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 A295499 Sum of lengths of longest unbordered subword, over all binary words of length n. 0
 2, 6, 18, 48, 124, 302, 720, 1672, 3828, 8624, 19222, 42402, 92824, 201730, 435848, 936548, 2003292, 4267162, 9056642, 19158430, 40409800, 85006554, 178392666, 373546760, 780628626, 1628332454, 3390841918, 7050048360, 14636882444, 30347358688, 62842024406 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS By "subword" we mean a contiguous substring within a word. A string x is "bordered" if it has some nonempty prefix (other than x) that is also a suffix, and "unbordered" otherwise. REFERENCES Pawel Gawrychowski, Gregory Kucherov, Benjamin Sach, and Tatiana Starikovskaya, "Computing the Longest Unbordered Substring", in C. Iliopoulos et al. (Eds.): SPIRE 2015, LNCS 9309, pp. 246-257, 2015. LINKS Pawel Gawrychowski, Gregory Kucherov, Benjamin Sach, and Tatiana Starikovskaya, Computing the Longest Unbordered Substring, in C. Iliopoulos et al. (Eds.): SPIRE 2015, LNCS 9309, 2015. EXAMPLE For n = 3 the longest unbordered subwords of 000,111 are of length 1; of 010,101 are of length 2; and for all others are of length 3. So a(3) = 1+1+2+2+3+3+3+3 = 18. PROG (Python) from itertools import product def bordered(b): for i in range(len(b)-1, 0, -1): if b[:i] == b[-i:]: return True return False def m(b): for i in range(len(b), 0, -1): for j in range(len(b)-i+1): if not bordered(b[j:j+i]): return i return 0 def a(n): return 2*sum(m("0"+"".join(b)) for b in product("01", repeat=n-1)) print([a(n) for n in range(1, 19)]) # Michael S. Branicky, Mar 19 2022 CROSSREFS Sequence in context: A217526 A018027 A218759 * A059413 A256828 A197055 Adjacent sequences: A295496 A295497 A295498 * A295500 A295501 A295502 KEYWORD nonn AUTHOR Jeffrey Shallit, Feb 17 2018 EXTENSIONS a(19) and beyond from Michael S. Branicky, Mar 19 2022 STATUS approved

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Last modified February 2 06:48 EST 2023. Contains 360000 sequences. (Running on oeis4.)