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 1-3 of 3 results.

A032279 Number of bracelets (turnover necklaces) of n beads of 2 colors, 5 of them black.

Original entry on oeis.org

1, 1, 3, 5, 10, 16, 26, 38, 57, 79, 111, 147, 196, 252, 324, 406, 507, 621, 759, 913, 1096, 1298, 1534, 1794, 2093, 2421, 2793, 3199, 3656, 4152, 4706, 5304, 5967, 6681, 7467, 8311, 9234, 10222, 11298, 12446, 13691, 15015, 16445
Offset: 5

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Comments

From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of 5 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in case k=5. The full solution was given by H. Gupta (1979); I gave a short proof of Gupta's result and showed an equivalence of this problem and every one of the following problems: enumerating the bracelets of n beads of 2 colors, k of them black, and enumerating the necklaces of k beads each of them painted by one of n colors.
a(n) is an essentially unimprovable upper estimate for the number of distinct values of the permanent in (0,1)-circulants of order n with five 1's in every row. (End)
a(n+5) is the number of symmetry-allowed, linearly-independent terms at n-th order in the series expansion of the T_1 X h vibronic perturbation matrix, H(Q) (cf. Dunn & Bates). - Bradley Klee, Jul 20 2015
From Petros Hadjicostas, Jul 17 2018: (Start)
Let (c(n): n >= 1) be a sequence of nonnegative integers and let C(x) = Sum_{n>=1} c(n)*x^n be its g.f. Let k be a positive integer. Let a_k = (a_k(n): n >= 1) be the output sequence of the DIK[k] transform of sequence (c(n): n >= 1), and let A_k(x) = Sum_{n>=1} a_k(n)*x^n be its g.f. See Christian G. Bower's web link below. It can be proved that, when k is odd, A_k(x) = ((1/k)*Sum_{d|k} phi(d)*C(x^d)^(k/d) + C(x^2)^((k-1)/2)*C(x))/2.
For this sequence, k = 5, c(n) = 1 for all n >= 1, and C(x) = x/(1-x). Thus, a(n) = a_5(n) for all n >= 1. Since a_k(n) = 0 for 1 <= n <= k-1, the offset of this sequence is n = k = 5. Applying the formula for the g.f. of DIK[5] of (c(n): n >= 1) with C(x) = x/(1-x) and k = 5, we get A(x) = A_5(x) = x^5*((1/5)*Sum_{d|5} phi(d)*(1-x^d)^(-5/d) + (1+x)/(1-x^2)^3)/2, which obviously equals the g.f. in the formula section below.
The g.f. is also a special case of Herbert Kociemba's formula that is valid for both even and odd k: A_k(x) = x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^Floor[(k+2)/2])/2.
Here, a(n) is defined to be the number of n-bead bracelets of two colors with 5 black beads and n-5 white beads. But it is also the number of dihedral compositions of n with 5 positive parts. (This statement is equivalent to Vladimir Shevelev's statement above that a(n) is the "number of non-equivalent necklaces of 5 beads each of them painted by one of n colors." By "necklaces" he means "turnover necklaces". See paragraph (2) of Section 2 in his 2004 paper in the Indian Journal of Pure and Applied Mathematics.)
Two cyclic compositions of n (with k = 5 parts) belong to the same equivalence class corresponding to a dihedral composition of n if and only if one can be obtained from the other by a rotation or reversal of order. (End)

Examples

			From _Petros Hadjicostas_, Jul 17 2018: (Start)
Every n-bead bracelet of two colors such that 5 beads are black and n-5 are white can be transformed into a dihedral composition of n with 5 positive parts in the following way. Start with one B bead and go in one direction (say clockwise) until you reach the next B bead. Continue this process until you come back to the original B bead.
Let b_i be the number of beads from B bead i until you reach the last W bead before B bead i+1 (or B bead 1). Here, b_i = 1 iff there are no W beads between B bead i and B bead i+1 (or B bead 5 and B bead 1). Then b_1 + b_2 + b_3 + b_4 + b_5 = n, and we get a dihedral composition of n. (Of course, b_2 + b_3 + b_4 + b_5 + b_1 and b_5 + b_4 + b_3 + b_2 + b_1 belong to the same equivalence class of the dihedral composition b_1 + b_2 + b_3 + b_4 + b_5.)
For example, a(8) = 5, and we have the following bracelets with 5 B beads and 3 W beads. Next to the bracelets we list the corresponding dihedral compositions of n with k=5 parts (they must be viewed on a circle):
BBBBBWWW <-> 1+1+1+1+4
BBBBWBWW <-> 1+1+1+2+3
BBWBBBWW <-> 1+2+1+1+3
BWBBWBWB <-> 2+1+2+2+1
BWBWBWBB <-> 2+2+2+1+1
(End)
		

References

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

Crossrefs

Programs

  • Magma
    m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1-x+2*x^3-x^5+x^6)/((1-x)^2*(1-x^2)^2*(1-x^5))); // Vincenzo Librandi, Sep 07 2013
  • Maple
    seq(floor(n^4/240 + n^3/24 + 5*n^2/24 + 25*n/48 + 1 + (-1)^n*n/16), n=0..100); # Robert Israel, Jul 22 2015
  • Mathematica
    k = 5; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    CoefficientList[Series[(1 - x + 2 x^3 - x^5 + x^6) / ((1 - x)^2 (1 - x^2)^2 (1 - x^5)), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 07 2013 *)
    k=5 (* Number of black beads in bracelet problem *); CoefficientList[Series[x^k*(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)
  • PARI
    a(n) = round((n^4 -10*n^3 +50*n^2 -(110+30*(1-n%2))*n)/240 +3/5) \\ Washington Bomfim, Jul 17 2008
    

Formula

"DIK[ 5 ]" (necklace, indistinct, unlabeled, 5 parts) transform of 1, 1, 1, 1, ...
G.f.: x^5*(1-x+2*x^3-x^5+x^6)/((1-x)^2*(1-x^2)^2*(1-x^5)). - corrected for offset 5 by Robert Israel, Jul 22 2015
From Vladimir Shevelev, Apr 23 2011: (Start)
Put s(n,k,d)=1, if n == k (mod d), and 0, otherwise. Then
a(n) = (2/5)*s(n,0,5) + (n-1)*(n-3)*((n-2)*(n-4) + 15)/240, if n is odd >= 5;
a(n) = (2/5)*s(n,0,5) + (n-2)*(n-4)*((n-1)*(n-3) + 15)/240, if n is even >= 5. (End)
a(n+5) = floor(n^4/240 + n^3/24 + 5*n^2/24 + 25*n/48 + 1 + (-1)^n*n/16). - Robert Israel, Jul 22 2015
a(n) = (A008646(n-5) + A119963(n, 5))/2 = (A008646(n-5) + C(floor((n-1)/2), 2))/2 for n >= 5. - Petros Hadjicostas, Jul 17 2018

A005514 Number of n-bead bracelets (turnover necklaces) with 8 red beads and n-8 black beads.

Original entry on oeis.org

1, 1, 5, 10, 29, 57, 126, 232, 440, 750, 1282, 2052, 3260, 4950, 7440, 10824, 15581, 21879, 30415, 41470, 56021, 74503, 98254, 127920, 165288, 211276, 268228, 337416, 421856, 523260, 645456, 790704, 963793, 1167645, 1408185
Offset: 8

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From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of 8 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in case k=8 (see our comment at A032279). (End)
From Petros Hadjicostas, Jul 14 2018: (Start)
Let (c(n): n >= 1) be a sequence of nonnegative integers and let C(x) = Sum_{n>=1} c(n)*x^n be its g.f. Let k be a positive integer. Let a_k = (a_k(n): n >= 1) be the output sequence of the DIK[k] transform of sequence (c(n): n >= 1), and let A_k(x) = Sum_{n>=1} a_k(n)*x^n be its g.f. See Bower's web link below. It can be proved that, when k is even, A_k(x) = ((1/k)*Sum_{d|k} phi(d)*C(x^d)^(k/d) + (1/2)*C(x^2)^((k/2)-1)*(C(x)^2 + C(x^2)))/2.
For this sequence, k=8, c(n) = 1 for all n >= 1, and C(x) = x/(1-x). Thus, a(n) = a_8(n) for all n >= 1. Since a_k(n) = 0 for 1 <= n <= k-1, the offset of this sequence is n = k = 8. Applying the formula for the g.f. of DIK[8] of (c(n): n >= 1) with C(x) = x/(1-x) and k=8, we get Herbert Kociemba's formula below.
Here, a(n) is defined to be the number of n-bead bracelets of two colors with 8 red beads and n-8 black beads. But it is also the number of dihedral compositions of n with 8 parts. (This statement is equivalent to Vladimir Shevelev's statement above that a(n) is the "number of non-equivalent necklaces of 8 beads each of them painted by one of n colors." By "necklaces", he means "turnover necklaces". See the second paragraph of Section 2 in his 2004 paper in the Indian Journal of Pure and Applied Mathematics.)
Two cyclic compositions of n (with k = 8 parts) belong to the same equivalence class corresponding to a dihedral composition of n if and only if one can be obtained from the other by a rotation or reversal of order. (End)

Examples

			From _Petros Hadjicostas_, Jul 14 2018: (Start)
Every n-bead bracelet of two colors such that 8 beads are red and n-8 are black can be transformed into a dihedral composition of n with 8 parts in the following way. Start with one R bead and go in one direction (say clockwise) until you reach the next R bead. Continue this process until you come back to the original R bead.
Let b_i be the number of beads from R bead i until you reach the last B bead before R bead i+1 (or R bead 1). Here, b_i = 1 iff there are no B beads between R bead i and R bead i+1 (or R bead 8 and R bead 1). Then b_1 + b_2 + ... + b_8 = n, and we get a dihedral composition of n. (Of course, b_2 + b_3 + ... + b_8 + b_1 and b_8 + b_7 + ... + b_1 belong to the same equivalence class of the dihedral composition b_1 + ... + b_8.)
For example, a(10) = 5, and we have the following bracelets with 8 R beads and 2 B beads. Next to the bracelets we list the corresponding dihedral compositions of n with k=8 parts (they must be viewed on a circle):
RRRRRRRRBB <-> 1+1+1+1+1+1+1+3
RRRRRRRBRB <-> 1+1+1+1+1+1+2+2
RRRRRRBRRB <-> 1+1+1+1+1+2+1+2
RRRRRBRRRB <-> 1+1+1+1+2+1+1+2
RRRRBRRRRB <-> 1+1+1+2+1+1+1+2
(End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Programs

  • Mathematica
    k = 8; Table[(Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n + Binomial[If[OddQ[n], n - 1, n - If[OddQ[k], 2, 0]]/2, If[OddQ[k], k - 1, k]/2])/2, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
    k=8;CoefficientList[Series[x^k*(1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])+(1+x)/(1-x^2)^Floor[(k+2)/2])/2,{x,0,50}],x] (* Herbert Kociemba, Nov 04 2016 *)

Formula

S. J. Cyvin et al. (1997) give a g.f. (See equation (18) on p. 870 of their paper. Their g.f. is the same as the one given by V. Jovovic below except for the extra x^8.) - Petros Hadjicostas, Jul 14 2018
G.f.: (x^8/16)*(1/(1 - x)^8 + 4/(1 - x^8) + 5/(1 - x^2)^4 + 2/(1 - x^4)^2 + 4/(1 - x)^2/(1 - x^2)^3) = x^8*(2*x^10 - 3*x^9 + 7*x^8 - 6*x^7 + 7*x^6 - 2*x^5 + 2*x^4 - 2*x^3 + 5*x^2 - 3*x + 1)/(1 - x)^8/(1 + x)^4/(1 + x^2)^2/(1 + x^4). - Vladeta Jovovic, Jul 17 2002
a(n) = ((n+4)/32)*s(n,0,8) + ((n-4)/32)*s(n,4,8) + (48*C(n-1,7) + (n+1)*(n-2)*(n-4)*(n-6))/768, if n is even >= 8; a(n) = (48*C(n-1,7) + (n-1)*(n-3)*(n-5)*(n-7))/768, if n odd >= 8, where s(n,k,d)=1, if n == k (mod d), and 0 otherwise. - Vladimir Shevelev, Apr 23 2011
G.f.: k=8, x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^floor((k+2)/2))/2. - Herbert Kociemba, Nov 05 2016 [edited by Petros Hadjicostas, Jul 18 2018]
From Petros Hadjicostas, Jul 14 2018: (Start)
a(n) = (A032193(n) + A119963(n, 8))/2 = (A032193(n) + C(floor(n/2), 4))/2 for n >= 8.
The sequence (a(n): n >= 8) is the output sequence of Bower's "DIK[ 8 ]" (bracelet, indistinct, unlabeled, 8 parts) transform of 1, 1, 1, 1, ...
(End)

Extensions

Sequence extended and description corrected by Christian G. Bower
Name edited by Petros Hadjicostas, Jul 20 2018

A106369 Number of circular compositions of n such that no two adjacent parts are equal.

Original entry on oeis.org

1, 1, 2, 2, 3, 6, 7, 11, 18, 29, 42, 73, 111, 183, 299, 491, 796, 1333, 2188, 3652, 6073, 10155, 16959, 28500, 47813, 80508, 135621, 228967, 386749, 654535, 1108353, 1879478, 3189495, 5418556, 9212099, 15676275, 26694509, 45493327, 77580915
Offset: 1

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Author

Christian G. Bower, Apr 29 2005

Keywords

Comments

By "circular compositions" here we mean equivalence classes of compositions with parts on a circle such that two compositions are equivalent if one is a cyclic shift of the other. - Petros Hadjicostas, Oct 15 2017

Examples

			a(6) = 6 because the 6 circular compositions of 6: 6, 5+1, 4+2, 3+2+1, 3+1+2, 2+1+2+1.
		

Crossrefs

Programs

  • Mathematica
    nmax = 40; Rest[CoefficientList[Series[x/(1-x) - Sum[EulerPhi[s]/s*(Log[1 - Sum[x^(s*n)/(1 + x^(s*n)), {n, 1, nmax}]] + Sum[Log[1 + x^(s*n)], {n, 1, nmax}]), {s, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Sep 06 2017, after Petros Hadjicostas *)

Formula

CycleBG transform of (1, 1, 1, 1, ...).
CycleBG transform T(A) = invMOEBIUS(invEULER(Carlitz(A)) + A(x^2) - A) + A.
Carlitz transform T(A(x)) has g.f. 1/(1 - Sum_{k>0} (-1)^(k+1)*A(x^k)).
G.f.: x/(1-x) - Sum_{s>=1} (phi(s)/s)*f(x^s), where f(x) = log(1 - Sum_{n>=1} x^n/(1 + x^n)) + Sum_{n>=1} log(1 + x^n) and phi(s)=A000010 is Euler's totient function. - Petros Hadjicostas, Sep 06 2017
Conjecture: a(n) ~ A241902^n / n. - Vaclav Kotesovec, Sep 06 2017
General formula for the CycleBG transform: T(A)(x) = A(x) - Sum_{k>=0} A(x^(2k+1)) + Sum_{k>=1} (phi(k)/k)*log(Carlitz(A)(x^k)). For a proof, see the links above. (For this sequence, A(x) = x/(1-x).) - Petros Hadjicostas, Oct 08 2017
G.f.: -Sum_{s>=1} x^(2s+1)/(1-x^(2s+1)) - Sum_{s>=1} (phi(s)/s)*g(x^s), where g(x) = log(1 + Sum_{n>=1} (-x)^n/(1 - x^n)). (This formula can be proved from the general formula for the CycleBG transform given above.) - Petros Hadjicostas, Oct 10 2017

Extensions

Name clarified by Andrew Howroyd, Oct 12 2017
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