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 1-2 of 2 results.

A295925 Number of bilaterally asymmetric 8-hoops with n symbols.

Original entry on oeis.org

6, 336, 3795, 23520, 102795, 355656, 1039626, 2674440, 6223140, 13354440, 26807781, 50885016, 92095185, 159981360, 268161060, 435614256, 688255506, 1060829280, 1599170055, 2362871280, 3428409831, 4892775096, 6877654350
Offset: 2

Views

Author

Petros Hadjicostas, Nov 30 2017

Keywords

Comments

This is a corrected version of the entries in sequence A210769, which is copied from the third row of Table 2 (p. 381) in Williamson (1972). Apparently, in that row, there are probable typographical errors for the values of a(4) and a(7). The formula for a(n) can be obtained by letting z_1=z_2=...=z_8=n in equation (24) on p. 377 in Williamson (1972). In any case, to be certain, we provide a sketch of an independent derivation of the formula.
Bilaterally symmetric bracelets are also known as circular palindromes. This kind of necklaces was first studied by Sommerville (1909) in the context of circular compositions.
Consider sequence A081720, which contains the numbers T(n,k) that are the number of bracelets (turn over necklaces) with n beads each of which is colored with one of k colors. The g.f. for column k of that triangle is (1/2)*((k*x+k*(k+1)*x^2/2)/(1-k*x^2) - Sum_{n>=1} (phi(n)/n)*log(1-k*x^n)). The part of the g.f. that is the number of bilaterally symmetric bracelets with n beads of k colors is (k*x+k*(k+1)*x^2/2)/(1-k*x^2). Thus, for fixed k (= number of colors for sequence A081720), the g.f. of the number of bilaterally asymmetric bracelets with n beads of k colors is the difference between the two g.f.'s, that is, (1/2)*(-(k*x+k*(k+1)*x^2/2)/(1-k*x^2) - Sum_{n>=1} (phi(n)/n)*log(1-k*x^n)). The coefficient of x^8 in the Taylor expansion w.r.t. x (around x=0) for the latter g.f. gives the number of bilaterally asymmetric 8-hoops obtained using (up to) k symbols. Taking the 8th derivative w.r.t. x of the last g.f., evaluating at x=0, dividing by 8!, and replacing k with n, we get the formulae given below.

Examples

			From the A060560(2) = 30 8-hoops (i.e., from the total number of ways of coloring the vertices of an octagon using up to n=2 colors, allowing for rotations and reflections), there are A019583(2+1) = 24 that are circular palindromes (i.e., bilaterally symmetric bracelets). Hence, there are 30-24=6 bilaterally asymmetric 8-hoops using up to 2 colors. They are the following: 01001111, 01000111, 01000011, 00101011, 00110111 and 11001000. (To view these 6 asymmetric bracelets, the 0's and 1's must be placed on the vertices of a regular octagon inscribed in a circle as it is done in Fig. 4 on p. 379 in Williamson (1972), where 0 is replaced by a and 1 by b.)
		

Crossrefs

Programs

  • Mathematica
    Drop[#, 2] &@ CoefficientList[Series[3 (7 x^4 + 82 x^3 + 237 x^2 + 92 x + 2) (x + 1) x^2/(1 - x)^9, {x, 0, 24}], x] (* Michael De Vlieger, Dec 02 2017 *)

Formula

a(n) = (1/16)*(n^3-n^2-2)*(n^2+n+2)*(n+1)*(n-1)*n = (n^8-4*n^5-3*n^4+2*n^2+4*n)/16.
a(n) = A060560(n) - A019583(n+1) = (A054622(n) - A019583(n+1))/2. (Notice that the offsets of the sequences in these formulae are not necessarily the same as the offset of the current sequence.)
G.f.: 3*(7*x^4 + 82*x^3 + 237*x^2 + 92*x + 2)*(x + 1)*x^2/(1-x)^9.
Recurrence: (1-Delta)^9 a(n) = 0, where Delta^m a(n) = a(n-m). Hence, a(n) = 9*a(n-1)-36*a(n-2)+84*a(n-3)-126*a(n-4)+126*a(n-5)-84*a(n-6)+36*a(n-7)-9*a(n-8)+a(n-9).
E.g.f.: exp(x)*x^2*(48 + 848*x + 1658*x^2 + 1046*x^3 + 266*x^4 + 28*x^5 + x^6)/16. - Stefano Spezia, Feb 18 2024

A054621 Number of ways to color vertices of a heptagon using <= n colors, allowing only rotations.

Original entry on oeis.org

0, 1, 20, 315, 2344, 11165, 39996, 117655, 299600, 683289, 1428580, 2783891, 5118840, 8964085, 15059084, 24408495, 38347936, 58619825, 87460020, 127695979, 182857160, 257298381, 356336860, 486403655, 655210224, 871930825, 1147401476, 1494336195, 1927561240
Offset: 0

Views

Author

N. J. A. Sloane, Apr 16 2000

Keywords

Comments

Length-7 necklaces with n kinds of beads. - Vincenzo Librandi Apr 30 2012

Crossrefs

Cf. A054622. Row 7 of A075195.

Programs

  • Magma
    I:=[0, 1, 20, 315, 2344, 11165, 39996, 117655]; [n le 8 select I[n] else 8*Self(n-1)-28*Self(n-2)+56*Self(n-3)-70*Self(n-4)+56*Self(n-5)-28*Self(n-6)+8*Self(n-7)-Self(n-8): n in [1..30]]; // Vincenzo Librandi, Apr 30 2012
  • Maple
    a:=proc(n) option remember:
    if n=0 then 0 elif n=1 then 1 elif n=2 then 20 elif n=3 then 315 elif n=4 then 2344 elif n=5 then 11165 elif n=6 then 39996 elif n=7 then 117655 elif n=8 then 299600 else
    8*a(n-1)-28*a(n-2)+56*a(n-3)-70*a(n-4)+56*a(n-5)-28*a(n-6)+8*a(n-7)-a(n-8): fi: end: seq(a(n), n=0..50); # Wesley Ivan Hurt, Sep 15 2015
  • Mathematica
    CoefficientList[Series[x*(1+12*x+183*x^2+328*x^3+183*x^4+ 12*x^5+x^6)/(x-1)^8,{x,0,33}],x] (* Vincenzo Librandi, Apr 30 2012 *)

Formula

a(n) = (1/7) * Sum_{d|7} phi(d)*n^(7/d) = (1/7) * (n^7 + 6*n). [corrected by Klaus Wagner, Sep 15 2015]
G.f.: x*(1+12*x+183*x^2+328*x^3+183*x^4+12*x^5+x^6) / (1-x)^8. - R. J. Mathar, Aug 30 2011
a(n) = 8*a(n-1)-28*a(n-2)+56*a(n-3)-70*a(n-4)+56*a(n-5)-28*a(n-6)+8*a(n-7)-a(n-8), n>7. - Wesley Ivan Hurt, Sep 15 2015

Extensions

Edited by Christian G. Bower, Sep 07 2002
Showing 1-2 of 2 results.