

A075326


AntiFibonacci numbers: start with a(0) = 0, and extend by the rule that the next term is the sum of the two smallest numbers that are not in the sequence nor were used to form an earlier sum.


22



0, 3, 9, 13, 18, 23, 29, 33, 39, 43, 49, 53, 58, 63, 69, 73, 78, 83, 89, 93, 98, 103, 109, 113, 119, 123, 129, 133, 138, 143, 149, 153, 159, 163, 169, 173, 178, 183, 189, 193, 199, 203, 209, 213, 218, 223, 229, 233, 238, 243, 249, 253, 258, 263, 269, 273, 279, 283
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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 "antiFibonacci 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 nonterm 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 antitribonacci numbers, see A265389; for antitetranacci, see A299405.  Clark Kimberling, May 01 2018
We see the Fibonacci numbers 3, 13, 89 and 233 occur in this sequence of antiFibonacci 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



FORMULA

See Zaslavsky (2016) link.


MAPLE

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}];
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}}]
********
(* 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]]
********
(* 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]]


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
(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


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



