]], 3]]

(* Peter J. C. Moses, May 01 2018 *)

********

(* Sequence "a" via Hofstadter substitutions; see his 2014 link *)

morph = Rest[Nest[Flatten[#/.{1->{3}, 3->{1, 1, 3}}]&, {1}, 6]]

hoff = Accumulate[Prepend[Flatten[morph/.Thread[{1, 3}->{{6, 4, 5, 5}, {6, 4, 6, 4, 6, 4, 5, 5}}]], 3]]

(* Peter J. C. Moses, May 01 2018 *)

(Haskell)

import Data.List ((\\))

a075326 n = a075326_list !! n

a075326_list = 0 : f [1..] where

f ws@(u:v:_) = y : f (ws \\ [u, v, y]) where y = u + v

-- Reinhard Zumkeller, Oct 26 2014

(Python)

def aupton(nn):

alst, disallowed, mink = [0], {0}, 1

for n in range(1, nn+1):

nextk = mink + 1

while nextk in disallowed: nextk += 1

an = mink + nextk

alst.append(an)

disallowed.update([mink, nextk, an])

mink = nextk + 1

while mink in disallowed: mink += 1

return alst

print(aupton(57)) # Michael S. Branicky, Jan 31 2022

(Python)

def A075326(n): return 5*n-1-int((n|(~((m:=n-1>>1)+1)&m).bit_length())&1) if n else 0 # Chai Wah Wu, Sep 11 2024

Cf. A008585, A075325, A075327, A249031, A249032 (first differences), A000045.

Cf. also A079523, A131323, A249031, A249032, A249406.

Sequence in context: A240108 A163595 A088090 * A298870 A095234 A240240

Adjacent sequences: A075323 A075324 A075325 * A075327 A075328 A075329

Amarnath Murthy, Sep 16 2002

More terms from David Wasserman, Jan 16 2005

Entry revised (including the addition of an initial 0) by N. J. A. Sloane, Oct 26 2014 and Sep 26 2016 (following a suggestion from Thomas Zaslavsky)

approved