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A091975
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a(1) = 1; for n>1, a(n) = largest integer k such that the word a(1)a(2)...a(n-1) is of is of the form xy^(k_1)Y^(k_2)y^(k_3)Y^(k_4)...y^(k_(m-1))Y^(k_m) where y has positive length and Y=reverse(y) and k_1+k_2+k_3+...+k_m = k.
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8
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1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 4, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 4, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 2, 2, 3
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OFFSET
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1,3
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COMMENTS
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Here ^ denotes concatenation. This is similar to Gijswijt's sequence A090822 except that the 'y' block still counts when reversed. Thus 2 1 1 2 counts as the 2 blocks (21)(12)
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LINKS
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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].
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PROG
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(Python)
def k(s):
maxk = 1
for m in range(1, len(s)+1):
i, y, kk = 1, s[-m:], len(s)//m
if kk <= maxk: return maxk
yY = [y, y[::-1]]
while s[-(i+1)*m:-i*m] in yY: i += 1
maxk = max(maxk, i)
def aupton(terms):
alst = [1]
for n in range(2, terms+1):
alst.append(k(alst))
return alst
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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J. Taylor (integersfan(AT)yahoo.com), Mar 15 2004
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STATUS
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approved
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