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).
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, 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, 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
Offset: 1
Keywords
Links
- Giovanni Resta, Table of n, a(n) for n = 1..10000
- Simon Plouffe, On the values of the functions zeta and gamma, arXiv preprint arXiv:1310.7195 [math.NT], 2013.
- Fokko J. van de Bult, Dion C. Gijswijt, John P. Linderman, N. J. A. Sloane, and Allan R. Wilks, A Slow-Growing Sequence Defined by an Unusual Recurrence, J. Integer Sequences, Vol. 10 (2007), #07.1.2.
- Levi van de Pol, The Growth Rate of Gijswijt's Sequence, J. Int. Seq. (2025) Vol. 28, Art. No. 25.4.6. See p. 4.
- Index entries for sequences related to Gijswijt's sequence
Programs
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Mathematica
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 *)
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PARI
A091799(n,A=[])={while(#A
k||break; k=m);A=concat(A,k));A} \\ M. F. Hasler, Aug 07 2018
Formula
From M. F. Hasler, Sep 29 2018: (Start)
The first '5' appears at position 343 which in hexadecimal is 1x, with x = 57 (in base 16).
The first two consecutive '5's appear at positions 5760309085..5760309086, which is {-1..0} + 1y in hexadecimal, where y = xxxx + 7.
Then '55' occurs again at positions {-1..0} + 1yy in hexadecimal, etc.
The first three consecutive '5's appear at positions 1z + {-2..0} in hexadecimal, where z = yyyy + 20 (in base 16).
Then '555' occurs again at positions {-2..0} + 1zz in hexadecimal, etc.
The first occurrence of '5555' is at positions {-4..-1} + 1w in hexadecimal, where w = zzzz + 98 (in base 16).
'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.
Then '5555' occurs again at positions {-4..-1} + 1w'w in hexadecimal, where 'w = 'zzzz + 98 (in base 16).
'555' also occurs again at positions {-2..0} + 1w'w'z in hexadecimal, etc.
'5555' also occurs again at positions {-3..0} + 1w'w'w in hexadecimal, and at positions {-4..-1} + 1w'w'w'w in hexadecimal.
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)
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