A091579 Lengths of suffix blocks associated with A090822.

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, 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, 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

Offset: 1

Author

N. J. A. Sloane, Mar 05 2004

Keywords

• nonn

Comments

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

Examples

			From _M. F. Hasler_, Aug 09 2018: (Start)
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).
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)
		

Links

• Dion Gijswijt, Table of n, a(n) for n = 1..2000
• Fokko J. van de Bult, Dion C. Gijswijt, John P. Linderman, N. J. A. Sloane, and Allan R. Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), #07.1.2.
• Levi van de Pol, The first occurrence of a number in Gijswijt's sequence, arXiv:2209.04657 [math.CO], 2022.
• Levi van de Pol, The Growth Rate of Gijswijt's Sequence, J. Int. Seq. (2025) Vol. 28, Art. No. 25.4.6.
• Index entries for sequences related to Gijswijt's sequence

Crossrefs

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

Cf. A091787 for the concatenation of the glue strings.

Programs

Python

# compute curling number of L
def curl(L):
    n = len(L)
    m = 1 #max nr. of repetitions at the end
    k = 1 #length of repeating block
    while(k*(m+1) <= n):
        good = True
        i = 1
        while(i <= k and good):
            for t in range(1, m+1):
                if L[-i-t*k] != L[-i]:
                    good = False
            i = i+1
        if good:
            m = m+1
        else:
            k = k+1
    return m
# compute lengths of first n glue strings
def A091579_list(n):
    Promote = [1] #Keep track of promoted elements
    L = [2]
    while len(Promote) <= n:
        c = curl(L)
        if c < 2:
            Promote = Promote+[len(L)+1]
            c = 2
        L = L+[c]
    return [Promote[i+1]-Promote[i] for i in range(n)]
# Dion Gijswijt, Oct 08 2015