A004937 -id:A004937 - 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.

A004938 a(n) = round(n*phi^3), where phi is the golden ratio, A001622.

Original entry on oeis.org

0, 4, 8, 13, 17, 21, 25, 30, 34, 38, 42, 47, 51, 55, 59, 64, 68, 72, 76, 80, 85, 89, 93, 97, 102, 106, 110, 114, 119, 123, 127, 131, 136, 140, 144, 148, 152, 157, 161, 165, 169, 174, 178, 182, 186, 191, 195, 199

Offset: 0

N. J. A. Sloane

nonn

Nearest integer to n*phi^3.

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A004937, A004939 - A004955.

Cf. A001622.

[Round((2+Sqrt(5))*n): n in [0..80]]; // G. C. Greubel, Oct 17 2023

Round[GoldenRatio^3*Range[0, 80]] (* G. C. Greubel, Oct 17 2023 *)

a(n) = round(n*(((sqrt(5)+1))/2)^3); \\ Michel Marcus, Aug 28 2016

[round(golden_ratio^3*n) for n in range(81)] # G. C. Greubel, Oct 17 2023

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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Formula

A004938 a(n) = round(n*phi^3), where phi is the golden ratio, A001622.

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Magma

Mathematica

PARI

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