A093305
Number of binary necklaces of length n with no subsequence 000.
Original entry on oeis.org
1, 2, 3, 4, 5, 9, 11, 19, 29, 48, 75, 132, 213, 369, 627, 1083, 1857, 3244, 5619, 9844, 17205, 30229, 53115, 93701, 165313, 292464, 517831, 918578, 1630933, 2900109, 5161443, 9197251, 16402841, 29283026, 52319379, 93558968, 167427845, 299846737, 537358107, 963651447, 1729192433
Offset: 1
- Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 500.
- Michael De Vlieger, Table of n, a(n) for n = 1..500
- P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
- P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
- Petros Hadjicostas, Cyclic Compositions of a Positive Integer with Parts Avoiding an Arithmetic Sequence, Journal of Integer Sequences, 19 (2016), #16.8.2.
- Silvana Ramaj, New Results on Cyclic Compositions and Multicompositions, Master's Thesis, Georgia Southern Univ., 2021. See p. 57.
- L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 89-96.
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Table[1/n * Sum[EulerPhi[n/d] (d Sum[Sum[Binomial[j, d - 3 k + 2 j] Binomial[k, j], {j, d - 3 k, k}]/k, {k, d}]), {d, Divisors@ n}], {n, 41}] (* Michael De Vlieger, Dec 28 2016, after Vladimir Joseph Stephan Orlovsky at A001644 *)
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N=66; x='x+O('x^N);
B(x)=x*(1+x+x^2);
A=sum(k=1, N, eulerphi(k)/k*log(1/(1-B(x^k))));
Vec(A)
/* Joerg Arndt, Aug 06 2012 */
A280218
Number of binary necklaces of length n with no subsequence 0000.
Original entry on oeis.org
1, 2, 3, 5, 6, 11, 15, 27, 43, 75, 125, 228, 391, 707, 1262, 2285, 4119, 7525, 13691, 25111, 46033, 84740, 156123, 288529, 533670, 989305, 1835983, 3412885, 6351031, 11834623, 22074821, 41222028, 77048131, 144148859, 269913278, 505826391, 948652695, 1780473001, 3343960175, 6284560319, 11818395345
Offset: 1
a(5)=6 because we have six binary cyclic sequences of length 5 that avoid four consecutive zeros: 00011, 00101, 00111, 01101, 01111, 11111.
- Alois P. Heinz, Table of n, a(n) for n = 1..3521
- P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
- P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
- Silvana Ramaj, New Results on Cyclic Compositions and Multicompositions, Master's Thesis, Georgia Southern Univ., 2021. See p. 57.
- F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc.
- F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]
- F. Ruskey, Necklaces, Lyndon words, De Bruijn sequences, etc. [Cached copy, with permission, pdf format only]
- L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 89-96.
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Table[(1/n) Sum[EulerPhi[n/d] SeriesCoefficient[(4 - 3 x - 2 x^2 - x^3)/(1 - x - x^2 - x^3 - x^4), {x, 0, d}], {d, Divisors@ n}], {n, 41}] (* Michael De Vlieger, Dec 30 2016 *)
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N=44; x='x+O('x^N);
B(x)=x*(1+x+x^2+x^3);
Vec(sum(k=1, N, eulerphi(k)/k * log(1/(1-B(x^k))))) \\ Joerg Arndt, Dec 29 2016
A322057
Array read by upwards antidiagonals: T(i,n) is the number of binary necklaces of length n that avoid 00...0 (i 0's).
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 3, 1, 1, 2, 3, 4, 3, 1, 1, 2, 3, 5, 5, 5, 1, 1, 2, 3, 5, 6, 9, 5, 1, 1, 2, 3, 5, 7, 11, 11, 8, 1, 1, 2, 3, 5, 7, 12, 15, 19, 10, 1, 1, 2, 3, 5, 7, 13, 17, 27, 29, 15, 1, 1, 2, 3, 5, 7, 13, 18, 31, 43, 48, 19, 1
Offset: 1
The first few antidiagonals are:
1;
1, 1;
1, 2, 1;
1, 2, 2, 1;
1, 2, 3, 3, 1;
1, 2, 3, 4, 3, 1;
1, 2, 3, 5, 5, 5, 1;
1, 2, 3, 5, 6, 9, 5, 1;
1, 2, 3, 5, 7, 11, 11, 8, 1;
1, 2, 3, 5, 7, 12, 15, 19, 10, 1;
...
From _Petros Hadjicostas_, Jan 16 2019: (Start)
In the above triangle (first few antidiagonals, read upwards), the j-th row corresponds to T(j,1), T(j-1,2), T(j-2,3), ..., T(1,j).
This, however, is not the j-th row of the square array (see the scanned page above).
For example, the sixth row of the square array is as follows:
T(6,1) = 1, T(6,2) = 2, T(6,3) = 3, T(6,4) = 5, T(6, 5) = 7, T(6, 6) = 13, ...
To generate these numbers, we use T(6, n) = (1/n)*Sum_{d|n} phi(n/d)*L(6,d), where
L(6,1) = 1, L(6,2) = 3, L(6,3) = 7, L(6,4) = 15, L(6,5) = 31, L(6,6) = 63, ...
See the sixth row of A125127. See also the Sage program below by Freddy Barrera.
(End)
- Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 520.
- Freddy Barrera, Rows n = 1..100 of triangle, flattened.
- Miklos Bona, Handbook of Enumerative Combinatorics, CRC Press, 2015, annotated scan of page 520.
- P. Flajolet and M. Soria, The Cycle Construction, SIAM J. Discr. Math., vol. 4 (1), 1991, pp. 58-60.
- L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 89-96.
-
T(i,n) = {my(p=1/(1 - x*(1 - x^i)/(1 - x))); polcoef(sum(d=1, n, eulerphi(d)*log(subst(p + O(x*x^(n\d)), x, x^d))/d), n)} \\ Andrew Howroyd, Jan 08 2024
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# uses the L method from A125127
def T(i, n):
return sum(euler_phi(n//d)*L(i, d) for d in n.divisors()) // n
[T(i, n) for d in (1..12) for i, n in zip((d..1, step=-1), (1..d))] # Freddy Barrera, Jan 15 2019
Showing 1-3 of 3 results.
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