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 A091844 a(1) = 4. 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,4). 8
 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 5, 5, 4, 4, 4, 4, 4, 5 (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. The first '6' occurs at a(3908). - Sergio Pimentel, Jul 13 2015 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. 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 [pdf, ps]. MATHEMATICA maxBlockLength = 100; a[1] = 4; a[n_] := a[n] = Module[{rev = Reverse[Array[a, n - 1]]}, blockCount[blockLength_] := Module[{par, p1, k}, par = Partition[rev, blockLength]; If[par == {}, Return[1]]; p1 = First[par]; k = 1; While[k <= Length[par], If[par[[k]] != p1, Break[], k++]]; k - 1]; Max[Max[Array[blockCount, maxBlockLength]], 4]]; Array[a, 100] (* Michael De Vlieger, Jul 13 2015, after Jean-François Alcover at A091799 *) PROG (PARI) {A091844(Nmax, L=1, A=List(), f(A, m=3, L=0)=while( #A>=(m+1)*L++, while( A[#A-L*m+1..#A]==A[#A-L*(m+1)+1..#A-L], (m+++1)*L>#A&& break)); m, t) = while(#A

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Last modified February 20 06:26 EST 2019. Contains 320332 sequences. (Running on oeis4.)