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.

A032198 "CIK" (necklace, indistinct, unlabeled) transform of 1,2,3,4,...

Original entry on oeis.org

1, 3, 6, 13, 25, 58, 121, 283, 646, 1527, 3601, 8678, 20881, 50823, 124054, 304573, 750121, 1855098, 4600201, 11442085, 28527446, 71292603, 178526881, 447919418, 1125750145, 2833906683, 7144450566, 18036423973
Offset: 1

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Examples

			From _Petros Hadjicostas_, Jan 07 2018: (Start)
We give some examples to illustrate the theory of C. G. Bower about transforms given in the weblink above. We assume we have boxes of different sizes and colors that we place on a circle to form a necklace. Two boxes of the same size and same color are considered identical (indistinct and unlabeled). We do, however, change the roles of the sequences (a(n): n>=1) and (b(n): n>=1) that appear in the weblink above. We assume (a(n): n>=1) = CIK((b(n): n>=1)).
Since b(1) = 1, b(2) = 2, b(3) = 3, etc., a box that can hold 1 ball only can be of 1 color only, a box that can hold 2 balls only can be one of 2 colors only, a box that can hold 3 balls can be one of 3 colors, and so on.
To prove that a(3) = 6, we consider three cases. In the first case, we have a single box that can hold 3 balls, and thus we have 3 possibilities for the 3 colors the box can be. In the second case, we have a box that can hold 2 balls and a box that can hold 1 ball. Here, we have 2 x 1 = 2 possibilities. In the third case, we have 3 identical boxes, each of which can hold 1 ball. This gives rise to 1 possibility. Hence, a(3) = 3 + 2 + 1 = 6.
To prove that a(4) = 13, we consider 5 cases: a box with 4 balls (4 possibilities), one box with 3 balls and one box with 1 ball (3 possibilities), two identical boxes each with 2 balls (3 possibilities), one box with 2 balls and two identical boxes each with 1 ball (2 possibilities), and four identical boxes each with 1 ball (1 possibility). Thus, a(4) = 4 + 3 + 3 + 2 + 1 = 13.
(End)
		

Crossrefs

Programs

  • Mathematica
    nmax = 30;
    f[x_] = Sum[n*x^n, {n, 1, nmax}];
    gf = Sum[(EulerPhi[n]/n)*Log[1/(1 - f[x^n])] + O[x]^nmax, {n, 1, nmax}]/x;
    CoefficientList[gf, x] (* Jean-François Alcover, Jul 29 2018, after Joerg Arndt *)
  • PARI
    N = 66;  x = 'x + O('x^N);
    f(x)=sum(n=1, N, n*x^n );
    gf = sum(n=1, N, eulerphi(n)/n*log(1/(1-f(x^n)))  );
    v = Vec(gf)
    /* Joerg Arndt, Jan 21 2013 */

Formula

a(n) = 1/n*Sum_{d divides n} phi(n/d)*A004146(d). - Vladeta Jovovic, Feb 15 2003
From Petros Hadjicostas, Jan 07 2018: (Start)
a(n) = -2 + (1/n)*Sum_{d|n} phi(n/d)*A005248(d) = -2 + (1/n)*Sum_{d|n} phi(n/d)*L(2*d), where L(n) = A000032(n) is the usual Lucas sequence.
G.f.: -Sum_{n >= 1} (phi(n)/n)*log(1 - B(x^n)), where B(x) = x + 2*x^2 + 3*x^3 + 4*x^4 + ... = x/(1-x)^2.
G.f.: -2*x/(1 - x) - Sum_{n>=1} (phi(n)/n)*log(1 - 3*x^n + x^(2*n)).
(End)
From Petros Hadjicostas, Jun 19 2019: (Start)
According to Gibson et al. (2018), a(n) is the number of m-color cyclic compositions of n where each part of size m has m possible colors. This is nothing else than the CIK transform of the sequence 1, 2, 3, 4, ...
Using the theory of Flajolet and Soria (1991), Gibson et al. (2018, Eq. (1.1)) proved that the g.f. of a(n) is Sum_{s >= 1} (phi(s)/s) * log((1 - x^s)^2/(1 - 3*x^s + x^(2*s))), which is exactly the same g.f. as the ones above.
Gibson et al. (2018, p. 3210) also proved that a(n) ~ (2/(3-sqrt(5)))^n/n for large n. See also Chapter 3 in Gibson (2017).
(End)

A032287 "DIK" (bracelet, indistinct, unlabeled) transform of 1,2,3,4,...

Original entry on oeis.org

1, 3, 6, 13, 24, 51, 97, 207, 428, 946, 2088, 4831, 11209, 26717, 64058, 155725, 380400, 936575, 2314105, 5744700, 14300416, 35708268, 89359536, 224121973, 563126689, 1417378191, 3572884062, 9019324297, 22797540648
Offset: 1

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From Petros Hadjicostas, Jun 21 2019: (Start)
Under Bower's transforms, the input sequence c = (c(m): m >= 1) describes how each part of size m in a composition is colored. In a composition (ordered partition) of n >= 1, a part of size m is assumed to be colored with one of c(m) colors.
Under the DIK transform, we are dealing with "dihedral compositions" of n >= 1. These are equivalence classes of ordered partitions of n such that two such ordered partitions are equivalent if one can be obtained from the other by rotation or reflection.
If the input sequence is c = (c(m): m >= 1), denote the output sequence under the DIK transform by b = (b(n): n >= 1); i.e., b(n) = (DIK c)(n) for n >= 1. If C(x) = Sum_{m >= 1} c(m)*x^m is the g.f. of the input sequence c, then the g.f. of b = DIK c is Sum_{n >= 1} b(n)*x^n = -(1/2) * Sum_{d >= 1} (phi(d)/d) * log(1 - C(x^d)) + (1 + C(x))^2/(4 * (1 - C(x^2))) - (1/4).
For the current sequence (a(n): n >= 1), the input sequence is c(m) = m for all m >= 1. That is, we are dealing with the so-called "m-color dihedral compositions". Here, a(n) is the number of dihedral compositions of n where each part of size m may be colored with one of m colors. For the linear and cyclic versions of such m-color compositions, see Agarwal (2000), Gibson (2017), and Gibson et al. (2018).
Since C(x) = x/(1 - x)^2, we have Sum_{n >= 1} a(n) * x^n = (1/2) * Sum_{d >= 1} (phi(d)/d) * log((1 - x^d)^2 / (1 - 3*x^d + x^(2*d))) + (1/2) * x * (1 + x - 2*x^2 + x^3 + x^4)/((1 - x)^2 * (1 + x - x^2) * (1 - x - x^2)), which is the g.f. given by Andrew Howroyd in the PARI program below.
Note that -Sum_{d >= 1} (phi(d)/d) * log (1 - C(x^d)) = Sum_{d >= 1} (phi(d)/d) * log((1 - x^d)^2 / (1 - 3*x^d + x^(2*d))) is the g.f. of the "m-color cyclic compositions" that appear in Gibson (2017) and Gibson et al. (2018). See sequence A032198, which is the CIK transform of sequence (c(m): m >= 1) = (m: m >= 1).
(End)

Crossrefs

Programs

  • Maple
    DIK := proc(L::list)
        local  x,cidx,ncyc,d,gd,g,g2,n ;
        n := nops(L) ;
        g := add(op(i,L)*x^i,i=1..n) ;
        # wrap into the cycle index of the cyclic group C_n
        cidx := 0 ;
        for ncyc from 1 to n do
            for d in numtheory[divisors](ncyc) do
                gd := subs(x=x^d,g) ;
                cidx := cidx+1/ncyc*numtheory[phi](d)*gd^(ncyc/d) ;
            end do:
        end do:
        # cycle index is half of th eone for the cyclic group plus two
        # different branches or D_n with even or odd n
        cidx := cidx/2 ;
        g2 := subs(x=x^2,g) ;
        for ncyc from 1 to n do
            if type(ncyc,'odd') then
                cidx := cidx+ g*g2^((ncyc-1)/2)/2 ;
            else
                cidx := cidx+ (g^2*g2^((ncyc-2)/2) + g2^(ncyc/2))/4 ;
            end if;
        end do:
        taylor(cidx,x=0,nops(L)) ;
        gfun[seriestolist](%) ;
    end proc:
    A032287_list := proc(n)
            local ele ;
            ele := [seq(i,i=1..40)] ;
            DIK(ele) ;
    end proc:
    A032287_list(50) ; # R. J. Mathar, Feb 14 2025
  • Mathematica
    seq[n_] := x(1 + x - 2 x^2 + x^3 + x^4)/((1 - x)^2 (1 - x - x^2)(1 + x - x^2)) + Sum[EulerPhi[d]/d Log[(1 - x^d)^2/(1 - 3 x^d + x^(2d)) + O[x]^(n+1)], {d, 1, n}] // CoefficientList[#, x]& // Rest // #/2&;
    seq[30] (* Jean-François Alcover, Sep 17 2019, from PARI *)
  • PARI
    seq(n)={Vec(x*(1 + x - 2*x^2 + x^3 + x^4)/((1 - x)^2*(1 - x - x^2)*(1 + x - x^2)) + sum(d=1, n, eulerphi(d)/d*log((1 - x^d)^2/(1 - 3*x^d + x^(2*d)) + O(x*x^n))))/2} \\ Andrew Howroyd, Jun 20 2018

Formula

From Petros Hadjicostas, Jun 21 2019: (Start)
a(n) = ( F(n+4) + (-1)^n * F(n-4) - 2 * (n + 1) + (1/n) * Sum_{d|n} phi(n/d) * L(2*d) )/2 for n >= 4, where F(n) = A000045(n) and L(n) = A000032(n) are the usual n-th Fibonacci and n-th Lucas numbers, respectively.
a(n) = (A032198(n) + A308747(n))/2 for n >= 1.
G.f.: (1/2) * Sum_{d >= 1} (phi(d)/d) * log((1 - x^d)^2 / (1 - 3*x^d + x^(2*d))) + (1/2) * x * (1 + x - 2*x^2 + x^3 + x^4)/((1 - x)^2 * (1 + x - x^2) * (1 - x - x^2)).
(End)

A308747 Number of achiral m-color cyclic compositions of n (that is, number of cyclic compositions of n with reflection symmetry where each part of size m can be colored with one of m colors).

Original entry on oeis.org

1, 3, 6, 13, 23, 44, 73, 131, 210, 365, 575, 984, 1537, 2611, 4062, 6877, 10679, 18052, 28009, 47315, 73386, 123933, 192191, 324528, 503233, 849699, 1317558, 2224621, 3449495, 5824220, 9030985, 15248099, 23643522, 39920141, 61899647, 104512392, 162055489, 273617107
Offset: 1

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Petros Hadjicostas, Jun 21 2019

Keywords

Comments

Cyclic compositions of a positive integer n are equivalence classes of ordered partitions of n such that two such partitions are equivalent if one can be obtained from the other by rotation. These were first studied by Sommerville (1909).
Symmetric cyclic compositions or circular palindromes or achiral cyclic compositions are those cyclic compositions that have at least one axis of symmetry. They were also studied by Sommerville (1909, pp. 301-304).
Let (c(m): m >= 1) be the input sequence and let b = (b(n): n >= 1) be the output sequence under the CPAL (circular palindrome) transform of c; that is, b(n) = (CPAC c)n for n >= 1. Hence, b(n) is the number of symmetric cyclic compositions of n where a part of size m can be colored with one of c(m) colors. If C(x) = Sum{m >= 1} c(m)*x^m is the g.f. of the input sequence c, then the g.f. of b = (CPAL c) is Sum_{n >= 1} b(n)*x^n = (1 + C(x))^2/(2 * (1 - C(x^2))) - (1/2).
For the current sequence, the input sequence is c(m) = m for m >= 1, and we are dealing with the so-called "m-color" compositions. m-color linear compositions were studied by Agarwal (2000), whereas m-color cyclic compositions were studied by Gibson (2017) and Gibson et al. (2018).
Thus, for the current sequence, a(n) is the number of symmetric (achiral) cyclic compositions of n where a part of size m may be colored with one of m colors (for each m >= 1).
The function A(x) = (exp(Pi*(x + 1)*I)*phi^(-x - 4) - exp(2*I*Pi*x)*phi^(4 - x) + exp(Pi*x*I)*phi^(x - 4) + phi^(x + 4))/sqrt(5) - 2*x, where phi is the golden ratio, shows that the sequence can be easily extended to all integers. - Peter Luschny, Aug 09 2020

Examples

			We have a(1) = 1 because we only have one symmetric cyclic composition of n = 1, namely 1_1 (and a part of size 1 can be colored with only one color).
We have a(2) = 3 because we have the following colored achiral cyclic compositions of n = 2: 2_1, 2_2, 1_1 + 1_1.
We have a(3) = 6 because we have the following colored achiral cyclic compositions of n = 3: 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 1_1 + 1_1 + 1_1.
We have a(4) = 13 because we have the following colored achiral cyclic compositions of n = 4: 4_1, 4_2, 4_3, 4_4, 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 1_1 + 1_1 + 1_1 + 1_1.
We have a(5) = 23 because we have the following colored achiral cyclic compositions of n = 5:
(i) with one part: 5_1, 5_2, 5_3, 5_4, 5_5;
(ii) with two parts: 1_1 + 4_1, 1_1 + 4_2, 1_1 + 4_3, 1_1 + 4_4, 2_1 + 3_1, 2_1 + 3_2, 2_1 + 3_3, 2_2 + 3_1, 2_2 + 3_2, 2_2 + 3_3;
(iii) with three parts: 1_1 + 3_1 + 1_1, 1_1 + 3_2 + 1_1, 1_1 + 3_3 + 1_1, 2_1 + 1_1 + 2_1, 2_2 + 1_1 + 2_2;
(iv) with four parts: 1_1 + 1_1 + 2_1 + 1_1, 1_1 + 1_1 + 2_2 + 1_1 (here, the axis of symmetry passes through one of the 1's and through 2);
(v) with five parts: 1_1 + 1_1 + 1_1 + 1_1 + 1_1.
		

Crossrefs

Formula

CPAL (circular palindrome) transform of 1, 2, 3, 4, ...
a(n) = 2*a(n - 1) + 2*a(n - 2) - 6*a(n - 3) + 2*a(n - 4) + 2*a(n - 5) - a(n - 6) for n >= 7 with a(1) = 1, a(2) = 3, a(3) = 6, a(4) = 13, a(5) = 23, and a(6) = 44.
a(n) = 3*a(n - 2) - a(n - 4) + 2*(n - 2) for n >= 5 with a(1) = 1, a(2) = 3, a(3) = 6, and a(4) = 13.
a(n) = Fib(n + 4) + (-1)^n * Fib(n - 4) - 2*n for n >= 4, where Fib(n) = A000045(n).
G.f.: x * (1 + x - 2*x^2 + x^3 + x^4)/((1 - x)^2 * (1 - x - x^2) * (1 + x - x^2)).
Showing 1-3 of 3 results.