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
The sequence of differences between primes splits into the following runs: (1), (2,2), (4), (2), (4), (2), (4), (6), (2), (6), (4), (2), (4), (6,6), (2), (6), (4), (2), (6), (4), (6).
Accumulate[Length/@Split[Differences[Array[Prime,100]],#1==#2&]]//Most - or - Select[Range[100],Prime[#+1]-Prime[#]!=Prime[#+2]-Prime[#+1]&]
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
a078649 n = a078649_list !! (n-1) a078649_list = map (+ 1) $ filter ((== 0) . a054354) [1..] -- Reinhard Zumkeller, Aug 03 2013
isA078649 := proc(n) if A000002(n) = A000002(n+1) then true; else false; end if; end proc: A078649 := proc(n) option remember; if n = 1 then 2; else for a from procname(n-1)+1 do if isA078649(a) then return a; end if; end do: end if; end proc: seq(A078649(n),n=1..50) ; # R. J. Mathar, Nov 15 2014
a2 = {1, 2, 2}; Do[ a2 = Join[a2, {1+Mod[n-1, 2]}], {n, 3, 80}, {a2[[n]]}]; a3 = Accumulate[a2]; Complement[ Range[ Last[a3]], a3] (* Jean-François Alcover, Jun 18 2013 *)
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]
a2 = {1, 2, 2}; Do[ a2 = Join[a2, {1 + Mod[n - 1, 2]}], {n, 3, 60}, {i, 1, a2[[n]]}]; a3 = Accumulate[a2]; 2*Range[Length[a3]] - a3 (* Jean-François Alcover, Jun 18 2013 *)
A2 = {1, 2, 2}; Do[If[Mod[n, 10^5] == 0, Print["n = ", n]]; m = 1 + Mod[n - 1, 2]; an = A2[[n]]; A2 = Join[A2, Table[m, {an}]], {n, 3, 10^6}]; A054353 = Accumulate[A2]; Clear[a]; a[0] = 1; a[n_] := a[n] = A054353[[a[n - 1]]] + 1; Table[a[n], {n, 0, 33}] (* Jean-François Alcover, Oct 30 2014, after Jean-Christophe Hervé *)
def aupton(nn): alst, A054353, idx = [1], 0, 1 K = Kolakoski() # using Kolakoski() in A000002 for n in range(2, nn+1): target = alst[-1] while idx <= target: A054353 += next(K) idx += 1 alst.append(A054353 + 1) # a(n) = A054353(a(n-1))+1 return alst print(aupton(36)) # Michael S. Branicky, Jan 12 2021
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 *)
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