This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
Start with a(1) = 1. By definition of the sequence, this says that the first run has length 1, so it must be a single 1, and a(2) = 2. Thus, the second run (which starts with this 2) must have length 2, so the third term must be also be a(3) = 2, and the fourth term can't be a 2, so must be a(4) = 1. Since a(3) = 2, the third run must have length 2, so we deduce a(5) = 1, a(6) = 2, and so on. The correction I made was to change a(4) to a(5) and a(5) to a(6). - _Labos Elemer_, corrected by _Graeme McRae_
a = 1:2: drop 2 (concat . zipWith replicate a . cycle $ [1,2]) -- John Tromp, Apr 09 2011
M := 100; s := [ 1,2,2 ]; for n from 3 to M do for i from 1 to s[ n ] do s := [ op(s),1+((n-1)mod 2) ]; od: od: s; A000002 := n->s[n]; # alternative implementation based on the Cloitre formula: A000002 := proc(n) local ksu,k ; option remember; if n = 1 then 1; elif n <=3 then 2; else for k from 1 do ksu := add(procname(i),i=1..k) ; if n = ksu then return (3+(-1)^k)/2 ; elif n = ksu+ 1 then return (3-(-1)^k)/2 ; end if; end do: end if; end proc: # R. J. Mathar, Nov 15 2014
a[steps_] := Module[{a = {1, 2, 2}}, Do[a = Append[a, 1 + Mod[(n - 1), 2]], {n, 3, steps}, {i, a[[n]]}]; a]
a[ n_] := If[ n < 3, Max[ 0, n], Module[ {an = {1, 2, 2}, m = 3}, While[ Length[ an] < n, an = Join[ an, Table[ Mod[m, 2, 1], { an[[ m]]} ]]; m++]; an[[n]]]] (* Michael Somos, Jul 11 2011 *)
n=8; Prepend[ Nest[ Flatten[ Partition[#, 2] /. {{2, 2} -> {2, 2, 1, 1}, {2, 1} -> {2, 2, 1}, {1, 2} -> {2, 1, 1}, {1, 1} -> {2, 1}}] &, {2, 2}, n], 1] (* Birkas Gyorgy, Jul 10 2012 *)
KolakoskiSeq[n_Integer] := Block[{a = {1, 2, 2}}, Fold[Join[#1, ConstantArray[Mod[#2, 2, 1], #1[[#2]]]] &, a, Range[3, n]]]; KolakoskiSeq[999] (* Mikk Heidemaa, Nov 01 2024 *) (* Corrected by Giorgos Kalogeropoulos, May 09 2025 *)
my(a=[1,2,2]); for(n=3,80, for(i=1,a[n],a=concat(a,2-n%2))); a
{a(n) = local(an=[1, 2, 2], m=3); if( n<1, 0, while( #an < n, an = concat( an, vector(an[m], i, 2-m%2)); m++); an[n])};
# For explanation see link.
def Kolakoski():
x = y = -1
while True:
yield [2,1][x&1]
f = y &~ (y+1)
x ^= f
y = (y+1) | (f & (x>>1))
K = Kolakoski()
print([next(K) for in range(100)]) # _David Eppstein, Oct 15 2016
a054353 n = a054353_list !! (n-1) a054353_list = scanl1 (+) a000002_list -- Reinhard Zumkeller, Aug 03 2013
a2 = {1, 2, 2}; Do[ a2 = Join[a2, {1+Mod[n-1, 2]}], {n, 3, 50}, {a2[[n]] } ]; Accumulate[a2] (* Jean-François Alcover, Jun 18 2013 *)
from itertools import accumulate def alst(nn): K = Kolakoski() # using Kolakoski() in A000002 return list(accumulate(next(K) for i in range(1, nn+1))) print(alst(66)) # Michael S. Branicky, Jan 12 2021
The sequence A000002 starts 1, 2, 2, 1, 1, 2, ..., so the sixth partial sum is 1 + 2 + 2 + 1 + 1 + 2 = 9, and therefore a(6) = 3*6 - 2*9 = 0. - _Michael B. Porter_, Jul 08 2016
a054354 n = a054354_list !! (n-1) a054354_list = zipWith (-) (tail a000002_list) a000002_list -- Reinhard Zumkeller, Aug 03 2013
a2 = {1, 2, 2}; Do[ a2 = Join[a2, {1 + Mod[n - 1, 2]}], {n, 3, 70}, {a2[[n]]}]; Differences[a2] (* Jean-François Alcover, Jun 18 2013 *)
The Kolakoski sequence (A000002) is: 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, ... with first differences (A054354): 1, 0, -1, 0, 1, -1, 1, 0, -1, 1, 0, -1, 0, 1, -1, 0, 1, 0, -1, 1, -1, 0, 1, -1, ... with first differences (A376604): -1, -1, 1, 1, -2, 2, -1, -1, 2, -1, -1, 1, 1, -2, 1, 1, -1, -1, 2, -2, 1, 1, -2, ...
kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],q[[-2]],Last[q]},{1,1,2},1,{1,2,1},2,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]]
kol[n_]:=Nest[kolagrow,{1},n-1];
Differences[kol[100],2]
The Kolakoski sequence is 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, ...; the partial sums are 1, 3, 5, 6, 7, 9, ..., so the sequence is 1-1=0, 3-2=1, 5-3=2, 6-4=2, 7-5=2, 9-6=3, ... .
a2 = {1, 2, 2}; Do[ a2 = Join[a2, {1 + Mod[n - 1, 2]}], {n, 3, 50}, {a2[[n]]}]; a3 = Accumulate[a2]; a3 - Range[Length[a3]] (* Jean-François Alcover, Jun 18 2013 *)
The first 32 entries of the Kolakoski sequence, A000002, are 12211212212211211221211212211211. From this we see that a(5)=15-17=-2, since among the first 2^5 letters, 15 of them are twos and 17 of them are ones.
There are five 1's among the first twelve terms of A000002 = 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 2, 2, ..., so a(12) = nearest integer to 100*5/12 = 41.666... rounded = 42.
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