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.

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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)

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)

A092481 Number of different sets of n-gons labeled 1...n such that all members of each set contain equivalent paths with increasing labels; i.e., the number of isotemporal classes of n-gons.

Original entry on oeis.org

1, 3, 3, 8, 9, 20, 29, 60, 93, 189, 315, 618, 1095, 2114, 3855, 7414, 13797, 26478, 49939, 95838, 182361, 350572, 671091, 1292604, 2485533, 4797616, 9256395, 17903928, 34636833, 67125304, 130150587, 252677904, 490853415, 954502948
Offset: 3

Views

Author

Benjamin de Bivort (bivort(AT)fas.harvard.edu), Apr 03 2004

Keywords

References

  • B. de Bivort, Isotemporal classes of n-gons, preprint, 2004.
  • B. de Bivort, An introduction to temporal networks, preprint, 2004.

Links

Crossrefs

Cf. A000031, A000029, A027671.

Programs

  • Mathematica
    f[n_] := Block[{d = Divisors[n], c = Divisors[n/2]}, Switch[ Mod[n, 4], 0, (Plus @@ (2^(n/d - 1)EulerPhi[d]) - Plus @@ (2^(n/(2c) - 1)EulerPhi[2c]))/n + 2^((n - 4)/2) + 2^((n - 8)/4) - 2^(Ceiling[(n - 4)/8] - 1), 1, (Plus @@ ((2^(n/d - 1) - 1)EulerPhi[d]))/n, 2, (Plus @@ (2^(n/d - 1)EulerPhi[d]) - Plus @@ (2^(n/(2c) - 1)EulerPhi[2c]))/n + 2^((n - 4)/2), 3, (Plus @@ ((2^(n/d - 1) - 1)EulerPhi[d]))/n]]; Table[ f[n], {n, 3, 36}]

Formula

If n odd: (1/n) sum_{d|n} (2^(n/d-1)-1) phi(d).
If n = 4k + 2: (1/n) {sum_{d|n} (2^(n/d-1) phi(d)) - sum_{c|n/2} (2^(n/2c-1) phi(2c)} + 2^(n-4)/2
If n = 4k: (1/n) {sum_{d|n} (2^(n/d-1) phi(d)) - sum_{c|n/2} (2^(n/2c-1) phi(2c))} + 2^(n-4)/2 + 2^(n-8)/4 - 2^(ceiling[(n-4)/8]-1).

Extensions

Edited by Robert G. Wilson v, Apr 09 2004

A214310 a(n) is the number of all three-color bracelets (necklaces with turning over allowed) with n beads and the three colors are from a repertoire of n distinct colors, for n >= 3.

Original entry on oeis.org

1, 24, 180, 1120, 5145, 23016, 91056, 357480, 1327095, 4893680, 17525508, 62254920, 217457695, 753332160, 2581110000, 8779264032, 29624681763, 99350001360, 331159123260, 1098168382080, 3624003213369, 11908069219816, 38972450763000, 127087400895000
Offset: 3

Views

Author

Wolfdieter Lang, Jul 31 2012

Keywords

Comments

This is the third column (m=3) of triangle A214306.
Each 3 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]], with p[1] >= p[2] >= p[3] >= 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,3)= A008284(n,3) partitions of n with 3 parts are considered. The color repertoire for a bracelet with n beads is [c[1], ..., c[n]].
Compare this with A027671 where also single color bracelets are included, and the color repertoire is only [c[1], c[2], c[3]] for all n.

Examples

			a(5) = A213941(5,4) + A213941(5,5) = 60 + 120 = 180 from the bracelet (with colors j for c[j], j=1, 2, ..., 5) 11123 and 11213, both taken cyclically, each representing a class of order A035206(5,4)= 30 (if all 5 colors are used), and 11223, 11232, 12123 and 12213, all taken cyclically, each representing a class of order A035206(5,5)= 30. For example, cyclic(11322) becomes equivalent to cyclic(11223) by turning over or reflection. The multiplicity A035206 depends only on the color signature.

Links

Crossrefs

Cf. A213941, A214306, A214307 (m=3, representative bracelets), A214312 (m=4).

Formula

a(n) = A214306(n,3), n >= 3.
a(n) = sum(A213941(n,k), k = A214314(n,3).. (A214314(n,3) - 1 + A008284(n,3))), n >= 3.
a(n) = binomial(n,3) * A056343(n). - Andrew Howroyd, Mar 25 2017

Extensions

a(26) from Andrew Howroyd, Mar 25 2017

A032294 Number of aperiodic bracelets (turnover necklaces) with n beads of 3 colors.

Original entry on oeis.org

3, 3, 7, 15, 36, 79, 195, 477, 1209, 3168, 8415, 22806, 62412, 172887, 481552, 1351485, 3808080, 10780653, 30615351, 87226932, 249144506, 713378655, 2046856563, 5884468110, 16946569332, 48883597728, 141217159239
Offset: 1

Views

Author

Christian G. Bower

Keywords

Links

Crossrefs

Column 3 of A276550.

Programs

  • Mathematica
    mx=40;gf[x_,k_]:=Sum[ MoebiusMu[n]*(-Log[1-k*x^n]/n+Sum[Binomial[k,i]x^(n i),{i,0,2}]/( 1-k x^(2n)))/2,{n,mx}]; CoefficientList[Series[gf[x,3],{x,0,mx}],x] (* Herbert Kociemba, Nov 28 2016 *)
  • PARI
    a(x, k) = sum(n=1, 40, moebius(n) * (-log(1 - k*x^n )/n + sum(i=0, 2, binomial(k, i) * x^(n*i)) / (1 - k* x^(2*n)))/2);
Vec(a(x, 3) + O(x^41)) \\ Indranil Ghosh, Mar 29 2017

Formula

MOEBIUS transform of A027671.
From Herbert Kociemba, Nov 28 2016: (Start)
More generally, gf(k) is the g.f. for the number of bracelets with primitive period n and beads of k colors.
gf(k): Sum_{n>=1} mu(n)*( -log(1-k*x^n)/n + Sum_{i=0..2} binomial(k,i)x^(n*i)/(1-k*x^(2*n)) )/2. (End)
Previous Showing 11-15 of 15 results.

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