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 A119462 Triangle read by rows: T(n,k) is the number of circular binary words of length n having k occurrences of 01 (0 <= k <= floor(n/2)). 2
 1, 2, 2, 2, 2, 6, 2, 12, 2, 2, 20, 10, 2, 30, 30, 2, 2, 42, 70, 14, 2, 56, 140, 56, 2, 2, 72, 252, 168, 18, 2, 90, 420, 420, 90, 2, 2, 110, 660, 924, 330, 22, 2, 132, 990, 1848, 990, 132, 2, 2, 156, 1430, 3432, 2574, 572, 26, 2, 182, 2002, 6006, 6006, 2002, 182, 2, 2, 210, 2730 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Row n contains 1 + floor(n/2) terms. Sum of entries in row n is 2^n (A000079). 2*binomial(n-1,2k) is also the number of permutations avoiding both 123 and 132 with k valleys, i.e., positions with w[i]>w[i+1]= 1; T(0,0) = 1. T(n,k) = 2*T(n-1,k) - T(n-2,k) + T(n-2,k-1) for n >= 3. G.f.: (1 - z^2 + t*z^2)/(1 - 2*z + z^2 - t*z^2). T(n,0) = 2 for n >= 1. T(n,1) = 2*binomial(n,2) = A002378(n-1). T(n,2) = 2*binomial(n,4) = A034827(n). T(n,k) = 2*A034839(n-1,k) for n >= 1. [Corrected by Georg Fischer, May 28 2023] Sum_{k=0..floor(n/2)} k*T(n,k) = A057711(n). EXAMPLE T(3,1) = 6 because we have 001, 010, 011, 100, 101 and 110. Triangle starts: 1; 2; 2, 2; 2, 6; 2, 12, 2; 2, 20, 10; 2, 30, 30, 2; ... MAPLE T:=proc(n, k) if n=0 and k=0 then 1 else 2*binomial(n, 2*k) fi end: for n from 0 to 15 do seq(T(n, k), k=0..floor(n/2)) od; # yields sequence in triangular form PROG (GAP) Concatenation([1], Flat(List([1..15], n->List([0..Int(n/2)], k->2*Binomial(n, 2*k))))); # Muniru A Asiru, Dec 20 2018 CROSSREFS Cf. A000079, A002378, A034827, A034839, A057711. Sequence in context: A080400 A351031 A328236 * A293221 A334512 A096625 Adjacent sequences: A119459 A119460 A119461 * A119463 A119464 A119465 KEYWORD nonn,tabf AUTHOR Emeric Deutsch, May 21 2006 STATUS approved

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Last modified July 17 09:00 EDT 2024. Contains 374363 sequences. (Running on oeis4.)