OFFSET

0,2

COMMENTS

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

LINKS

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

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

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

FORMULA

See Zaslavsky (2016) link.

MAPLE

# 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))];

MATHEMATICA

(* 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 *)

PROG

(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

CROSSREFS

KEYWORD

nonn

AUTHOR

Amarnath Murthy, Sep 16 2002

EXTENSIONS

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)

STATUS

approved