A105672 -id:A105672 - cp's OEIS frontend

A105774 A "fractal" transform of the Fibonacci numbers: a(1)=1; then if F(n) < k <= F(n+1), a(k) = F(n+1) - a(k - F(n)) where F(n) = A000045(n).

1, 1, 2, 4, 4, 7, 7, 6, 12, 12, 11, 9, 9, 20, 20, 19, 17, 17, 14, 14, 15, 33, 33, 32, 30, 30, 27, 27, 28, 22, 22, 23, 25, 25, 54, 54, 53, 51, 51, 48, 48, 49, 43, 43, 44, 46, 46, 35, 35, 36, 38, 38, 41, 41, 40, 88, 88, 87, 85, 85, 82, 82, 83, 77, 77, 78, 80, 80, 69, 69, 70, 72, 72

Offset: 1

Benoit Cloitre, May 04 2005

nonn

Let tau = (1+sqrt(5))/2; then the missing numbers 3,5,8,10,13,16,18,21,... are given by round(tau^2*k) for k > 0 (A004937).
Indices n such that a(n) = a(n+1) are given by floor(tau^2*k) - 1 for k > 0 (A003622).
Numbers n such that a(n) differs from a(n+1) are given by floor(tau*k+1/tau) for k > 0 (A022342).
Indices n giving isolated terms (a(n) differs from a(n-1) and a(n+1)) are given by floor(tau*floor(tau^2*k)) for k > 0 (A003623).
Remove 0's from the first differences of sorted values; then you get a version of the infinite Fibonacci word (A001468). I.e., sorted values are 1,1,2,4,4,6,7,7,9,9,11,12,12,..., first differences are 0,1,2,0,2,1,0,2,0,2,1,0,2,0,1,...; removing 0's gives 1,2,2,1,2,2,1,2,1,2,2,1,2,1,2,2,1,2,... #{ k : a(k)=k}=infty.

			For 1 = F(2) < k <= F(3) = 2 the rule gives a(2) = 2 - a(1) = 1 ... if 5 = F(5) < k <= F(6) = 8 the rule forces a(6) = 8 - a(6-5) = 8 - a(1) = 7; a(7) = 8 - a(2) = 7; a(8) = 8 - a(3) = 6.

Antti Karttunen, Table of n, a(n) for n = 1..10946
Benoit Cloitre and Jeffrey Shallit, Some Fibonacci-Related Sequences, arXiv:2312.11706 [math.CO], 2023.
Jeffrey Shallit, Using automata to prove theorems about sequences, One FLAT World Seminar, 2024.

Cf. A000045, A001468, A003622, A003623, A004937, A006498, A022342, A072649, A085002, A105669, A105670, A105672, A093347, A093348, A283766.

a(A000045(n)) = A006498(n-1) for n >= 1. - Typo corrected by Antti Karttunen, Mar 17 2017
limsup a(n)/n = tau and liminf a(n)/n = (tau+2)/5 where tau = (1+sqrt(5))/2. - Corrected by Jeffrey Shallit, Dec 17 2023
a(n) mod 2 = A085002(n) - Benoit Cloitre, May 10 2005
a(1) = 1; for n > 1, a(n) = A000045(2+A072649(n-1)) - a(n-A000045(1 + A072649(n-1))). - Antti Karttunen, Mar 17 2017

A105669 A "fractal" transform of the Fibonacci numbers F(n)=A000045(n): a(1)=1, then for n>1 if F(n) < k < F(n+1) we have a(k) = F(n+1)-a(k-F(n)) and when k = F(n+1) we force a(F(n+1)) = F(n+1) + (1+(-1)^n)*F(n).

Original entry on oeis.org

1, 2, 2, 4, 7, 7, 6, 6, 12, 11, 11, 9, 20, 20, 19, 19, 17, 14, 14, 15, 15, 33, 32, 32, 30, 27, 27, 28, 28, 22, 23, 23, 25, 54, 54, 53, 53, 51, 48, 48, 49, 49, 43, 44, 44, 46, 35, 35, 36, 36, 38, 41, 41, 40, 40, 88, 87, 87, 85, 82, 82, 83, 83, 77, 78, 78, 80, 69, 69, 70, 70, 72

Offset: 1

Benoit Cloitre, May 03 2005

nonn

Let b denote the sequence of n such that a(n)=a(n+1), then b(n)=floor(tau^2*n) where tau=(1+sqrt(5))/2.
Missing numbers are the nearest integer to tau^2*n, n>=0 (cf. A004937).
#{k>0 : a(k) = k} = infinity.
This kind of "fractal" transform can be applied to any increasing monotonic sequence giving true fractal properties for sequences = (m^n)_{n>0} with m integer >=2, specially when m is odd (cf. A093347, A093348).

			For 5 = F(5) < k <= F(6) = 8 we get a(6) = 8-a(6-5) = 8-a(1) = 7.
a(7) = 8-a(7-5) = 8-a(2) = 6.
a(8) = 8-a(8-5) = 8-a(3) = 6.

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

Cf. A000045, A001622, A004937, A105670, A105672, A093347, A093348.

a[n_] := a[n] = If[n <= 1, 1, Fibonacci[(k = Floor[Log[Sqrt[5]*n]/Log[GoldenRatio]]) + 1] - a[n - Fibonacci[k]]]; Array[a, 100] (* Amiram Eldar, Jun 17 2022 *)

f=(1+sqrt(5))/2; a(n)=if(n<2,1,fibonacci(floor(log(sqrt(5)*n)/log(f))+1)-a(n-fibonacci(floor(log(sqrt(5)*n)/log(f)))))

F(2n) = F(2n+1) - F(n+1)^2 + F(n)*F(n-1) for n>0.
a(F(2n-1)) = F(2n)-1 for n>1.
1/tau < a(n)/n < tau.

A105670 a(1)=1 then bracketing n by powers of 2 as f(t)=2^t for f(t) < n <= f(t+1), a(n) = f(t+1) - a(n-f(t)).

Original entry on oeis.org

1, 1, 3, 3, 7, 7, 5, 5, 15, 15, 13, 13, 9, 9, 11, 11, 31, 31, 29, 29, 25, 25, 27, 27, 17, 17, 19, 19, 23, 23, 21, 21, 63, 63, 61, 61, 57, 57, 59, 59, 49, 49, 51, 51, 55, 55, 53, 53, 33, 33, 35, 35, 39, 39, 37, 37, 47, 47, 45, 45, 41, 41, 43, 43, 127, 127, 125, 125, 121, 121, 123

Offset: 1

Benoit Cloitre, May 03 2005

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A006068, A105669, A105672, A093347, A093348.

A062383 := proc(n)
        ceil(log(n)/log(2)) ;
        2^% ;
end proc:
A105670 := proc(n)
        option remember;
        if n = 1 then
                1;
        else
                fn1 := A062383(n) ;
                fn := fn1/2 ;
                fn1-procname(n-fn) ;
        end if;
end proc:
seq(A105670(n),n=1..80) ; # R. J. Mathar, Nov 06 2011

t[0] = 0; t[1] = 1; t[n_?EvenQ] := t[n] = t[n/2]; t[n_?OddQ] := t[n] = 1 - t[(n-1)/2]; a[1] = 1; a[n_?EvenQ] := a[n] = a[n - 1]; a[n_] := a[n] = 2*a[Ceiling[n/2]] - 1 + 2*t[Ceiling[n/2] - 1]; Table[a[n], {n, 1, 71}] (* Jean-François Alcover, Aug 13 2013 *)

b(n,m)=if(n<2,1,m*m^floor(log(n-1)/log(m))-b(n-m^floor(log(n-1)/log(m)),m))

a(2n-1) = a(2n).
a(n) = 2*a(ceiling(n/2)) -1 + 2*t(ceiling(n/2)-1) where t(n) = A010060(n) is the Thue-Morse sequence.
a(2n-1) = a(2n) = 2*A006068(n-1)+1. - Jeffrey Shallit, Mar 15 2025

Typo in data corrected by Jean-François Alcover, Aug 13 2013

A106026 A fractal transform of Pell numbers : a(1)=1 then if b(n)A000129(n).

Original entry on oeis.org

1, 1, 4, 4, 1, 11, 11, 8, 8, 11, 1, 1, 28, 28, 25, 25, 28, 18, 18, 21, 21, 18, 28, 28, 1, 1, 4, 4, 1, 69, 69, 66, 66, 69, 59, 59, 62, 62, 59, 69, 69, 42, 42, 45, 45, 42, 52, 52, 49, 49, 52, 42, 42, 69, 69, 66, 66, 69, 1, 1, 4, 4, 1, 11, 11, 8, 8, 11, 1, 1, 168, 168, 165, 165, 168, 158

Offset: 1

Benoit Cloitre, May 05 2005

nonn

Graphs of a(k) for k=1 up to A000129(n) and n=1,2,3,... present fractal aspects.

Cf. A105669 (fractal transform of Fibonacci's numbers), A105670 (fractal transform of powers of 2), A105672(fractal transform of powers of 3).

Among many properties a(A000129(n))=1

cp's OEIS Frontend

A105774 A "fractal" transform of the Fibonacci numbers: a(1)=1; then if F(n) < k <= F(n+1), a(k) = F(n+1) - a(k - F(n)) where F(n) = A000045(n).

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A105669 A "fractal" transform of the Fibonacci numbers F(n)=A000045(n): a(1)=1, then for n>1 if F(n) < k < F(n+1) we have a(k) = F(n+1)-a(k-F(n)) and when k = F(n+1) we force a(F(n+1)) = F(n+1) + (1+(-1)^n)*F(n).

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A105670 a(1)=1 then bracketing n by powers of 2 as f(t)=2^t for f(t) < n <= f(t+1), a(n) = f(t+1) - a(n-f(t)).

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A106026 A fractal transform of Pell numbers : a(1)=1 then if b(n)A000129(n).

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