

A119458


Triangle read by rows: T(n,k) is the number of (marked) circular binary words of length n having k occurrences of 00 (0 <= k <= n).


6



1, 1, 1, 3, 0, 1, 4, 3, 0, 1, 7, 4, 4, 0, 1, 11, 10, 5, 5, 0, 1, 18, 18, 15, 6, 6, 0, 1, 29, 35, 28, 21, 7, 7, 0, 1, 47, 64, 60, 40, 28, 8, 8, 0, 1, 76, 117, 117, 93, 54, 36, 9, 9, 0, 1, 123, 210, 230, 190, 135, 70, 45, 10, 10, 0, 1, 199, 374, 440, 396, 286, 187, 88, 55, 11, 11, 0, 1
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OFFSET

0,4


COMMENTS

Sum of entries in row n is 2^n (A000079).
In Carlitz and Scoville (p. 252) the first term is 2.
From Petros Hadjicostas, Jan 05 2019: (Start)
Note that T(n,k) is the number of marked circular binary words of length n having k occurrences of 00 (0 <= k <= n). Let W(n,k) be the number of binary necklaces (= unmarked circular binary words) of length n with exactly k occurrences of the pattern 00 provided wrapping around the circle is allowed when n=1. More precisely, when n=1, we allow the string to wrap around itself on the circle to form a circular string of length 2. We have W(n, k) = A320341(n, k) for 0 <= k <= n.
Fortunately, since T(n=1, k=0) = 1 = T(n=1, k=1), the author of the sequence (indirectly) assumes that the string 0 has one occurrence of the pattern 00 (if allowed to wrap around itself on the circle only once), while the string 1 has zero occurrences of the pattern 00.
It makes sense to define T(n,k) = 0 when k > n. Note that G(z,t) = Sum_{n,k >=0} T(n,k)*z^n*t^k = 1  z*(d(1A(z,t))/dz)/(1A(z,t)), where A(z,t) = z*(t+1) + z^2*(1t) is the "FlajoletSoria polynomial (g.f.)" in z of some (difficult to find) linear combinatorial structure that is defined very abstractly at the beginning of their paper and in the book by Flajolet and Sedgewick (2009). (Since the series here converge for z and t close to 0, we have that 1t is positive for all t in a "small" interval around 0.)
Using the theory in Flajolet and Soria (1991) and the one in Hadjicostas and Zhang (2018), we can prove that the g.f. of the numbers W(n,k) is F(z,t) = Sum_{n >= 1, k >= 0} W(n,k)*z^n*t^k = Sum_{d>=1} (phi(d)/d)*log(1A(z^d,t^d)). We can also prove that W(n,k) = (1/n)*Sum_{dgcd(n,k)} phi(d)*T(n/d, k/d) for n>=1 and 0 <= k <= n. (We omit the details.)
For k=0, we get that T(n, k=0) = A000204(n) and W(n, k=0) = A000358(n) for n >= 1, and the FlajoletSoria polynomial is A(z,t=0) = z + z^2 (which was discovered by Joerg Arndt).
To get univariate g.f.'s of the sequences (T(n,k): n >= 1) and (W(n, k): n >= 1) when k >= 1, we have to differentiate the previous two g.f.'s k times with respect to t, set t=0, and divide by k!. (Obviously, the log now disappears.)
For k=1, we get T(n, k=1) = A006490(n) and W(n, k=1) = A212804(n1) = A006490(n)/n for n >= 1.
For k=2, we get T(n, k=2) = A006491(n1) for n >= 1 (with A006491(0) := 0) and W(n, k=2) = (1/n)*(T(n,2) + I(2n)*T(n/2,1)) for n >= 1, where I(2n) = 1 if 2n, and 0 otherwise.
For k=3, we get T(n, k=3) = A006492(n2) for n >=1 (with A006492(m) = 0 for m = 1,2) and W(n, k=3) = (1/n)*(T(n,3) + 2*I(3n)*T(n/3, 1)) for n >= 1.
This theory can be generalized for any pattern P of 0's and 1's provided one discovers the correct "FlajoletSoria" polynomial A_P(z,t). In other words, if P is a finite pattern of zeros and ones of length L, and we let T_P(n,k) be the number of (marked) circular binary words of length n having k occurrences of P (0 <= k <= n) and we allow strings of length n, with 1 <= n < L, to wrap around themselves on the circle to form a string of length n*ceiling(L/n), then (most probably) we can express the g.f. of T_p(n,k), say G_p(z,t), in the form 1z*(d(1A_P(z,t))/dz)/(1A_P(z,t)). In such a case, if W_P(n,k) is the number of binary necklaces (= unmarked circular binary words) of length n with exactly k occurrences of the pattern P, we have that its generating function is Sum_{n >= 1, k >= 0} W_P(n,k)*z^n*t^k = Sum_{d>=1} (phi(d)/d)*log(1A_P(z^d,t^d)). We can also prove that W_P(n,k) = (1/n)*Sum_{dgcd(n,k)} phi(d)*T_P(n/d, k/d) for n>=1 and 0 <= k <= n.
One final note: it seems that the theory of Flajolet and Soria (1991) applies only to the case k=0 and to the case we consider all k simultaneously. (For fixed k >= 2, W_P(n,k) depends not only on T_P(n,k), but also on all T(n/d, k/d) where d ranges over the common divisors of n and k. Thus, for fixed k >= 2, it seems there is no linear combinatorial structure whose list of cycles of elements corresponds to the collection of necklaces we want.) See also Flajolet and Sedgewick (2009), pp. 8485 and 729730.
(End)


LINKS

Table of n, a(n) for n=0..77.
L. Carlitz and R. Scoville, Zeroone sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246254.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009.
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 5860.
P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 5860.
P. Hadjicostas and L. Zhang, On cyclic strings avoiding a pattern, Discrete Mathematics, 341 (2018), 16621674.
L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 8996.


FORMULA

T(n,k) = T(n1,k) + T(n2,k) + T(n1,k1)  T(n2,k1) for n >= 3 and k >= 1.
G.f.: G(z,t) = Sum_{n,k >=0} T(n,k)*z^n*t^k = (1 + z^2  t*z^2)/(1  z  z^2  t*z + t*z^2). [edited by Petros Hadjicostas, Jan 05 2019]
T(n,0) = A000204(n) for n >= 1 (Lucas numbers).
T(n,1) = A006490(n) for n >= 1.
T(n,2) = A006491(n1) for n >= 1 (with A006491(0):=0).
Sum_{k=0..n} k*T(n,k) = A057711(n).


EXAMPLE

T(5,3) = 5 because we have 10000, 01000, 00100, 00010 and 00001.
Triangle for T(n,k) begins:
n=0: 1;
n=1: 1, 1;
n=2: 3, 0, 1;
n=3: 4, 3, 0, 1;
n=4: 7, 4, 4, 0, 1;
n=5: 11, 10, 5, 5, 0, 1
n=6: 18, 18, 15, 6, 6, 0, 1;
n=7: 29, 35, 28, 21, 7, 7, 0, 1;
n=8: 47, 64, 60, 40, 28, 8, 8, 0, 1;
n=9: 76, 117, 117, 93, 54, 36, 9, 9, 0, 1;
n=10: 123, 210, 230, 190, 135, 70, 45, 10, 10, 0, 1;
...
From Petros Hadjicostas, Jan 06 2019: (Start)
If we take the Taylor expansion of g.f. G(z,t) of T(n,k) around z=0, we get G(z,t) = 1 + (1+t)*z + (3+0*t+t^2)*z^2 + (4+3*t+0*t^2+t^3)*z^3 + (7+4*t+4*t^2+0*t^3+t^4)*z^4 +(11+10*t+5*t^2+5*t^3+0*t^4+t^5)*z^5 + ...
One the other hand, if we take the Taylor expansion of the g.f. F(z,t) of W(n,k) = A320341(n, k) around z=0, we get F(z,t) = (1+t)*z + (2+0*t+t^2)*z^2 + (2+t+0*t^2+t^3)*z^3 + (3+t+t^2+0*t^3+t^4)*z^4 + (3+2*t+t^2+t^3+0*t^4+t^5)*z^5 + ...
Triangle for W(n,k) begins:
n=1: 1, 1;
n=2: 2, 0, 1;
n=3: 2, 1, 0, 1;
n=4: 3, 1, 1, 0, 1;
n=5: 3, 2, 1, 1, 0, 1;
n=6: 5, 3, 3, 1, 1, 0, 1;
n=7: 5, 5, 4, 3, 1, 1, 0, 1;
n=8: 8, 8, 8, 5, 4, 1, 1, 0, 1;
n=9: 10, 13, 13, 11, 6, 4, 1, 1, 0, 1;
n=10: 15, 21, 24, 19, 14, 7, 5, 1, 1, 0, 1;
...
For example, for n=4, we have the following marked and unmarked circular binary words (the square brackets denote equivalence classes):
k=0: [1111], [1110,1101,1011,0111], [1010,0101], T(4,0) = 7 and W(4,0) = 3;
k=1: [1100,1001,0011,0110], T(4,1) = 4 and W(4,1) = 1;
k=2: [0001,0010,0100,1000], T(4,2) = 4 and W(4,2) = 1;
k=3: none, T(4,3) = 0 = W(4,3);
k=4: [0000], T(4,4) = 1 = W(4,4).
(End)


MAPLE

T:=proc(n, k): if k>n or k<0 then 0 elif n=0 and k=0 then 1 elif n=1 then 1 elif n=2 and k=0 then 3 elif n=2 and k=1 then 0 else T(n1, k)+T(n2, k)+T(n1, k1)T(n2, k1) fi end: for n from 0 to 12 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form


MATHEMATICA

T[n_, k_] := T[n, k] = Which[k > n  k < 0, 0, n == 0 && k == 0, 1, n == 1, 1, n == 2 && k == 0, 3, n == 2 && k == 1, 0, True, T[n1, k] + T[n2, k] + T[n1, k1]  T[n2, k1]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* JeanFrançois Alcover, Jan 11 2019 *)


CROSSREFS

Cf. A000079, A000204, A000358, A006490, A006491, A006492, A057711, A212804, A320341.
Sequence in context: A197126 A256987 A048963 * A106356 A091613 A039727
Adjacent sequences: A119455 A119456 A119457 * A119459 A119460 A119461


KEYWORD

nonn,tabl


AUTHOR

Emeric Deutsch, May 20 2006


EXTENSIONS

Name was edited by Petros Hadjicostas, Jan 06 2019
Values of T(9,5) and T(9,6) were corrected in the example by Petros Hadjicostas, Jan 06 2019


STATUS

approved



