|
|
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.
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.
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
|
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).
|
|
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:
|
|
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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,nice,changed
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|