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A090822 Gijswijt's sequence: a(1) = 1; for n>1, a(n) = largest integer k such that the word a(1)a(2)...a(n-1) is of the form xy^k for words x and y (where y has positive length), i.e., the maximal number of repeating blocks at the end of the sequence so far. 84
1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
Here xy^k means the concatenation of the words x and k copies of y.
The name "Gijswijt's sequence" is due to N. J. A. Sloane, not the author!
Fix n and suppose a(n) = k. Let len_y(n) = length of shortest y for this k and let len_x = n-1 - k*len_y(n) = corresponding length of x. A091407 and A091408 give len_y and len_x. For the subsequence when len_x = 0 see A091410 and A091411.
The first 4 occurs at a(220) (see A091409).
The first 5 appears around term 10^(10^23).
We believe that for all N >= 6, the first time N appears is at about position 2^(2^(3^(4^(5^...^(N-1))))). - N. J. A. Sloane and Allan Wilks, Mar 14 2004
For a similar formula, see p. 6 of Levi van de Pol article. - Levi van de Pol, Feb 06 2023
In the first 100000 terms the fraction of [1's, 2's, 3's, 4's] seems to converge, to about [.287, .530, .179, .005] respectively. - Allan Wilks, Mar 04 2004
When k=12 is reached, say, it is treated as the number 12, not as 1,2. This is not a base-dependent sequence.
Does this sequence have a finite average? Does anyone know the exact value? - Franklin T. Adams-Watters, Jan 23 2008
Answer: Given that "...the fraction of [1's, 2's, 3's, 4's] seems to converge, to about [.287, .530, .179, .005]..." that average should be the dot product of these vectors, i.e., about 1.904. - M. F. Hasler, Jan 24 2008
Second answer: The asymptotic densities of the numbers exist, and the average is the dot product. See pp. 56-59 of Levi van de Pol article. - Levi van de Pol, Feb 06 2023
Which is the first step with two consecutive 4's? Or the shortest run found so far between two 4's? - Sergio Pimentel, Oct 10 2016
Answer: The first x such that the x-th and (x+1)-th element are 4, is 255895648634818208370064452304769558261700170817472823... ...398081655524438021806620809813295008281436789493636144. See p. 55 of Levi van de Pol article. - Levi van de Pol, Feb 06 2023
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
Andreas Abel and Andres Loeh, Haskell program for A090822
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.
Benjamin Chaffin and N. J. A. Sloane, The Curling Number Conjecture, preprint.
D. C. Gijswijt, Krulgetallen, Pythagoras, 55ste Jaargang, Nummer 3, Jan 2016, (Article in Dutch about this sequence, see pages 10-13, cover and back cover).
Levi van de Pol, The first occurrence of a number in Gijswijt's sequence, arXiv:2209.04657 [math.CO], 2022.
N. J. A. Sloane, Seven Staggering Sequences.
N. J. A. Sloane, Confessions of a Sequence Addict (AofA2017), slides of invited talk given at AofA 2017, Jun 19 2017, Princeton. Mentions this sequence.
MAPLE
K:= proc(L)
local n, m, k, i, b;
m:= 0;
n:= nops(L);
for k from 1 do
if k*(m+1) > n then return(m) fi;
b:= L[-k..-1];
for i from 1 while i*k <= n and L[-i*k .. -(i-1)*k-1] = b do od:
m:= max(m, i-1);
od:
end proc:
A[1]:= 1:
for i from 2 to 220 do
A[i]:= K([seq(A[j], j=1..i-1)])
od:
seq(A[i], i=1..220); # Robert Israel, Jul 02 2015
MATHEMATICA
ClearAll[a]; reversed = {a[2]=1, a[1]=1}; blocs[len_] := Module[{bloc1, par, pos}, bloc1 = Take[reversed, len]; par = Partition[ reversed, len]; pos = Position[par, bloc_ /; bloc != bloc1, 1, 1]; If[pos == {}, Length[par], pos[[1, 1]] - 1]]; a[n_] := a[n] = Module[{an}, an = Table[{blocs[len], len}, {len, 1, Quotient[n-1, 2]}] // Sort // Last // First; PrependTo[ reversed, an]; an]; A090822 = Table[a[n], {n, 1, 99}] (* Jean-François Alcover, Aug 13 2012 *)
PROG
(Haskell) see link.
(PARI) A090822(n, A=[])={while(#A<n, my(k=1, 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, Aug 08 2018
(Python)
def k(s):
maxk = 1
for m in range(1, len(s)+1):
i, y, kk = 1, s[-m:], len(s)//m
if kk <= maxk: return maxk
while s[-(i+1)*m:-i*m] == y: i += 1
maxk = max(maxk, i)
def aupton(terms):
alst = [1]
for n in range(2, terms+1):
alst.append(k(alst))
return alst
print(aupton(99)) # Michael S. Branicky, Mar 28 2022
CROSSREFS
A091412 gives lengths of runs. A091413 gives partial sums.
Generalizations: A094781, A091975, A091976, A092331-A092335.
Sequence in context: A053633 A216460 A156755 * A091975 A091976 A267825
KEYWORD
nonn,nice
AUTHOR
Dion Gijswijt, Feb 27 2004
STATUS
approved

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Last modified February 27 08:19 EST 2024. Contains 370367 sequences. (Running on oeis4.)