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Lengths of suffix blocks associated with A090822.
16

%I #36 Nov 15 2022 17:09:29

%S 1,3,1,9,4,24,1,3,1,9,4,67,1,3,1,9,4,24,1,3,1,9,4,196,3,1,9,4,24,1,3,

%T 1,9,4,68,3,1,9,4,24,1,3,1,9,4,581,3,1,9,4,25,3,1,9,4,67,1,3,1,9,4,24,

%U 1,3,1,9,4,196,3,1,9,4,24,1,3,1,9,4,68,3,1,9,4,24,1,3,1,9,4,1731,3,1,9,4,24

%N Lengths of suffix blocks associated with A090822.

%C The suffix blocks are what is called "glue string" in the paper by Gijswijt et al (2007). Roughly speaking, these are the terms >= 2 appended before the sequence (A090822) goes on with a(n+1) = 1 followed by all other initial terms a(2..n), cf. Example. The concatenation of these glue strings yields A091787. - _M. F. Hasler_, Aug 08 2018

%H Dion Gijswijt, <a href="/A091579/b091579.txt">Table of n, a(n) for n = 1..2000</a>

%H F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Sloane/sloane55.html">A Slow-Growing Sequence Defined by an Unusual Recurrence</a>, J. Integer Sequences, Vol. 10 (2007), #07.1.2.

%H Levi van de Pol, <a href="https://arxiv.org/abs/2209.04657">The first occurrence of a number in Gijswijt's sequence</a>, arXiv:2209.04657 [math.CO], 2022.

%H <a href="/index/Ge#Gijswijt">Index entries for sequences related to Gijswijt's sequence</a>

%e From _M. F. Hasler_, Aug 09 2018:

%e In sequence A090822, after the initial (1, 1) follows the first suffix block or glue string (2) of length a(1) = 1. This is followed by A090822(4) = 1 which indicates that the suffix block has ended, and the whole sequence A090822(1..3) up to and including this suffix block is repeated: A090822(4..6) = A090822(1..3).

%e Then A090822 goes on with (2, 2, 3, 1, ...), which tells that the second suffix block is A090822(7..9) = (2, 2, 3) of length a(2) = 3, whereafter the sequence starts over again: A090822(10..18) = A090822(1..9). (End)

%o (Python)

%o # compute curling number of L

%o def curl(L):

%o n = len(L)

%o m = 1 #max nr. of repetitions at the end

%o k = 1 #length of repeating block

%o while(k*(m+1) <= n):

%o good = True

%o i = 1

%o while(i <= k and good):

%o for t in range(1, m+1):

%o if L[-i-t*k] != L[-i]:

%o good = False

%o i = i+1

%o if good:

%o m = m+1

%o else:

%o k = k+1

%o return m

%o # compute lengths of first n glue strings

%o def A091579_list(n):

%o Promote = [1] #Keep track of promoted elements

%o L = [2]

%o while len(Promote) <= n:

%o c = curl(L)

%o if c < 2:

%o Promote = Promote+[len(L)+1]

%o c = 2

%o L = L+[c]

%o return [Promote[i+1]-Promote[i] for i in range(n)]

%o # _Dion Gijswijt_, Oct 08 2015

%Y Cf. A090822, A091587 (records). For a smoothed version see A091839.

%Y Cf. A091787 for the concatenation of the glue strings.

%K nonn

%O 1,2

%A _N. J. A. Sloane_, Mar 05 2004