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 1-4 of 4 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

Views

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

Links

Crossrefs

Columns 1-6: A057427, A056342, A056343, A056344, A056345, A056346.
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)

A056491 Number of periodic palindromes using exactly five different symbols.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 60, 120, 960, 1800, 9300, 16800, 71400, 126000, 480060, 834120, 2968560, 5103000, 17355300, 29607600, 97567800, 165528000, 533274060, 901020120, 2855012160, 4809004200, 15050517300, 25292030400, 78417448200, 131542866000, 404936532060
Offset: 1

Views

Author

Marks R. Nester

Keywords

Examples

			For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome.
There are 120 permutations of the five letters used in ABACDEDC.  These 120 arrangements can be paired up with a half turn (e.g., ABACDEDC-DEDCABAC) to arrive at the 60 arrangements for n=8. - _Robert A. Russell_, Sep 26 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]

Links

Crossrefs

Cf. A056456.
Column 5 of A305540.

Programs

  • GAP
    a:=[0,0,0,0,0,0,0,60,120];; for n in [10..35] do a[n]:=a[n-1]+14*a[n-2]-14*a[n-3]-71*a[n-4]+71*a[n-5]+154*a[n-6]-154*a[n-7]-120*a[n-8]+120*a[n-9]; od; a; # Muniru A Asiru, Sep 26 2018
  • Magma
    m:=50; R:=PowerSeriesRing(Integers(), m); [0, 0, 0, 0, 0, 0, 0] cat Coefficients(R!(-60*x^8*(x+1)/((x-1)*(2*x-1)*(2*x+1)*(2*x^2-1)*(3*x^2-1)*(5*x^2-1)))); // G. C. Greubel, Oct 13 2018
  • Maple
    with(combinat):  a:=n->(factorial(5)/2)*(Stirling2(floor((n+1)/2),5)+Stirling2(ceil((n+1)/2),5)): seq(a(n),n=1..35); # Muniru A Asiru, Sep 26 2018
  • Mathematica
    k = 5; 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, 14, -14, -71, 71, 154, -154, -120, 120}, {0, 0,
0, 0, 0, 0, 0, 60, 120}, 40] (* Robert A. Russell, Sep 29 2018 *)
  • PARI
    a(n) = my(k=5); (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*A056345(n) - A056285(n).
G.f.: -60*x^8*(x+1)/((x-1)*(2*x-1)*(2*x+1)*(2*x^2-1)*(3*x^2-1)*(5*x^2-1)). - Colin Barker, Jul 08 2012
a(n) = (k!/2)*(S2(floor((n+1)/2),k) + S2(ceiling((n+1)/2),k)), with k=5 different colors used and where S2(n,k) is the Stirling subset number A008277. - Robert A. Russell, Jun 05 2018
a(n) = a(n-1) + 14*a(n-2) - 14*a(n-3) - 71*a(n-4) + 71*a(n-5) + 154*a(n-6) - 154*a(n-7) - 120*a(n-8) + 120*a(n-9). - Muniru A Asiru, Sep 26 2018

A214313 a(n) is the number of all five-color bracelets (necklaces with turning over allowed) with n beads and the four colors are from a repertoire of n distinct colors, for n >= 5.

Original entry on oeis.org

12, 900, 25200, 442680, 5846400, 64420272, 622175400, 5466166200, 44611306740, 343916472900, 2531921456064, 17956666859040, 123458676825120, 827056125453600, 5419508203393200, 34847210197637424, 220424306985639540, 1374479672119161300, 8463477229726134000, 51536194734146965920, 310706598354410079360
Offset: 5

Views

Author

Wolfdieter Lang, Aug 08 2012

Keywords

Comments

This is the fifth column (m=5) of triangle A214306.
Each 5 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[5]], with p[1] >= p[2] >= .. >= p[5] >= 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,5)= A008284(n,5) partitions of n with 5 parts are considered. The color repertoire for a bracelet with n beads is [c[1], ..., c[n]].
It appears that this sequence is divisible by 12, producing 1, 75, 2100, 36890, 487200, 5368356, 51847950, 455513850, ...
Compare this with A056345 where only 5 colors are used for all n >= 5.

Examples

			a(6) = A213941(6,10) = 900 from the bracelet with color signature [2,1,1,1,1] and color repertoire [c[j], j=1, 2, ..., 6]. There are A213939(6,10) = 30 bracelets with representative color multinomials c[1]^2 c[2] c[3] c[4] c[5]. If the colors c[j] are taken as j, e.g., 112345, 112354, 112435, 112453, 112534, 112543, 113245, 113254, 113425, (113452 is equivalent to 112543 by turning over), 113524, (113542 ==112453), 114235, ..., 121345, ... (all taken cyclically). Each of these 30 bracelets represents a class of A035206(6,10) = 30 bracelets when all six colors are used. Thus a(6) = 30*30 = 900 = 12*75.

Links

Crossrefs

Cf. A213941, A214306, A214311 (m=5, representative bracelets), A214312 (m=4).

Formula

a(n) = A214306(n,5), n >= 5.
a(n) = sum(A213941(n,k),k = A214314(n,5) .. (A214314(n,5) - 1 + A008284(n,5))), n >= 5.
a(n) = binomial(n,5) * A056345(n). - Andrew Howroyd, Mar 25 2017

A056351 Number of primitive (period n) bracelets using exactly five different colored beads.

Original entry on oeis.org

0, 0, 0, 0, 12, 150, 1200, 7905, 46400, 255624, 1346700, 6901575, 34663020, 171785250, 843130676, 4110950625, 19951305240, 96528446150, 466073976900, 2247626821095, 10832193570260, 52194107870250
Offset: 1

Views

Author

Marks R. Nester

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. A056290.

Formula

sum mu(d)*A056345(n/d) where d|n.
Showing 1-4 of 4 results.

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