A117248 -id:A117248 - cp's OEIS frontend

A079051 Recamán variation: a(0) = 0; for n >= 1, a(n) = a(n-1) - f(n) if that number is positive and not already in the sequence, otherwise a(n) = a(n-1) + f(n), where f(n) = floor(sqrt(n)) (A000196).

0, 1, 2, 3, 5, 7, 9, 11, 13, 10, 13, 16, 19, 22, 25, 28, 24, 20, 24, 28, 32, 36, 40, 44, 48, 43, 38, 33, 38, 43, 48, 53, 58, 63, 68, 73, 67, 61, 55, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 96, 89, 82, 75, 82, 89, 96, 103, 110, 117, 124, 131, 138, 145, 152, 144, 136, 128, 120

Offset: 0

Benoit Cloitre, Feb 02 2003

nonn

N. J. A. Sloane and Allan Wilks, On sequences of Recaman type, paper in preparation, 2006.

Ivan Neretin, Table of n, a(n) for n = 0..10000
Nick Hobson, Python program for this sequence

Cf. A000196, A005132. Numbers missed are in A117247.

Cf. A117248, A117516, A117518.

Fold[Append[#1, If[MemberQ[#1, (a = #1[[-1]]) - (r = Floor@Sqrt@#2)], a + r, a - r]] &, {0, 1}, Range[2, 70]] (* Ivan Neretin, Apr 22 2018 *)

lista(nn) = {va = vector(nn+1); last = 0; for (n=1, nn, new = last - sqrtint(n); if ((new <= 0) || vecsearch(vecsort(va,,8), new), new = last + sqrtint(n)); va[n+1] = new; last = new;); va;} \\ Michel Marcus, Apr 23 2018

Conjecture: for n>100, 1/2 < a(n)/(n*log(n)) < 1.

The conjecture is false. In fact, a(n) = n^(3/2)/6 + O(n). - N. J. A. Sloane, Apr 29 2006

A352745 a(n) is the number of Lyndon factors of the Fibonacci string of length n.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 4, 6, 5, 8, 6, 10, 7, 12, 8, 14, 9, 16, 10, 18, 11, 20, 12, 22, 13, 24, 14, 26, 15, 28, 16, 30, 17, 32, 18, 34, 19, 36, 20, 38, 21, 40, 22, 42, 23, 44

Offset: 0

Peter Luschny, Apr 06 2022

The Fibonacci string of length n is defined Fibonacci(n) = cat(Fibonacci(n - 1), Fibonacci(n - 2)) for 1 < n and the initial conditions Fibonacci(0) = "1" and Fibonacci(1) = "0", where 'cat' is the operation of concatenating strings. The length of Fibonacci(n) is A352744(1, n). The sequence starts: "1", "0", "01", "010", "01001", "01001010", ...

Apart from the first four terms seems to be identical with A117248.

			The Lyndon factorization of the Fibonacci strings of length n = 0..9.
[0] ["1"]
[1] ["0"]
[2] ["01"]
[3] ["01", "0"]
[4] ["01", "001"]
[5] ["01", "00101", "0"]
[6] ["01", "00101", "001", "001"]
[7] ["01", "00101", "0010010100101", "0"]
[8] ["01", "00101", "0010010100101", "00100101", "001", "001"]
[9] ["01", "00101", "0010010100101", "0010010100100101001010010010100101", "0"]

Guy Melançon, Lyndon factorization of infinite words, STACS 96 (Grenoble, 1996), 147-154, Lecture Notes in Comput. Sci., 1046, Springer, Berlin, 1996.
Wikipedia, Lyndon word

Cf. A000045, A014707, A074650, A117248, A211100, A352746, A352744.

with(StringTools): A352745 := n -> nops(LyndonFactors(Fibonacci(n))):
seq(A352745(n), n = 0..12);

cp's OEIS Frontend

A079051 Recamán variation: a(0) = 0; for n >= 1, a(n) = a(n-1) - f(n) if that number is positive and not already in the sequence, otherwise a(n) = a(n-1) + f(n), where f(n) = floor(sqrt(n)) (A000196).

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