cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A210769 Erroneous version of A295925.

Original entry on oeis.org

6, 336, 3995, 23520, 102795, 335656, 1039626, 2674440, 6223140
Offset: 2

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Author

Keywords

Comments

Included in accordance with OEIS policy of including incorrect sequences to serve as pointers to the correct versions.

Links

A060560 Number of ways to color the vertices of an octagon using <= n colors, allowing rotations and reflections.

Original entry on oeis.org

0, 1, 30, 498, 4435, 25395, 107331, 365260, 1058058, 2707245, 6278140, 13442286, 26942565, 51084943, 92383305, 160386360, 268718116, 436365945, 689252778, 1062132490, 1600850055, 2365010571, 3431103775, 4896133188, 6881801550, 9539306725, 13054804776, 17655599430
Offset: 0

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Author

N. J. A. Sloane, Apr 12 2001

Keywords

Comments

In Williamson's terminology, this is "Number of 8-hoops with n symbols."

Links

Crossrefs

Cf. A081720, A210766, A210768, A295925.

Programs

  • PARI
    a(n) = { (n^8 + 4*n^5 + 5*n^4 + 2*n^2 + 4*n)/16 } \\ Harry J. Smith, Jul 07 2009

Formula

a(n) = n*(n + 1)*(n^6 - n^5 + n^4 + 3*n^3 + 2*n^2 - 2*n + 4)/16.
G.f.: x*(1 + 21*x + 264*x^2 + 949*x^3 + 1014*x^4 + 258*x^5 + 13*x^6)/(1 - x)^9. - Colin Barker, Jan 29 2012

A210766 Number of 8-hoops with n symbols and no a-rooted trees.

Original entry on oeis.org

1, 13, 141, 1204, 7570, 36021, 136948, 436696, 1212399, 3013165, 6843991, 14431308, 28591576, 53726869, 96473910, 166535536, 277726093, 449264781, 707353489, 1087078180
Offset: 1

Views

Author

N. J. A. Sloane, Mar 25 2012

Keywords

Links

Crossrefs

Cf. A060560, A210768, A295925.

Programs

  • Maple
    A210766 := proc(n)
        (n^8-8*n^7+36*n^6-92*n^5+155*n^4-148*n^3 +88*n^2-16*n)/16 ;
end proc: # R. J. Mathar, Mar 27 2012

Formula

G.f.: x*(1+x)*(1+3*x+57*x^2+262*x^3+582*x^4+321*x^5+34*x^6)/(1-x)^9. [Colin Barker, Apr 17 2012]

A210768 Number of bilaterally asymmetric 8-hoops with n symbols and no a-rooted trees.

Original entry on oeis.org

1, 3, 918, 6669, 33665, 131616, 425866, 1192178, 2977857
Offset: 2

Views

Author

N. J. A. Sloane, Mar 25 2012

Keywords

Comments

It appears that table 2 of the Williamson article contains typographical errors and the sequence was intended to read as in A210699.

Links

Crossrefs

Cf. A210766, A060560, A295925.

A308401 Number of bracelets (turnover necklaces) of length n that have no reflection symmetry and consist of 6 white beads and n-6 black beads.

Original entry on oeis.org

3, 6, 16, 30, 56, 91, 150, 224, 336, 477, 672, 912, 1233, 1617, 2112, 2700, 3432, 4290, 5340, 6552, 8008, 9678, 11648, 13888, 16503, 19448, 22848, 26658, 31008, 35853, 41346, 47424, 54264, 61803, 70224, 79464, 89733, 100947, 113344, 126840, 141680, 157780, 175416, 194480, 215280, 237708
Offset: 9

Views

Author

Petros Hadjicostas, May 24 2019

Keywords

Comments

Bracelets that have no reflection symmetry are also known as chiral bracelets.
Here, for n >= 6, a(n) is also the number of dihedral compositions of n with 6 parts that have no reflection symmetry. Taking the MacMahon conjugates of these dihedral compositions, we see that a(n) is also the number of dihedral compositions of n into n-6 parts that have no reflection symmetry.
A cyclic composition b_1 + b_2 + ... + b_k of n into k parts is an equivalent class of (linear) compositions of n into k parts (placed on a circle) such that two such (linear) compositions are equivalent iff one can be obtained from the other by a rotation. Such compositions were first studied extensively by Sommerville (1909).
A dihedral composition b_1 + b_2 + ... + b_k of n into k parts is an equivalent class of (linear) compositions of n into k parts (placed on a circle) such that two such (linear) compositions are equivalent iff one can be obtained from the other by a rotation or a reversal of order. Such compositions were studied, for example, by Knopfmacher and Robbins (2013).
Given a bracelet of length n with k white beads and n-k black beads, we may get the corresponding dihedral composition using MacMahon's correspondence: start with a white bead and count that bead and the black beads that follow (in one direction), and call that b_1; then start with the next white bead and count that one and the black beads that follow, and call that b_2; repeat this process until you reach the k-th white bead and count that one and the black beads that follow, and call that b_k. The corresponding dihedral composition is b_1 + b_2 + ... + b_k.
If in the previous paragraph (given a bracelet of length n with k white beads and n-k black beads), we replace the white beads with black beads and the black beads with white beads, we get a dihedral composition of n into n-k parts: c_1 + c_2 + ... + c_{n-k}. These two dihedral compositions (which correspond to the same bracelet) are called "conjugate" compositions. See p. 273 in Sommerville (1909) for an explanation of "conjugate" compositions in the context of cyclic compositions.
Symmetric cyclic compositions of a positive integer n were first studied by Sommerville (1909, pp. 301-304). It can be proved that the study of necklaces with reflection symmetry using beads of two colors is equivalent to the study of symmetric cyclic compositions of a positive integer. Clearly all the necklaces with reflection symmetry are all the bracelets (turnover necklaces) with reflection symmetry. See also the comments for sequences A119963, A292200, and A295925.

Examples

			Using Frank Ruskey's website (listed above) to generate bracelets of fixed content (6, 3) with string length n = 9 and alphabet size 2, we get the following A005513(n = 9) = 7 bracelets: (1) WWWWWWBBB, (2) WWWWWBWBB, (3) WWWWBWWBB, (4) WWWWBWBWB, (5) WWWBWWWBB, (6) WWWBWWBWB, and (7) WWBWWBWWB. From these, bracelets 1, 4, 5, and 7 have reflection symmetry, while bracelets 2, 3 and 6 have no reflection symmetry (and thus, a(9) = 3).
Starting with a black bead, we count that bead and how many white beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence to get the following dihedral compositions of n = 9 into 3 parts: (1) 1 + 7 + 1, (2) 1 + 2 + 6, (3) 1 + 3 + 5, (4) 2 + 5 + 2, (5) 4 + 1 + 4, (6) 2 + 3 + 4, and (7) 3 + 3 + 3. Again, dihedral compositions 1, 4, 5, and 7 are symmetric (have reflection symmetry), while dihedral compositions 2, 3, and 6 are not symmetric (and thus, a(9) = 3).
We may also start with a white bead and count that bead and how many black beads follow (in one direction), and continue this process until we count all beads around the circle. We thus use MacMahon's correspondence again to get the following (conjugate) dihedral compositions of n = 9 into 6 parts: (1) 1 + 1 + 1 + 1 + 1 + 4, (2) 1 + 1 + 1 + 1 + 2 + 3, (3) 1 + 1 + 1 + 2 + 1 + 3, (4) 1 + 1 + 1 + 2 + 2 + 2, (5) 1 + 1 + 2 + 1 + 1 + 3, (6) 1 + 1 + 2 + 1 + 2 + 2, and (7) 1 + 2 + 1 + 2 + 1 + 2. Again, dihedral compositions 1, 4, 5, and 7 have reflection symmetries, while dihedral compositions 2, 3, and 6 do not have reflection symmetries (and thus, a(9) = 3). For example, dihedral composition 1 is symmetric because we can draw an axis of symmetry through one of the 1s and 4. In addition, dihedral composition 5 is symmetric because we may draw an axis of symmetry through the numbers 2 and 3.

Links

Crossrefs

Cf. A005513, A008804, A032246, A032247, A032248, A032249, A058187, A119963, A292200, A295925. Column k = 6 of A180472.

Programs

  • PARI
    a(n) = (1/12)* (sumdiv(gcd(n, 6), d,  eulerphi(d)*binomial((n/d) - 1, (6/d) - 1))) - (1/2)*binomial(floor(n/2), 3); \\ Michel Marcus, May 28 2019
  • PARI
    Vec(x^9*(3 + x^2 + x^3 + x^4) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x + x^2)^2) + O(x^50)) \\ Colin Barker, Jun 02 2019

Formula

G.f.: (x^k/2) * (-(1 + x)/(1 - x^2)^floor((k/2) + 1) + (1/k) * Sum_{m|k} phi(m)/(1 - x^m)^(k/m)) with k = 6. (This formula is due to Herbert Kociemba.)
a(n) = A005513(n) - A058187(n-6) = A005513(n) - binomial(floor(n/2), 3) for n >= 6.
a(n) = -(1/2)*binomial(floor(n/2), 3) + (1/12)* Sum_{d|gcd(n, 6)} phi(d)*binomial((n/d) - 1, (6/d) - 1) for n >= 6. (This is a modification of formulas found in Gupta (1979) and Shevelev (2004).)
From Colin Barker, May 26 2019: (Start)
G.f.: x^9*(3 + x^2 + x^3 + x^4) / ((1 - x)^6*(1 + x)^3*(1 - x + x^2)*(1 + x + x^2)^2).
a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) - a(n-4) + a(n-5) + 4*a(n-6) - 3*a(n-7) - 3*a(n-8) + 4*a(n-9) + a(n-10) - a(n-11) - 3*a(n-12) + a(n-13) + 2*a(n-14) - a(n-15) for n > 23. (End)
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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