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 1-10 of 12 results. Next

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

Views

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.

Links

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

A283846 Number of n-gonal inositol homologs with 2 kinds of achiral proligands.

Original entry on oeis.org

2, 6, 10, 31, 68, 226, 650, 2259, 7542, 27036, 96350, 352786, 1294652, 4806366, 17912120, 67160083, 252710672, 954641186, 3617076710, 13744708060, 52358745532, 199914446106, 764881848410, 2932043941394, 11259015845684, 43303894193076, 166800053312630
Offset: 1

Views

Author

N. J. A. Sloane, Apr 01 2017

Keywords

Comments

Counts A032275 up to paired color permutation (equivalent to full color permutation on the 2-tuples of two subcolors, e.g., convert quaternary beads 0 1 2 3 to dibit beads 00 01 10 11). - Travis Scott, Jan 09 2023

Links

Crossrefs

The 8 sequences in Table 8 of Fujita (2017) are A053656, A000011, A256216, A256217, A123045, A283846, A283847, A283848.

Programs

  • Maple
    f:=  proc(m) uses numtheory;
  if m::even then 1/(4*m)*add(phi(d)*4^(m/d)*`if`(d::even,2,1), d = divisors(m))
+ 3*2^(m-2)
  else
1/(4*m)*add(phi(d)*4^(m/d),d=divisors(m))+2^(m-1)
  fi
end proc:
map(f, [$1..100]); # Robert Israel, Aug 21 2018
  • Mathematica
    f[m_] := If[EvenQ[m], 1/(4m)*Sum[EulerPhi[d]*4^(m/d)*If[EvenQ[d], 2, 1], {d, Divisors[m]}]+ 3*2^(m-2), 1/(4m)*Sum[EulerPhi[d]*4^(m/d), {d, Divisors[m]}] + 2^(m-1)];
f /@ Range[1, 25] (* Jean-François Alcover, Feb 26 2019, after Robert Israel *)

Formula

From Robert Israel, Aug 21 2018 after Fujita (2017), Eq. (99)(set n=2, m=n): (Start)
if n is even, a(n) = (4*n)^(-1)*(Sum_{d|n} phi(d)*4^(n/d) + Sum_{d|n, d even} phi(d)*4^(n/d)) + 3*2^(n-2).
if n is odd, a(n) = 2^(n-1) + (4*n)^(-1)*Sum_{d|n} phi(d)*4^(n/d). (End)

Extensions

a(1)-a(2) prepended by Travis Scott, Jan 09 2023

A051137 Table T(n,k) read by antidiagonals: number of necklaces allowing turnovers (bracelets) with n beads of k colors.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 6, 10, 10, 5, 1, 1, 8, 21, 20, 15, 6, 1, 1, 13, 39, 55, 35, 21, 7, 1, 1, 18, 92, 136, 120, 56, 28, 8, 1, 1, 30, 198, 430, 377, 231, 84, 36, 9, 1, 1, 46, 498, 1300, 1505, 888, 406, 120, 45, 10, 1
Offset: 0

Views

Author

Alford Arnold

Keywords

Comments

Unlike A075195 and A284855, antidiagonals go from bottom-left to top-right.

Examples

			Table begins with T[0,1]:
1  1    1     1      1       1        1        1         1         1
1  2    3     4      5       6        7        8         9        10
1  3    6    10     15      21       28       36        45        55
1  4   10    20     35      56       84      120       165       220
1  6   21    55    120     231      406      666      1035      1540
1  8   39   136    377     888     1855     3536      6273     10504
1 13   92   430   1505    4291    10528    23052     46185     86185
1 18  198  1300   5895   20646    60028   151848    344925    719290
1 30  498  4435  25395  107331   365260  1058058   2707245   6278140
1 46 1219 15084 110085  563786  2250311  7472984  21552969  55605670
1 78 3210 53764 493131 3037314 14158228 53762472 174489813 500280022

Links

Crossrefs

Columns 2-6 are A000029, A027671, A032275, A032276, and A056341.
Rows 2-7 are A000217, A000292, A002817, A060446, A027670, and A060532.
Cf. A000031.
Cf. A081720, A081721.
T(n,k) = (A075195(n,k) + A284855(n,k)) / 2.

Programs

  • Mathematica
    b[n_, k_] := DivisorSum[n, EulerPhi[#]*k^(n/#) &] / n;
c[n_, k_] := If[EvenQ[n], (k^(n/2) + k^(n/2+1))/2, k^((n+1)/2)];
T[0, ] = 1; T[n, k_] := (b[n, k] + c[n, k])/2;
Table[T[n, k-n], {k, 1, 11}, {n, k-1, 0, -1}] // Flatten
(* Robert A. Russell, Sep 21 2018 after Jean-François Alcover *)

Formula

T(n, k) = (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/(2*n)) * Sum_{d divides n} phi(d) * k^(n/d). - Robert A. Russell, Sep 21 2018
G.f. for column k: (kx/4)*(kx+x+2)/(1-kx^2) - Sum_{d>0} phi(d)*log(1-kx^d)/2d. - Robert A. Russell, Sep 28 2018
T(n, k) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i). (See A075195 formulas.) - Richard L. Ollerton, May 04 2021

A214309 a(n) is the number of representative four-color bracelets (necklaces with turning over allowed) with n beads, for n >= 4.

Original entry on oeis.org

3, 6, 26, 93, 424, 1180, 4844, 16165, 66953, 216804, 852822, 2949804, 12119134, 40886724, 160826008, 572457489, 2331396595, 8104270828, 32043699894, 115995102806, 471268872328, 1674576998468, 6641876380417, 24364816845446, 98894256728960, 357006263815751
Offset: 4

Views

Author

Wolfdieter Lang, Jul 31 2012

Keywords

Comments

This is the fourth column (m=4) of triangle A213940.
The relevant p(n,4)= A008284(n,4) representative color multinomials have exponents (signatures) from the 4 part partitions of n, written with nonincreasing parts. E.g., n=6: [3,1,1,1] and [2,2,1,1] (p(6,4)=2). The corresponding representative bracelets have the four-color multinomials c[1]^3*c[2]*c[3]*c[4] and c[1]^2*c[2]^2*c[3]*c[4].
Compare this with A032275 where also bracelets with less than four colors are included, and not only representatives are counted.
Number of n-length bracelets w over a 4-ary alphabet {a1,a2,...,a4} such that #(w,a1) >= #(w,a2) >= ... >= #(w,a4) >= 1, where #(w,x) counts the letters x in word w (bracelet analog of A226883). The number of 4 color bracelets up to permutations of colors is given by A056359. - Andrew Howroyd, Sep 26 2017

Examples

			a(4) = A213939(4,5) = 3 from the representative bracelets (with colors  j for c[j], j=1, 2, ..., 4) 1234, 1342 and 1423, all taken cyclically. The necklace cyclic(1324), for example, becomes equivalent to cyclic(1423) under the dihedral D_4 turning over (or reflection) operation.
a(6) = A213939(6, 8) = A213939(6, 9) =  10 + 16 = 26. See the comment above for the representative color multinomials for each case.

Links

Crossrefs

Cf. A213939, A213940, A214311 (m=5), A214312 (m=4, all bracelets).
Cf. A056359, A226883.

Programs

  • PARI
    Cyc(v)={my(g=fold(gcd,v), s=vecsum(v)); sumdiv(g, d, eulerphi(d)*(s/d)!/prod(i=1, #v, (v[i]/d)!))/s}
CPal(v)={my(odds=#select(t->t%2,v), s=vecsum(v));  if(odds>2, 0, ((s-odds)/2)!/prod(i=1, #v, (v[i]\2)!))}
a(n)={my(t=0); forpart(p=n, t+=Cyc(Vec(p))+CPal(Vec(p)), [1,n], [4,4]); t/2} \\ Andrew Howroyd, Sep 26 2017

Formula

a(n) = A213940(n,4), n >= 4.
a(n) = sum(A213939(n,k),k=(2+floor(n/2) + p(n,3))..(p(n,4)+1+floor(n/2)+p(n,3))), n>=4, with p(n,m) = A008284(n,m) the number of partitions of n with m parts.

Extensions

Terms a(26) and beyond from Andrew Howroyd, Sep 26 2017

A056344 Number of bracelets of length n using exactly four different colored beads.

Original entry on oeis.org

0, 0, 0, 3, 24, 136, 612, 2619, 10480, 41388, 159780, 614058, 2341920, 8919816, 33905188, 128907279, 490213680, 1866127840, 7111777860, 27140369148, 103721218000, 396974781456, 1521577377012, 5840547488954
Offset: 1

Views

Author

Marks R. Nester

Keywords

Comments

Turning over will not create a new bracelet.

Examples

			For a(4)=3, the arrangements are ABCD, ABDC, and ACBD, all chiral, their reverses being ADCB, ACDB, and ADBC respectively.

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Crossrefs

Column 4 of A273891.
Cf. A056284 (oriented), A056490 (achiral), A305543 (chiral).

Programs

  • 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)]);
T[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
a[n_] := T[n, 4];
Array[a, 24] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=4; Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,30}] (* Robert A. Russell, Sep 27 2018 *)

Formula

a(n) = A032275(n) - 4*A027671(n) + 6*A000029(n) - 4.
From Robert A. Russell, Sep 27 2018: (Start)
a(n) = (k!/4) * (S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/2n) * Sum_{d|n} phi(d) * S2(n/d,k), where k=4 is the number of colors and S2 is the Stirling subset number A008277.
G.f.: (k!/4) * x^(2k-2) * (1+x)^2 / Product_{i=1..k} (1-i x^2) - Sum_{d>0} (phi(d)/2d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=4 is the number of colors.
a(n) = (A056284(n) + A056490(n)) / 2 = A056284(n) - A305543(n) = A305543(n) + A056490(n). (End)

A321791 Table read by descending antidiagonals: T(n,k) is the number of unoriented cycles (bracelets) of length n using up to k available colors.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 6, 1, 0, 1, 6, 15, 20, 21, 8, 1, 0, 1, 7, 21, 35, 55, 39, 13, 1, 0, 1, 8, 28, 56, 120, 136, 92, 18, 1, 0, 1, 9, 36, 84, 231, 377, 430, 198, 30, 1, 0
Offset: 0

Views

Author

Robert A. Russell, Dec 18 2018

Keywords

Examples

			Table begins with T(0,0):
  1 1  1    1     1      1       1        1        1         1         1 ...
  0 1  2    3     4      5       6        7        8         9        10 ...
  0 1  3    6    10     15      21       28       36        45        55 ...
  0 1  4   10    20     35      56       84      120       165       220 ...
  0 1  6   21    55    120     231      406      666      1035      1540 ...
  0 1  8   39   136    377     888     1855     3536      6273     10504 ...
  0 1 13   92   430   1505    4291    10528    23052     46185     86185 ...
  0 1 18  198  1300   5895   20646    60028   151848    344925    719290 ...
  0 1 30  498  4435  25395  107331   365260  1058058   2707245   6278140 ...
  0 1 46 1219 15084 110085  563786  2250311  7472984  21552969  55605670 ...
  0 1 78 3210 53764 493131 3037314 14158228 53762472 174489813 500280022 ...
For T(3,3)=10, the unoriented cycles are 9 achiral (AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, CCC) and 1 chiral pair (ABC-ACB).

Links

Crossrefs

Cf. A075195 (oriented), A293496(chiral), A284855 (achiral).
Cf. A051137 (ascending antidiagonals).
Columns 0-6 are A000007, A000012, A000029, A027671, A032275, A032276, and A056341.
Rows 0-7 are A000012, A001477, A000217, A000292, A002817, A060446, A027670, and A060532.
Main diagonal gives A081721.

Programs

  • Mathematica
    Table[If[k>0, DivisorSum[k, EulerPhi[#](n-k)^(k/#)&]/(2k) + ((n-k)^Floor[(k+1)/2]+(n-k)^Ceiling[(k+1)/2])/4, 1], {n, 0, 12}, {k, 0, n}] // Flatten

Formula

T(n,k) = [n==0] + [n>0] * (k^floor((n+1)/2) + k^ceiling((n+1)/2)) / 4 + (1/(2*n)) * Sum_{d|n} phi(d) * k^(n/d).
T(n,k) = (A075195(n,k) + A284855(n,k)) / 2.
T(n,k) = A075195(n,k) - A293496(n,k) = A293496(n,k) + A284855(n,k).
Linear recurrence for row n: T(n,k) = Sum_{j=0..n} -binomial(j-n-1,j+1) * T(n,k-1-j) for k >= n + 1.
O.g.f. for column k >= 0: Sum_{n>=0} T(n,k)*x^n = 3/4 + (1 + k*x)^2/(4*(1 - k*x^2)) - (1/2) * Sum_{d >= 1} (phi(d)/d) * log(1 - k*x^d). - Petros Hadjicostas, Feb 07 2021

A056345 Number of bracelets of length n using exactly five different colored beads.

Original entry on oeis.org

0, 0, 0, 0, 12, 150, 1200, 7905, 46400, 255636, 1346700, 6901725, 34663020, 171786450, 843130688, 4110958530, 19951305240, 96528492700, 466073976900, 2247627076731, 10832193571460, 52194109216950
Offset: 1

Views

Author

Marks R. Nester

Keywords

Comments

Turning over will not create a new bracelet.

Examples

			For a(5)=12, pair up the 24 permutations of BCDE, each with its reverse, such as BCDE-EDCB.  Precede the first of each pair with an A, such as ABCDE.  These are the 12 arrangements, all chiral.  If we precede the second of each pair with an A, such as AEDCB, we get the chiral partner of each. - _Robert A. Russell_, Sep 27 2018

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Crossrefs

Column 5 of A273891.
Equals (A056285 + A056491) / 2 = A056285 - A305544 = A305544 + A056491.

Programs

  • 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)]);
T[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
a[n_] := T[n, 5];
Array[a, 22] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=5; Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,30}] (* Robert A. Russell, Sep 27 2018 *)

Formula

a(n) = A032276(n) - 5*A032275(n) + 10*A027671(n) - 10*A000029(n) + 5.
From Robert A. Russell, Sep 27 2018: (Start)
a(n) = (k!/4) * (S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/2n) * Sum_{d|n} phi(d) * S2(n/d,k), where k=5 is the number of colors and S2 is the Stirling subset number A008277.
G.f.: (k!/4) * x^(2k-2) * (1+x)^2 / Product_{i=1..k} (1-i x^2) - Sum_{d>0} (phi(d)/2d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=5 is the number of colors. (End)

A214312 a(n) is the number of all four-color bracelets (necklaces with turning over allowed) with n beads and the four colors are from a repertoire of n distinct colors, for n >= 4.

Original entry on oeis.org

3, 120, 2040, 21420, 183330, 1320480, 8691480, 52727400, 303958710, 1674472800, 8928735816, 46280581620, 234611247780, 1166708558400, 5710351190400, 27565250985360, 131495088522060, 620771489730000, 2903870526350640, 13473567673441260, 62061657617625204, 283995655732351200
Offset: 4

Views

Author

Wolfdieter Lang, Jul 31 2012

Keywords

Comments

This is the fourth column (m=4) of triangle A214306.
Each 4 part partition of n, with the parts written in nonincreasing order, defines a color signature. For a given color signature, say [p[1], p[2], p[3], p[4]], with p[1] >= p[2] >= p[3] >= p[4] >= 1, there are A213941(n,k)= A035206(n,k)*A213939(n,k) bracelets if this signature corresponds (with the order of the parts reversed) to the k-th partition of n in Abramowitz-Stegun (A-St) order. See A213941 for more details. Here all p(n,4)= A008284(n,4) partitions of n with 4 parts are considered. The color repertoire for a bracelet with n beads is [c[1], ..., c[n]].
Compare this with A032275 where also bracelets with less than four colors are included, and the color repertoire is only [c[1], c[2], c[3], c[4]] for all n.

Examples

			a(5) = A213941(5,6) = 120 from the bracelet (with colors j for c[j], j=1, 2, ..., 5) 11234, 11243, 11324, 12134, 13124 and 14123, all six taken cyclically, each representing a class of order A035206(5,6) = 20 (if all 5 colors are used). For example, cyclic(11342) becomes equivalent to cyclic(11243) by turning over or reflection. The multiplicity 20 depends only on the color signature.

Links

Crossrefs

Cf. A213941, A214306, A214309 (m=4, representative bracelets), A214313 (m=5).

Programs

  • 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)]);
a56344[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
a[n_] := Binomial[n, 4]*a56344[n, 4];
Table[a[n], {n, 4, 25}] (* Jean-François Alcover, Jul 02 2018, after Andrew Howroyd *)

Formula

a(n) = A214306(n,4), n >= 4.
a(n) = sum(A213941(n,k),k = A214314(n,4) .. (A214314(n,4) - 1 + A008284(n,4))), n >= 4.
a(n) = binomial(n,4) * A056344(n). - Andrew Howroyd, Mar 25 2017

A278640 Number of pairs of orientable necklaces with n beads and up to 4 colors; i.e., turning the necklace over does not leave it unchanged. The turned-over necklace is not included in the count.

Original entry on oeis.org

0, 0, 0, 4, 15, 72, 270, 1044, 3795, 14060, 51204, 188604, 694130, 2572920, 9567090, 35758704, 134137875, 505159200, 1908554190, 7233104844, 27486506268, 104713296760, 399817262550, 1529746919604, 5864041395730, 22517964582504, 86607602546220, 333599838189804, 1286742419927070, 4969488707124120, 19215357085867800
Offset: 0

Views

Author

Herbert Kociemba, Nov 24 2016

Keywords

Comments

Number of chiral bracelets of n beads using up to four different colors.

Examples

			Example: For 3 beads and the colors A, B, C and D the 4 orientable necklaces are ABC, ABD, ACD and BCD. The turned-over necklaces ACB, ADB, ADC and BDC are not included in the count.

Crossrefs

Column 4 of A293496.
Cf. A059076 (2 colors), A278639 (3 colors).
Equals (A001868 - A056486) / 2 = A001868 - A032275 = A032275 - A056486.

Programs

  • Mathematica
    mx=40;f[x_,k_]:=(1-Sum[EulerPhi[n]*Log[1-k*x^n]/n,{n,1,mx}]-Sum[Binomial[k,i]*x^i,{i,0,2}]/(1-k*x^2))/2;CoefficientList[Series[f[x,4],{x,0,mx}],x]
k=4; Prepend[Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) - (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}], 0] (* Robert A. Russell, Sep 24 2018 *)

Formula

G.f.: k=4, (1 - Sum_{n>=1} phi(n)*log(1 - k*x^n)/n - Sum_{i=0..2} Binomial[k,i]*x^i / ( 1-k*x^2) )/2.
For n > 0, a(n) = -(k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/2n)* Sum_{d|n} phi(d)*k^(n/d), where k=4 is the maximum number of colors. - Robert A. Russell, Sep 24 2018

A056346 Number of bracelets of length n using exactly six different colored beads.

Original entry on oeis.org

0, 0, 0, 0, 0, 60, 1080, 11970, 105840, 821952, 5874480, 39713550, 258136200, 1631273220, 10096734312, 61536377700, 370710950400, 2213749658880, 13132080672480, 77509456944318, 455754569692680
Offset: 1

Views

Author

Marks R. Nester

Keywords

Comments

Turning over will not create a new bracelet.

Examples

			For a(6)=60, pair up the 120 permutations of BCDEF, each with its reverse, such as BCDEF-FEDCB.  Precede the first of each pair with an A, such as ABCDEF.  These are the 60 arrangements, all chiral.  If we precede the second of each pair with an A, such as AFEDCB, we get the chiral partner of each. - _Robert A. Russell_, Sep 27 2018

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Crossrefs

Column 6 of A273891.
Equals (A056286 + A056492) / 2 = A056286 - A305545 = A305545 + A056492.
Cf. A008277.

Programs

  • 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)]);
T[n_, k_] := Sum[(-1)^i*Binomial[k, i]*t[n, k - i], {i, 0, k - 1}];
a[n_] := T[n, 6];
Array[a, 21] (* Jean-François Alcover, Nov 05 2017, after Andrew Howroyd *)
k=6; Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,30}] (* Robert A. Russell, Sep 27 2018 *)
  • PARI
    a(n) = my(k=6); (k!/4) * (stirling(floor((n+1)/2),k,2) + stirling(ceil((n+1)/2),k,2)) + (k!/(2*n))*sumdiv(n, d, eulerphi(d)*stirling(n/d,k,2)); \\ Michel Marcus, Sep 29 2018

Formula

a(n) = A056341(n) - 6*A032276(n) + 15*A032275(n) - 20*A027671(n) + 15*A000029(n) - 6.
From Robert A. Russell, Sep 27 2018: (Start)
a(n) = (k!/4) * (S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)) + (k!/2n) * Sum_{d|n} phi(d) * S2(n/d,k), where k=6 is the number of colors and S2 is the Stirling subset number A008277.
G.f.: (k!/4) * x^(2k-2) * (1+x)^2 / Product_{i=1..k} (1-i x^2) - Sum_{d>0} (phi(d)/2d) * Sum_{j} (-1)^(k-j) * C(k,j) * log(1-j x^d), where k=6 is the number of colors. (End)
Showing 1-10 of 12 results. Next

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

Content is available under The OEIS End-User License Agreement.