A091787 - OEIS

A091787

a(1) = 2. To get a(n+1), write the string a(1)a(2)...a(n) as xy^k for words x and y (where y has positive length) and k is maximized, i.e., k = the maximal number of repeating blocks at the end of the sequence so far. Then a(n+1) = max(k,2).

2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 3, 3, 4, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 3, 3, 4, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Here xy^k means the concatenation of the words x and k copies of y.

a(77709404388415370160829246932345692180) = 5 is the first time 5 appears.

This is also the concatenation of the glue strings of A090822, whose respective lengths are given in A091579. - M. F. Hasler, Oct 04 2018

This sequence is called the level-2 Gijswijt sequence.

REFERENCES

N. J. A. Sloane, Seven Staggering Sequences, in Homage to a Pied Puzzler, E. Pegg Jr., A. H. Schoen and T. Rodgers (editors), A. K. Peters, Wellesley, MA, 2009, pp. 93-110.

LINKS

Giovanni Resta, Table of n, a(n) for n = 1..10000

F. J. van de Bult, D. C. Gijswijt, J. P. Linderman, N. J. A. Sloane and Allan Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), #07.1.2.

B. Chaffin, J. P. Linderman, N. J. A. Sloane and Allan Wilks, On Curling Numbers of Integer Sequences, arXiv:1212.6102 [math.CO], Dec 25 2012.

B. Chaffin, J. P. Linderman, N. J. A. Sloane and Allan Wilks, On Curling Numbers of Integer Sequences, Journal of Integer Sequences, Vol. 16 (2013), Article 13.4.3.

N. J. A. Sloane, Seven Staggering Sequences.

Index entries for sequences related to Gijswijt's sequence

Levi van de Pol, The first occurrence of a number in Gijswijt's sequence, arXiv:2209.04657 [math.CO], 2022.

EXAMPLE

To get a(2): a(1) = 2 = (2)^1, so k = 1, a(2) = 2.

To get a(3): a(1)a(2) = 22 = (2)^2, so a(3) = k = 2.

To get a(4): a(1)a(2)a(3) = 222 = (2)^3, so a(3) = k = 3.

PROG

(PARI) A091787(n, A=[])={while(#A<n, my(k=2, L=0, m=k); while((k+1)*(L+1)<=#A, for(N=L+1, #A/(m+1), A[-m*N..-1]==A[-(m+1)*N..-N-1]&&(m+=1)&&break); m>k||break; k=m); A=concat(A, k)); A} \\ M. F. Hasler, Oct 04 2018

(Python)

from itertools import islice

def c(w):

for k in range(len(w), 0, -1):

for l in range(1, len(w)//k + 1):

if w[-k*l:] == w[-l:]*k: return k

def agen(): # generator of terms

alst, an = [], 2

while True: yield an; alst.append(an); an = max(2, c(alst))

print(list(islice(agen(), 99))) # Michael S. Branicky, Sep 10 2022

CROSSREFS

Cf. A090822, A091799, A091840.