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A091799 a(1) = 3. 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,3).

%I A091799 #61 Aug 19 2025 09:42:32
%S A091799 3,3,3,3,4,3,3,3,3,4,3,3,3,3,4,3,3,3,3,4,4,3,3,3,3,4,3,3,3,3,4,3,3,3,
%T A091799 3,4,3,3,3,3,4,4,3,3,3,3,4,3,3,3,3,4,3,3,3,3,4,3,3,3,3,4,4,3,3,3,3,4,
%U A091799 3,3,3,3,4,3,3,3,3,4,3,3,3,3,4,4,4,3,3,3,3,4,3,3,3,3,4,3,3,3,3
%N A091799 a(1) = 3. 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,3).
%C A091799 Here xy^k means the concatenation of the words x and k copies of y.
%C A091799 The first '5' appears at step 343. At which step does the first '6' appear? - _Sergio Pimentel_, Jul 13 2015
%C A091799 The first '6' appears at about 4.33*10^616, see FORMULA for exact value. The positions where record lengths of strings 5 and 6 occur can be computed via sequence A091844, see PARI code there. - _M. F. Hasler_, Sep 29 2018
%C A091799 This sequence is the concatenation of the "glue strings" of sequence A091787, just like sequence A091844 is the concatenation of the glue strings of this sequence. - _M. F. Hasler_, Oct 04 2018
%H A091799 Giovanni Resta, <a href="/A091799/b091799.txt">Table of n, a(n) for n = 1..10000</a>
%H A091799 Simon Plouffe, <a href="http://arxiv.org/abs/1310.7195">On the values of the functions zeta and gamma</a>, arXiv preprint arXiv:1310.7195 [math.NT], 2013.
%H A091799 Fokko J. van de Bult, Dion C. Gijswijt, John P. Linderman, N. J. A. Sloane, and Allan R. Wilks, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Sloane/sloane55.html">A Slow-Growing Sequence Defined by an Unusual Recurrence</a>, J. Integer Sequences, Vol. 10 (2007), #07.1.2.
%H A091799 Levi van de Pol, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL28/Vandepol/vandepol5.html">The Growth Rate of Gijswijt's Sequence</a>, J. Int. Seq. (2025) Vol. 28, Art. No. 25.4.6. See p. 4.
%H A091799 <a href="/index/Ge#Gijswijt">Index entries for sequences related to Gijswijt's sequence</a>
%F A091799 From _M. F. Hasler_, Sep 29 2018: (Start)
%F A091799 The first '5' appears at position 343 which in hexadecimal is 1x, with x = 57 (in base 16).
%F A091799 The first two consecutive '5's appear at positions 5760309085..5760309086, which is {-1..0} + 1y in hexadecimal, where y = xxxx + 7.
%F A091799 Then '55' occurs again at positions {-1..0} + 1yy in hexadecimal, etc.
%F A091799 The first three consecutive '5's appear at positions 1z + {-2..0} in hexadecimal, where z = yyyy + 20 (in base 16).
%F A091799 Then '555' occurs again at positions {-2..0} + 1zz in hexadecimal, etc.
%F A091799 The first occurrence of '5555' is at positions {-4..-1} + 1w in hexadecimal, where w = zzzz + 98 (in base 16).
%F A091799 '555' also occurs again at positions {-2..0} + 1w'z in hexadecimal, where 'z = 'yyyy + 20, 'y = 'xxxx + 7 and 'x = x - 1 = 56 (in base 16), and also at positions {-2..0} + 1w'zz in hexadecimal, and at positions {-2..0} + 1w'zzz in hexadecimal.
%F A091799 Then '5555' occurs again at positions {-4..-1} + 1w'w in hexadecimal, where 'w = 'zzzz + 98 (in base 16).
%F A091799 '555' also occurs again at positions {-2..0} + 1w'w'z in hexadecimal, etc.
%F A091799 '5555' also occurs again at positions {-3..0} + 1w'w'w in hexadecimal, and at positions {-4..-1} + 1w'w'w'w in hexadecimal.
%F A091799 The first occurrence of a '6' (immediately after the first '55555', followed by a sixth '5') is at position 1w'w'w'w + 30E in hexadecimal, which equals 782 + 16^510*257 + (16^512-1)/(16^128-1)*(158 - 16^126 + (16^128-1)/(16^32-1)*(32 + (16^32-1)/(16^8-1)*(7 + (16^8-1)/255*87))). (End)
%t A091799 maxBlockLength = 21; a[1] = 3; 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]], 3]]; Array[a, 99] (* _Jean-François Alcover_, Nov 07 2013 *)
%o A091799 (PARI) A091799(n,A=[])={while(#A<n, my(k=3,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++&& break);m>k||break; k=m);A=concat(A,k));A} \\ _M. F. Hasler_, Aug 07 2018
%Y A091799 Cf. A090822, A091787, A091844.
%K A091799 nonn
%O A091799 1,1
%A A091799 _N. J. A. Sloane_, Mar 08 2004