A263655 Table T(m, n) of number of circular binary strings with m ones and n zeros without zigzags, read by antidiagonals (see reference for precise definition).
0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 4, 0, 1, 1, 0, 5, 5, 0, 1, 1, 0, 6, 6, 6, 0, 1, 1, 0, 7, 7, 7, 7, 0, 1, 1, 0, 8, 8, 12, 8, 8, 0, 1, 1, 0, 9, 9, 18, 18, 9, 9, 0, 1, 1, 0, 10, 10, 25, 30, 25, 10, 10, 0, 1, 1, 0, 11, 11, 33, 44, 44, 33, 11, 11, 0, 1
Offset: 0
Examples
Table starts: 0 1 1 1 1 1 1 1 1 1 1 1 1 ... 1 0 0 0 0 0 0 0 0 0 0 0 0 ... 1 0 4 5 6 7 8 9 10 11 12 13 14 ... 1 0 5 6 7 8 9 10 11 12 13 14 15 ... 1 0 6 7 12 18 25 33 42 52 63 75 88 ... 1 0 7 8 18 30 44 60 78 98 120 144 170 ... 1 0 8 9 25 44 70 104 147 200 264 340 429 ... 1 0 9 10 33 60 104 168 255 368 510 684 893 ... 1 0 10 11 42 78 147 255 412 629 918 1292 1765 ... 1 0 11 12 52 98 200 368 629 1014 1558 2300 3283 ... 1 0 12 13 63 120 264 510 918 1558 2514 3885 5786 ...
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..819
- E. Munarini and N. Z. Salvi, Circular Binary Strings without Zigzags, Integers: Electronic Journal of Combinatorial Number Theory 3 (2003), #A19.
Programs
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Mathematica
max = 11; U[r_, k_] := Binomial[r - k + 2*Floor[k/3], Floor[k/3]]; V[r_, k_] := Binomial[r - Ceiling[k/2] - 1, Floor[k/2]]; T[0, 0] = T[1, 1] = 0; T[0, ] = T[, 0] = 1; T[n_ /; n >= 2, m_] /; m <= n := T[n, m] = Switch[m, 1, 0, 2, n + 2, 3, n + 3, _, Sum[ U[m, k]*U[n, k] - 2*V[m, k]*V[n, k]*(-1)^k, {k, 0, max-3}]]; T[n_, m_] /; m > n := T[m, n]; Table[T[n - k, k], {n, 0, max}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 01 2018, after Andrew Howroyd *)
Formula
From Andrew Howroyd, Feb 26 2017: (Start)
T(n,m) = Sum_{k>=0} U(m,k)*U(n,k) - 2*V(m,k)*V(n,k)*(-1)^k
where U(r,k)=binomial(r-k+2*floor(k/3), floor(k/3)), V(r,k)=binomial(r-ceiling(k/2)-1, floor(k/2)).
T(n,0)=1 for n>=1, T(n,1)=0 for n>=1, T(n,2)=n+2 for n>=2, T(n,3)=n+3 for n>=2.
T(n,4)=(n-1)*(n+4)/2 for n>=3, T(n,5)=(n-2)*(n+5) for n>=3. (End)
Extensions
a(66)-a(77) from Andrew Howroyd, Feb 26 2017
Comments