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 1-4 of 4 results.

A081720 Triangle T(n,k) read by rows, giving number of bracelets (turnover necklaces) with n beads of k colors (n >= 1, 1 <= k <= n).

Original entry on oeis.org

1, 1, 3, 1, 4, 10, 1, 6, 21, 55, 1, 8, 39, 136, 377, 1, 13, 92, 430, 1505, 4291, 1, 18, 198, 1300, 5895, 20646, 60028, 1, 30, 498, 4435, 25395, 107331, 365260, 1058058, 1, 46, 1219, 15084, 110085, 563786, 2250311, 7472984, 21552969, 1, 78, 3210, 53764, 493131, 3037314
Offset: 1

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Author

N. J. A. Sloane, based on information supplied by Gary W. Adamson, Apr 05 2003

Keywords

Comments

From Petros Hadjicostas, Nov 29 2017: (Start)
The formula given below is clear from the programs given in the Maple and Mathematica sections, while the g.f. for column k can be obtained using standard techniques.
If we differentiate the column k g.f. m times, then we can get a formula for row m. (For this sequence, we only need to use this row m formula for 1 <= k <= m, but it is valid even for k>m.) For example, to get the formula for row 8, we have T(n=8,k) = d^8/dx^8 (column k g.f.)/8! evaluated at x=0. Here, "d^8/dx^8" means "8th derivative w.r.t. x" of the column k g.f. Doing so, we get T(n=8, k) = (k^6 - k^5 + k^4 + 3*k^3 + 2*k^2 - 2*k + 4)*(k + 1)*k/16, which is the formula given for sequence A060560. (Here, we use this formula only for 1 <= k <= 8.)
(End)

Examples

			1;                                                (A000027)
1,  3;                                            (A000217)
1,  4,  10;                                       (A000292)
1,  6,  21,   55;                                 (A002817)
1,  8,  39,  136,   377;                          (A060446)
1, 13,  92,  430,  1505,   4291;                  (A027670)
1, 18, 198, 1300,  5895,  20646,  60028;          (A060532)
1, 30, 498, 4435, 25395, 107331, 365260, 1058058; (A060560)
...
For example, when n=k=3, we have the following T(3,3)=10 bracelets of 3 beads using up to 3 colors: 000, 001, 002, 011, 012, 022, 111, 112, 122, and 222. (Note that 012 = 120 = 201 = 210 = 102 = 021.) _Petros Hadjicostas_, Nov 29 2017
		

References

  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Cf. A321791 (extension to n >= 0, k >= 0).
Cf. A081721 (diagonal), A081722 (row sums), column sequences k=2..6: A000029, A027671, A032275, A032276, A056341.

Programs

  • Maple
    A081720 := proc(n, k)
        local d, t1;
        t1 := 0;
        if n mod 2 = 0 then
            for d from 1 to n do
                if n mod d = 0 then
                    t1 := t1+numtheory[phi](d)*k^(n/d);
                end if;
            end do:
            (t1+(n/2)*(1+k)*k^(n/2)) /(2*n) ;
        else
            for d from 1 to n do
                if n mod d = 0 then
                    t1 := t1+numtheory[phi](d)*k^(n/d);
                end if;
            end do;
            (t1+n*k^((n+1)/2)) /(2*n) ;
        end if;
    end proc:
    seq(seq(A081720(n,k),k=1..n),n=1..10) ;
  • Mathematica
    t[n_, k_] := (For[t1 = 0; d = 1, d <= n, d++, If[Mod[n, d] == 0, t1 = t1 + EulerPhi[d]*k^(n/d)]]; If[EvenQ[n], (t1 + (n/2)*(1 + k)*k^(n/2))/(2*n), (t1 + n*k^((n + 1)/2))/(2*n)]); Table[t[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Sep 13 2012, after Maple, updated Nov 02 2017 *)
    Needs["Combinatorica`"]; Table[Table[NumberOfNecklaces[n,k,Dihedral],{k,1,n}],{n,1,8}]//Grid  (* Geoffrey Critzer, Oct 07 2012, after code by T. D. Noe in A027671 *)

Formula

See Maple code.
From Petros Hadjicostas, Nov 29 2017: (Start)
T(n,k) = ((1+k)*k^{n/2}/2 + (1/n)*Sum_{d|n} phi(n/d)*k^d)/2, if n is even, and = (k^{(n+1)/2} + (1/n)*Sum_{d|n} phi(n/d)*k^d)/2, if n is odd.
G.f. for column k: (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)) provided we chop off the Taylor expansion starting at x^k (and ignore all the terms x^n with n
(End)
2*n*T(n,k) = A054618(n,k)+n*(1+k)^(n/2)/2 if n even, = A054618(n,k)+n*k^((n+1)/2) if n odd. - R. J. Mathar, Jan 23 2022

Extensions

Name edited by Petros Hadjicostas, Nov 29 2017

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

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.)
		

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

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

Author

N. J. A. Sloane, Mar 25 2012

Keywords

Crossrefs

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

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.

Crossrefs

Showing 1-4 of 4 results.