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 1-5 of 5 results.

A273891 Triangle read by rows: T(n,k) is the number of n-bead bracelets with exactly k different colored beads.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 6, 3, 1, 6, 18, 24, 12, 1, 11, 56, 136, 150, 60, 1, 16, 147, 612, 1200, 1080, 360, 1, 28, 411, 2619, 7905, 11970, 8820, 2520, 1, 44, 1084, 10480, 46400, 105840, 129360, 80640, 20160, 1, 76, 2979, 41388, 255636, 821952, 1481760, 1512000, 816480, 181440
Offset: 1

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Author

Marko Riedel, Jun 02 2016

Keywords

Comments

For bracelets, chiral pairs are counted as one.

Examples

			Triangle begins with T(1,1):
1;
1,  1;
1,  2,    1;
1,  4,    6,     3;
1,  6,   18,    24,     12;
1, 11,   56,   136,    150,     60;
1, 16,  147,   612,   1200,   1080,     360;
1, 28,  411,  2619,   7905,  11970,    8820,    2520;
1, 44, 1084, 10480,  46400, 105840,  129360,   80640,  20160;
1, 76, 2979, 41388, 255636, 821952, 1481760, 1512000, 816480, 181440;
For T(4,2)=4, the arrangements are AAAB, AABB, ABAB, and ABBB, all achiral.
For T(4,4)=3, the arrangements are ABCD, ABDC, and ACBD, whose chiral partners are ADCB, ACDB, and ADBC respectively. - _Robert A. Russell_, Sep 26 2018
		

Crossrefs

Row sums give A019537.
Cf. A087854 (oriented), A305540 (achiral), A305541 (chiral).

Programs

  • Mathematica
    (* t = A081720 *) 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}]; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Oct 07 2017, after Andrew Howroyd *)
    Table[k! DivisorSum[n, EulerPhi[#] StirlingS2[n/#,k]&]/(2n) + k!(StirlingS2[Floor[(n+1)/2], k] + StirlingS2[Ceiling[(n+1)/2], k])/4, {n,1,10}, {k,1,n}] // Flatten (* Robert A. Russell, Sep 26 2018 *)

Formula

T(n,k) = Sum_{i=0..k-1} (-1)^i * binomial(k,i) * A081720(n,k-i). - Andrew Howroyd, Mar 25 2017
From Robert A. Russell, Sep 26 2018: (Start)
T(n,k) = (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 S2 is the Stirling subset number A008277.
G.f. for column k>1: (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).
T(n,k) = (A087854(n,k) + A305540(n,k)) / 2 = A087854(n,k) - A305541(n,k) = A305541(n,k) + A305540(n,k).
(End)

A056489 Number of periodic palindromes using exactly three different symbols.

Original entry on oeis.org

0, 0, 0, 3, 6, 21, 36, 93, 150, 345, 540, 1173, 1806, 3801, 5796, 11973, 18150, 37065, 55980, 113493, 171006, 345081, 519156, 1044453, 1569750, 3151785, 4733820, 9492213, 14250606, 28550361, 42850116, 85798533, 128746950, 257690505, 386634060, 773661333
Offset: 1

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Author

Keywords

Comments

For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome.

Examples

			For n=4, the three arrangements are ABAC, ABCB, and ACBC.
For n=5, the six arrangements are AABCB, AACBC, ABACC, ABBAC, ABCCB, and ACBBC.
		

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

Cf. A056454.
Column 3 of A305540.

Programs

  • GAP
    a:=[0,0,0,3,6];; for n in [6..40] do a[n]:=a[n-1]+5*a[n-2]-5*a[n-3]-6*a[n-4]+6*a[n-5]; od; a; # Muniru A Asiru, Sep 28 2018
  • Maple
    seq(coeff(series(3*x^4*(1+x)/((1-x)*(1-2*x^2)*(1-3*x^2)),x,n+1), x, n), n = 1..40); # Muniru A Asiru, Sep 28 2018
  • Mathematica
    k = 3; Table[(k!/2) (StirlingS2[Floor[(n + 1)/2], k] +
    StirlingS2[Ceiling[(n + 1)/2], k]), {n, 1, 40}] (* Robert A. Russell, Jun 05 2018 *)
    LinearRecurrence[{1, 5, -5, -6, 6}, {0, 0, 0, 3, 6}, 80] (* Vincenzo Librandi, Sep 27 2018 *)
  • PARI
    a(n) = my(k=3); (k!/2)*(stirling(floor((n+1)/2), k, 2) + stirling(ceil((n+1)/2), k, 2)); \\ Michel Marcus, Jun 05 2018
    

Formula

a(n) = 2 * A056343(n) - A056283(n).
G.f.: 3*x^4*(1+x)/((1-x)*(1-2*x^2)*(1-3*x^2)). - Colin Barker, May 06 2012
a(n) = (k!/2)*(S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)), with k=3 different colors used and where S2(n,k) is the Stirling subset number A008277. - Robert A. Russell, Jun 05 2018

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.
		

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

A326660 Number of n-bead asymmetric bracelets with exactly 3 different colored beads.

Original entry on oeis.org

0, 0, 1, 3, 12, 34, 111, 315, 933, 2622, 7503, 21033, 59472, 167118, 472120, 1332945, 3777600, 10720869, 30516447, 87032994, 248820704, 712743768, 2045784183, 5882367570, 16942974048, 48876558318, 141204944529, 408494941773, 1183247473872, 3431450670601
Offset: 1

Views

Author

Andrew Howroyd, Sep 12 2019

Keywords

Comments

Asymmetric bracelets are those primitive (period n) bracelets that when turned over are different from themselves.

Examples

			Case n = 4: There are 3 distinct asymmetric bracelets with exactly 3 colors which are aabc, abbc, abcc.
		

Crossrefs

Formula

a(n) = A032240(n) - 3*A032239(n) for n >= 3.
Moebius transform of A305542.

A056349 Number of primitive (period n) bracelets using exactly three different colored beads.

Original entry on oeis.org

0, 0, 1, 6, 18, 55, 147, 405, 1083, 2961, 8043, 22182, 61278, 170883, 477910, 1344825, 3795750, 10757763, 30572427, 87146139, 248991674, 713088309, 2046303339, 5883410760, 16944543792, 48879708297
Offset: 1

Views

Author

Keywords

Comments

Turning over will not create a new bracelet.

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

Cf. A056288.

Formula

sum mu(d)*A056343(n/d) where d|n.
Showing 1-5 of 5 results.