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A353047
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Number of length n words on alphabet {0,1,2} that contain each of the subwords 01, 02, 10, 12, 20, and 21 as (not necessarily contiguous) subwords.
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1
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12, 108, 600, 2664, 10404, 37476, 127920, 420768, 1348476, 4242204, 13169160, 40490712, 123635028, 375623892, 1137095520, 3433306896, 10347106860, 31141984140, 93639862200, 281372571720, 845074016772, 2537235316548, 7615933808400, 22856659795584, 68588501433564
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OFFSET
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5,1
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COMMENTS
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Let A be an alphabet containing m letters. Let S be the set of m^2-m ordered two-tuples of distinct letters in A. The generating function for the number of length n words on A that contain each two-tuple in S as a (not necessarily contiguous) subword is m*(m-1)!^2*x^(2*m-1)/((1-m*x)*Product_{k=1..m-1} (1-k*x)^2).
Appears to equal 12 times A222993, except that sequence only has a conjectured formula. - N. J. A. Sloane, Jun 17 2022
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LINKS
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Table of n, a(n) for n=5..29.
Index entries for linear recurrences with constant coefficients, signature (9,-31,51,-40,12).
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FORMULA
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G.f.: (12*x^5)/((1 - 2*x)^2*(1 - x)^2*(1 - 3*x)).
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EXAMPLE
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a(5) = 12 because we have: {0, 1, 2, 0, 1}, {0, 1, 2, 1, 0}, {0, 2, 1, 0, 2}, {0, 2, 1, 2, 0}, {1, 0, 2, 0, 1}, {1, 0, 2, 1, 0}, {1, 2, 0, 1, 2}, {1, 2, 0, 2, 1}, {2, 0, 1, 0, 2}, {2, 0, 1, 2, 0}, {2, 1, 0, 1, 2}, {2, 1, 0, 2, 1}.
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MATHEMATICA
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nn = 15; vertices = Level[Outer[ List, {a, b, c}, {d, e, f}, {h, i, j}, {k, l, m}, {n, o, p}, {q, r, s}], {6}]; x = {a -> b, d -> e, i -> j, o -> p}; y = {b -> c, h -> i, k -> l, r -> s}; z = {e -> f, l -> m, n -> o, q -> r}; replacementlist = Table[vertices[[kk]] -> kk, {kk, 1, 729}]; G= Normal[SparseArray[Flatten[Table[Normal[Merge[ Map[{mm, vertices[[mm]] /. # /. replacementlist} -> 1 &, {x, y, z}], Total]], {mm, 1, 729}]]]]; Iwg =
Inverse[IdentityMatrix[729] - w G]; CoefficientList[ Series[Iwg[[1, 729]], {w, 0, nn}], w]
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CROSSREFS
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Cf. A058809, A222993, A005803 (binary words).
Sequence in context: A154671 A321672 A241230 * A037972 A111990 A053469
Adjacent sequences: A353044 A353045 A353046 * A353048 A353049 A353050
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KEYWORD
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nonn,easy
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AUTHOR
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Geoffrey Critzer, Apr 19 2022
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STATUS
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approved
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