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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A001710 Order of alternating group A_n, or number of even permutations of n letters.

Original entry on oeis.org

1, 1, 1, 3, 12, 60, 360, 2520, 20160, 181440, 1814400, 19958400, 239500800, 3113510400, 43589145600, 653837184000, 10461394944000, 177843714048000, 3201186852864000, 60822550204416000, 1216451004088320000, 25545471085854720000, 562000363888803840000
Offset: 0

Views

Author

Keywords

Comments

For n >= 3, a(n-1) is also the number of ways that a 3-cycle in the symmetric group S_n can be written as a product of 2 long cycles (of length n). - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 14 2001
a(n) is the number of Hamiltonian circuit masks for an n X n adjacency matrix of an undirected graph. - Chad Brewbaker, Jan 31 2003
a(n-1) is the number of necklaces one can make with n distinct beads: n! bead permutations, divide by two to represent flipping the necklace over, divide by n to represent rotating the necklace. Related to Stirling numbers of the first kind, Stirling cycles. - Chad Brewbaker, Jan 31 2003
Number of increasing runs in all permutations of [n-1] (n>=2). Example: a(4)=12 because we have 12 increasing runs in all the permutations of [3] (shown in parentheses): (123), (13)(2), (3)(12), (2)(13), (23)(1), (3)(2)(1). - Emeric Deutsch, Aug 28 2004
Minimum permanent over all n X n (0,1)-matrices with exactly n/2 zeros. - Simone Severini, Oct 15 2004
The number of permutations of 1..n that have 2 following 1 for n >= 1 is 0, 1, 3, 12, 60, 360, 2520, 20160, ... . - Jon Perry, Sep 20 2008
Starting (1, 3, 12, 60, ...) = binomial transform of A000153: (1, 2, 7, 32, 181, ...). - Gary W. Adamson, Dec 25 2008
First column of A092582. - Mats Granvik, Feb 08 2009
The asymptotic expansion of the higher order exponential integral E(x,m=1,n=3) ~ exp(-x)/x*(1 - 3/x + 12/x^2 - 60/x^3 + 360/x^4 - 2520/x^5 + 20160/x^6 - 81440/x^7 + ...) leads to the sequence given above. See A163931 and A130534 for more information. - Johannes W. Meijer, Oct 20 2009
For n>1: a(n) = A173333(n,2). - Reinhard Zumkeller, Feb 19 2010
Starting (1, 3, 12, 60, ...) = eigensequence of triangle A002260, (a triangle with k terms of (1,2,3,...) in each row given k=1,2,3,...). Example: a(6) = 360, generated from (1, 2, 3, 4, 5) dot (1, 1, 3, 12, 60) = (1 + 2 + 9 + 48 + 300). - Gary W. Adamson, Aug 02 2010
For n>=2: a(n) is the number of connected 2-regular labeled graphs on (n+1) nodes (Cf. A001205). - Geoffrey Critzer, Feb 16 2011.
The Fi1 and Fi2 triangle sums of A094638 are given by the terms of this sequence (n>=1). For the definition of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Also [1, 1] together with the row sums of triangle A162608. - Omar E. Pol, Mar 09 2012
a(n-1) is, for n>=2, also the number of necklaces with n beads (only C_n symmetry, no turnover) with n-1 distinct colors and signature c[.]^2 c[.]^(n-2). This means that two beads have the same color, and for n=2 the second factor is omitted. Say, cyclic(c[1]c[1]c[2]c[3]..c[n-1]), in short 1123...(n-1), taken cyclically. E.g., n=2: 11, n=3: 112, n=4: 1123, 1132, 1213, n=5: 11234, 11243, 11324, 11342, 11423, 11432, 12134, 12143, 13124, 13142, 14123, 14132. See the next-to-last entry in line n>=2 of the representative necklace partition array A212359. - Wolfdieter Lang, Jun 26 2012
For m >= 3, a(m-1) is the number of distinct Hamiltonian circuits in a complete simple graph with m vertices. See also A001286. - Stanislav Sykora, May 10 2014
In factorial base (A007623) these numbers have a simple pattern: 1, 1, 1, 11, 200, 2200, 30000, 330000, 4000000, 44000000, 500000000, 5500000000, 60000000000, 660000000000, 7000000000000, 77000000000000, 800000000000000, 8800000000000000, 90000000000000000, 990000000000000000, etc. See also the formula based on this observation, given below. - Antti Karttunen, Dec 19 2015
Also (by definition) the independence number of the n-transposition graph. - Eric W. Weisstein, May 21 2017
Number of permutations of n letters containing an even number of even cycles. - Michael Somos, Jul 11 2018
Equivalent to Brewbaker's and Sykora's comments, a(n - 1) is the number of undirected cycles covering n labeled vertices, hence the logarithmic transform of A002135. - Gus Wiseman, Oct 20 2018
For n >= 2 and a set of n distinct leaf labels, a(n) is the number of binary, rooted, leaf-labeled tree topologies that have a caterpillar shape (column k=1 of A306364). - Noah A Rosenberg, Feb 11 2019
Also the clique covering number of the n-Bruhat graph. - Eric W. Weisstein, Apr 19 2019
a(n) is the number of lattices of the form [s,w] in the weak order on S_n, for a fixed simple reflection s. - Bridget Tenner, Jan 16 2020
For n > 3, a(n) = p_1^e_1*...*p_m^e_m, where p_1 = 2 and e_m = 1. There exists p_1^x where x <= e_1 such that p_1^x*p_m^e_m is a primitive Zumkeller number (A180332) and p_1^e_1*p_m^e_m is a Zumkeller number (A083207). Therefore, for n > 3, a(n) = p_1^e_1*p_m^e_m*r, where r is relatively prime to p_1*p_m, is also a Zumkeller number. - Ivan N. Ianakiev, Mar 11 2020
For n>1, a(n) is the number of permutations of [n] that have 1 and 2 as cycle-mates, that is, 1 and 2 are contained in the same cycle of a cyclic representation of permutations of [n]. For example, a(4) counts the 12 permutations with 1 and 2 as cycle-mates, namely, (1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3), (1 4 3 2), (1 2 3) (4), (1 3 2) (4), (1 2 4 )(3), (1 4 2)(3), (1 2)(3 4), and (1 2)(3)(4). Since a(n+2)=row sums of A162608, our result readily follows. - Dennis P. Walsh, May 28 2020

Examples

			G.f. = 1 + x + x^2 + 3*x^3 + 12*x^4 + 60*x^5 + 360*x^6 + 2520*x^7 + ...
		

References

  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 87-8, 20. (a), c_n^e(t=1).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n+1)= A046089(n, 1), n >= 1 (first column of triangle), A161739 (q(n) sequence).
Bisections are A002674 and A085990 (essentially).
Row 3 of A265609 (essentially).
Row sums of A307429.

Programs

  • Magma
    [1] cat [Order(AlternatingGroup(n)): n in [1..20]]; // Arkadiusz Wesolowski, May 17 2014
    
  • Maple
    seq(mul(k, k=3..n), n=0..20); # Zerinvary Lajos, Sep 14 2007
  • Mathematica
    a[n_]:= If[n > 2, n!/2, 1]; Array[a, 21, 0]
    a[n_]:= If[n<3, 1, n*a[n-1]]; Array[a, 21, 0]; (* Robert G. Wilson v, Apr 16 2011 *)
    a[ n_]:= If[n<0, 0, n! SeriesCoefficient[(2-x^2)/(2-2x), {x, 0, n}]]; (* Michael Somos, May 22 2014 *)
    a[ n_]:= If[n<0, 0, n! SeriesCoefficient[1 +Sinh[-Log[1-x]], {x, 0, n}]]; (* Michael Somos, May 22 2014 *)
    Numerator[Range[0, 20]!/2] (* Eric W. Weisstein, May 21 2017 *)
    Table[GroupOrder[AlternatingGroup[n]], {n, 0, 20}] (* Eric W. Weisstein, May 21 2017 *)
  • PARI
    {a(n) = if( n<2, n>=0, n!/2)};
    
  • PARI
    a(n)=polcoeff(1+x*sum(m=0,n,m^m*x^m/(1+m*x+x*O(x^n))^m),n) \\ Paul D. Hanna
    
  • PARI
    A001710=n->n!\2+(n<2) \\ M. F. Hasler, Dec 01 2013
    
  • Python
    from math import factorial
    def A001710(n): return factorial(n)>>1 if n > 1 else 1 # Chai Wah Wu, Feb 14 2023
    
  • SageMath
    def A001710(n): return (factorial(n) +int(n<2))//2
    [A001710(n) for n in range(31)] # G. C. Greubel, Sep 28 2024
  • Scheme
    ;; Using memoization-macro definec for which an implementation can be found in http://oeis.org/wiki/Memoization
    (definec (A001710 n) (cond ((<= n 2) 1) (else (* n (A001710 (- n 1))))))
    ;; Antti Karttunen, Dec 19 2015
    

Formula

a(n) = numerator(n!/2) and A141044(n) = denominator(n!/2).
D-finite with recurrence: a(0) = a(1) = a(2) = 1; a(n) = n*a(n-1) for n>2. - Chad Brewbaker, Jan 31 2003 [Corrected by N. J. A. Sloane, Jul 25 2008]
a(0) = 0, a(1) = 1; a(n) = Sum_{k=1..n-1} k*a(k). - Amarnath Murthy, Oct 29 2002
Stirling transform of a(n+1) = [1, 3, 12, 160, ...] is A083410(n) = [1, 4, 22, 154, ...]. - Michael Somos, Mar 04 2004
First Eulerian transform of A000027. See A000142 for definition of FET. - Ross La Haye, Feb 14 2005
From Paul Barry, Apr 18 2005: (Start)
a(n) = 0^n + Sum_{k=0..n} (-1)^(n-k-1)*T(n-1, k)*cos(Pi*(n-k-1)/2)^2.
T(n,k) = abs(A008276(n, k)). (End)
E.g.f.: (2 - x^2)/(2 - 2*x).
E.g.f. of a(n+2), n>=0, is 1/(1-x)^3.
E.g.f.: 1 + sinh(log(1/(1-x))). - Geoffrey Critzer, Dec 12 2010
a(n+1) = (-1)^n * A136656(n,1), n>=1.
a(n) = n!/2 for n>=2 (proof from the e.g.f). - Wolfdieter Lang, Apr 30 2010
a(n) = (n-2)! * t(n-1), n>1, where t(n) is the n-th triangular number (A000217). - Gary Detlefs, May 21 2010
a(n) = ( A000254(n) - 2* A001711(n-3) )/3, n>2. - Gary Detlefs, May 24 2010
O.g.f.: 1 + x*Sum_{n>=0} n^n*x^n/(1 + n*x)^n. - Paul D. Hanna, Sep 13 2011
a(n) = if n < 2 then 1, otherwise Pochhammer(n,n)/binomial(2*n,n). - Peter Luschny, Nov 07 2011
a(n) = Sum_{k=0..floor(n/2)} s(n,n-2*k) where s(n,k) are Stirling number of the first kind, A048994. - Mircea Merca, Apr 07 2012
a(n-1), n>=3, is M_1([2,1^(n-2)])/n = (n-1)!/2, with the M_1 multinomial numbers for the given n-1 part partition of n. See the second to last entry in line n>=3 of A036038, and the above necklace comment by W. Lang. - Wolfdieter Lang, Jun 26 2012
G.f.: A(x) = 1 + x + x^2/(G(0)-2*x) where G(k) = 1 - (k+1)*x/(1 - x*(k+3)/G(k+1)); (continued fraction). - Sergei N. Gladkovskii, Dec 26 2012.
G.f.: 1 + x + (Q(0)-1)*x^2/(2*(sqrt(x)+x)), where Q(k) = 1 + (k+2)*sqrt(x)/(1 - sqrt(x)/(sqrt(x) + 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 15 2013
G.f.: 1 + x + (x*Q(x)-x^2)/(2*(sqrt(x)+x)), where Q(x) = Sum_{n>=0} (n+1)!*x^n*sqrt(x)*(sqrt(x) + x*(n+2)). - Sergei N. Gladkovskii, May 15 2013
G.f.: 1 + x/2 + (Q(0)-1)*x/(2*(sqrt(x)+x)), where Q(k) = 1 + (k+1)*sqrt(x)/(1 - sqrt(x)/(sqrt(x) + 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 15 2013
G.f.: 1 + x + x^2*G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: 1+x + x^2*W(0), where W(k) = 1 - x*(k+3)/( x*(k+3) - 1/(1 - x*(k+1)/( x*(k+1) - 1/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
From Antti Karttunen, Dec 19 2015: (Start)
a(0)=a(1)=1; after which, for even n: a(n) = (n/2) * (n-1)!, and for odd n: a(n) = (n-1)/2 * ((n-1)! + (n-2)!). [The formula was empirically found after viewing these numbers in factorial base, A007623, and is easily proved by considering formulas from Lang (Apr 30 2010) and Detlefs (May 21 2010) shown above.]
For n >= 1, a(2*n+1) = a(2*n) + A153880(a(2*n)). [Follows from above.] (End)
Inverse Stirling transform of a(n) is (-1)^(n-1)*A009566(n). - Anton Zakharov, Aug 07 2016
a(n) ~ sqrt(Pi/2)*n^(n+1/2)/exp(n). - Ilya Gutkovskiy, Aug 07 2016
a(n) = A006595(n-1)*n/A000124(n) for n>=2. - Anton Zakharov, Aug 23 2016
a(n) = A001563(n-1) - A001286(n-1) for n>=2. - Anton Zakharov, Sep 23 2016
From Peter Bala, May 24 2017: (Start)
The o.g.f. A(x) satisfies the Riccati equation x^2*A'(x) + (x - 1)*A(x) + 1 - x^2 = 0.
G.f.: A(x) = 1 + x + x^2/(1 - 3*x/(1 - x/(1 - 4*x/(1 - 2*x/(1 - 5*x/(1 - 3*x/(1 - ... - (n + 2)*x/(1 - n*x/(1 - ... ))))))))) (apply Stokes, 1982).
A(x) = 1 + x + x^2/(1 - 2*x - x/(1 - 3*x/(1 - 2*x/(1 - 4*x/(1 - 3*x/(1 - 5*x/(1 - ... - n*x/(1 - (n+2)*x/(1 - ... ))))))))). (End)
H(x) = (1 - (1 + x)^(-2)) / 2 = x - 3*x^2/2! + 12*x^3/3! - ..., an e.g.f. for the signed sequence here (n!/2!), ignoring the first two terms, is the compositional inverse of G(x) = (1 - 2*x)^(-1/2) - 1 = x + 3*x^2/2! + 15*x^3/3! + ..., an e.g.f. for A001147. Cf. A094638. H(x) is the e.g.f. for the sequence (-1)^m * m!/2 for m = 2,3,4,... . Cf. A001715 for n!/3! and A001720 for n!/4!. Cf. columns of A094587, A173333, and A213936 and rows of A138533. - Tom Copeland, Dec 27 2019
From Amiram Eldar, Jan 08 2023: (Start)
Sum_{n>=0} 1/a(n) = 2*(e-1).
Sum_{n>=0} (-1)^n/a(n) = 2/e. (End)

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Aug 20 2001
Further terms from Simone Severini, Oct 15 2004

A173333 Triangle read by rows: T(n, k) = n! / k!, 1 <= k <= n.

Original entry on oeis.org

1, 2, 1, 6, 3, 1, 24, 12, 4, 1, 120, 60, 20, 5, 1, 720, 360, 120, 30, 6, 1, 5040, 2520, 840, 210, 42, 7, 1, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 3628800, 1814400, 604800, 151200, 30240, 5040, 720, 90, 10, 1
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 19 2010

Keywords

Comments

From Wolfdieter Lang, Jun 27 2012: (Start)
T(n-1,k), k=1,...,n-1, gives the number of representative necklaces with n beads (C_N symmetry) of n+1-k distinct colors, say c[1],c[2],...,c[n-k+1], corresponding to the color signature determined by the partition k,1^(n-k) of n. The representative necklaces have k beads of color c[1]. E.g., n=4, k=2: partition 2,1,1, color signature (parts as exponents) c[1]c[1]c[2]c[3], 3=T(3,2) necklaces (write j for color c[j]): cyclic(1123), cyclic(1132) and cyclic(1213). See A212359 for the numbers for general partitions or color signatures. (End)

Examples

			Triangle starts:
n\k      1       2      3      4     5    6   7  8  9 10 ...
1        1
2        2       1
3        6       3      1
4       24      12      4      1
5      120      60     20      5     1
6      720     360    120     30     6    1
7     5040    2520    840    210    42    7   1
8    40320   20160   6720   1680   336   56   8  1
9   362880  181440  60480  15120  3024  504  72  9  1
10 3628800 1814400 604800 151200 30240 5040 720 90 10  1
... - _Wolfdieter Lang_, Jun 27 2012
		

Crossrefs

Row sums give A002627.
Central terms give A006963:
T(2*n-1,n) = A006963(n+1).
T(2*n,n) = A001813(n).
T(2*n,n+1) = A001761(n).
1 < k <= n: T(n,k) = T(n,k-1) / k.
1 <= k <= n: T(n+1,k) = A119741(n,n-k+1).
1 <= k <= n: T(n+1,k+1) = A162995(n,k).
T(n,1) = A000142(n).
T(n,2) = A001710(n) for n>1.
T(n,3) = A001715(n) for n>2.
T(n,4) = A001720(n) for n>3.
T(n,5) = A001725(n) for n>4.
T(n,6) = A001730(n) for n>5.
T(n,7) = A049388(n-7) for n>6.
T(n,8) = A049389(n-8) for n>7.
T(n,9) = A049398(n-9) for n>8.
T(n,10) = A051431(n) for n>9.
T(n,n-7) = A159083(n+1) for n>7.
T(n,n-6) = A053625(n+1) for n>6.
T(n,n-5) = A052787(n) for n>5.
T(n,n-4) = A052762(n) for n>4.
T(n,n-3) = A007531(n) for n>3.
T(n,n-2) = A002378(n-1) for n>2.
T(n,n-1) = A000027(n) for n>1.
T(n,n) = A000012(n).

Programs

  • Haskell
    a173333 n k = a173333_tabl !! (n-1) !! (k-1)
    a173333_row n = a173333_tabl !! (n-1)
    a173333_tabl = map fst $ iterate f ([1], 2)
       where f (row, i) = (map (* i) row ++ [1], i + 1)
    -- Reinhard Zumkeller, Jul 04 2012
  • Mathematica
    Table[n!/k!, {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 01 2019 *)

Formula

E.g.f.: (exp(x*y) - 1)/(x*(1 - y)). - Olivier Gérard, Jul 07 2011
T(n,k) = A094587(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012

A213939 Partition array for the number of representative bracelets (dihedral symmetry D_n) with n beads, each available in n colors. Only the color type (signature) matters.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 1, 1, 2, 2, 4, 6, 12, 1, 1, 3, 3, 3, 6, 11, 10, 16, 30, 60, 1, 1, 3, 4, 3, 9, 10, 18, 15, 30, 48, 60, 90, 180, 360, 1, 1, 4, 5, 8, 4, 12, 19, 33, 38, 21, 54, 70, 108, 171, 105, 210, 318, 420, 630, 1260, 2520, 1, 1, 4, 7, 10, 4, 16, 28, 38, 48, 76, 94
Offset: 1

Views

Author

Wolfdieter Lang, Jul 20 2012

Keywords

Comments

The row lengths sequence is A000041(n), n >= 1.
The partitions are ordered like in Abramowitz-Stegun (A-St order). For the reference see A036036, where also a link to a work by C. F. Hindenburg from 1779 is found where this order has been used.
A bracelet with n beads (n-bracelet) has the dihedral D_n symmetry group of degree n (order 2n). In addition to cyclic C_n operations, also a turnover (in 3-space) or a reflection (in 2-space) is allowed. In the Harary-Palmer reference, p. 44, the term necklace is used instead of bracelet.
a(n,k) gives the number of representative n-bracelets, with up to n colors for each bead, belonging to the k-th partition of n in A-St order in the following way. Write this partition with nonincreasing parts (this is the reverse of the partition as given by A-St), e.g., [3,1^2], not [1^2,3], which is written as [3,1,1], a partition of n=5. In general a (reversed) partition of n is written as [p[1],p[2],...,p[m]], with p[1] >= p[2] >= ... >= p[m] >= 1, with m the number of parts. To each such partition of n corresponds an n-multiset obtained by 'exponentiation'. For more details see the W. Lang link in A213938 with more details as well as a list of multiset signatures and corresponding multiset representatives. For the given example the 5-multiset is {1^3,2^1,3^1}={1,1,1,2,3}. In general, {1^p[1],2^p[2],...,m^p[m]}. We will also use a list notation with square brackets for these multisets. Such an n-multiset representative (of a repetition class defined by the exponents, also called signature) encodes the representative n-bracelet color monomial by c[1]^p[1]*c[2]^p[2]*...*c[m]^p[m]. For the example one has c[1]^3*c[2]*c[3]. The number of 5-bracelets with this color assignment is a(5,4) because [3,1,1] is the 4th partition of 5 in A-St order. The a(5,4)=2 non-equivalent 5-bracelets with this color assignment are cyclic(c[1]c[1]c[1]c[2]c[3]) and cyclic(c[1]c[1]c[2]c[1]c[3]). For the necklace case c[1]c[1]c[1]c[3]c[2] and c[1]c[1]c[3]c[1]c[2] (both taken cyclically) also have to be counted, but due to a turn over (or a reflection) they become equivalent to the two given bracelets, respectively.
Such a set of a(n,k) n-bracelets for the given color signature stands for other sets of the same order when different colors from the repertoire {c[1],...,c[n]} are chosen. In the example, the partition [3,1,1] with the representative multiset [1^3,2,3] stands for all-together 5*binomial(4,2) = 30 such sets, each leading to 2 possible non-equivalent 5-bracelet arrangements. Thus one has all-together 30*2=60 5-bracelets with color signature determined from the partition [3,1,1]. See the partition array A213941 for these total bracelet numbers.
a(n,k) is computed from the cycle index Z(D_n) for the dihedral group (see A212355 and the link given there) after the variables x_j have been replaced by the j-th power sum sum(c[i]^j,i=1..n), abbreviated as Z(D_n,c_n) with c_n:=sum(c[i],i=1..n), n >= 1. The coefficient of the representative color multinomial determined by the k-th partition of n in A-St order, as explained above, is a(n,k). See the Harary-Palmer reference, p. 36, Theorem (PET) with A = D_n and p. 37 eq. (2.2.11) for the cycle index polynomial Z(D_n). See the W. Lang link for more details.
The row sums are given by A213943.

Examples

			n\k 1 2 3 4 5 6  7  8  9 10 11 12 13  14  15 ...
1   1
2   1 1
3   1 1 1
4   1 1 2 2 3
5   1 1 2 2 4 6 12
6   1 1 3 3 3 6 11 10 16 30 60
7   1 1 3 4 3 9 10 18 15 30 48 60 90 180 360
...
Row n = 8 is 1 1 4 5 8 4 12 19 33 38 21 54 70 108 171 105 210 318 420 630 1260 2520.
See the link for the rows n=1 to n=15, and the corresponding color polynomials for n=1 to n=10.
a(4,5) = 3 because the partition in question is [1^4]=[1,1,1,1], the corresponding representative color multinomial is c[1]*c[2]*c[3]*c[4] (all four colors are involved), and there are the 3 D_4 non-equivalent 4-bracelets (we use here j for color c[j]): 1234, 1324 and 1423 (all taken as cyclically). For this partition there is only one color choice. The necklace solutions 1243, 1342, 1432, taken cyclically, become equivalent to the given bracelets, respectively (for necklaces see A212359).
a(4,4) = 2 because the partition is [2,1^2]=[2,1,1], the color representative multinomial is c[1]^2*c[2]*c[3], and the bracelet arrangements are 1123 and 1213 (all taken cyclically). The necklace cyclic(1132) becomes equivalent to the first bracelet under reflection. In total, there are 4*binomial(3,2)=12 color multinomials of this signature (color type) in Z(D_4,c_4), each with a coefficient 2.
		

References

  • F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973

Crossrefs

Cf. A212355 (Z(D_n)), A213943(row sums), A213940 (triangle with entries for fixed m summed).

Formula

a(n,k) is the number of representative bracelet arrangements with n beads (respecting the dihedral D_n symmetry) with color assignment given by the multiset representative obtained uniquely from the k-th partition of n in A-St order. See the comment for more details and the A-St reference.

A213940 Triangle with entry a(n,m) giving the number of bracelets of n beads (dihedral D_n symmetry) with n colors available for each bead, but only m distinct fixed colors, say c[1],...,c[m], are present, with m from {1,...,n} and n>=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 2, 3, 1, 3, 6, 6, 12, 1, 7, 20, 26, 30, 60, 1, 8, 40, 93, 150, 180, 360, 1, 18, 106, 424, 633, 1050, 1260, 2520, 1, 22, 304, 1180, 3260, 5040, 8400, 10080, 20160, 1, 46, 731, 4844, 16212, 29244, 45360, 75600, 90720, 181440
Offset: 1

Views

Author

Wolfdieter Lang, Jul 20 2012

Keywords

Comments

This triangle is obtained from the partition array A213939 by summing in row n, for n>=1, all entries related to partitions of n with the same number of parts m.
a(n,m) is the number of bracelets of n beads (dihedral D_n symmetry) corresponding to the representative color multinomials obtained from all partitions of n with m parts by 'exponentiation', hence only m from the available n colors are present. As a representative multinomial of each of the p(n,m)=A008284(n,m) such m-color classes we take the one where the considered m part partition of n, [p[1],...,p[m]], written in nonincreasing order, is distributed as exponents on the color indices like c[1]^p[1]*...*c[m]^p[m]. That is only the first m colors from the n available ones are involved.
See the comments on A212359 for the Abramowitz-Stegun (A-St) order of partitions, and the 'exponentiation' to obtain multisets, used to encode color multinomials, from partitions.
The row sums of this triangle coincide with the ones of array A213939, and they are given by A213943.
Number of n-length bracelets w over a k-ary alphabet {a1,a2,...,ak} such that #(w,a1) >= #(w,a2) >= ... >= #(w,ak) >= 1, where #(w,x) counts the letters x in word w (bracelet analog of A226874). - Andrew Howroyd, Sep 26 2017

Examples

			n\m  1  2   3    4     5     6     7     8     9     10 ...
1    1
2    1  1
3    1  1   1
4    1  3   2    3
5    1  3   6    6    12
6    1  7  20   26    30    60
7    1  8  40   93   150   180   360
8    1 18 106  424   633  1050  1260  2520
9    1 22 304 1180  3260  5040  8400 10080 20160
10   1 46 731 4844 16212 29244 45360 75600 90720 181440
...
a(5,3) = 2 + 4 = 6, from A213939(5,4) + A213939(5,5), because k(5,3,1) = 4 and p(5,3) = 2.
a(2,1) = 1 because the partition [2] of n=2 with part number m=1 corresponds to the representative color multinomial (here monomial) c[1]^2 = c[1]*c[1], and there is one such representative bracelet. There is another bracelet color monomial in this class of n=2 colors where only m=1 color is active: c[2]*c[2]. See the triangle entry A213941(2,1)=2. The same holds for the necklace case.
a(3,1) = 1 from the color monomial representative c[1]^3. This class has 2 other members: c[2]^3 and c[3]^3. See A213941(3,1)=3. The same holds for the necklace case.
Like in the necklace case one has in general a(n,1)=1 and A213941(n,1) = n from the partition [n] providing the color signature and a representative c[1]^n.
a(3,2) = 1 from the representative color multinomial c[1]^2*c[2] (from the m=2 partition [2,1] of n=3) leading to just one representative bracelet (and necklace) cyclic(112) (when one uses j for color c[j]). The whole class consists of A213941(3,2)=6 bracelets (or necklaces): cyclic(112), cyclic(113), cyclic(221), cyclic(223), cyclic(331) and cyclic(332).
a(3,3) = 1. The representative color multinomial is c[1]*c[2]*c[3] (from the m=3 partition [1,1,1]). There is only one bracelet cyclic(1,2,3) which constitutes already the whole class (A213941(3,3)=1). The necklace cyclic(1,3,2) becomes equivalent under D_3.
a(4,2) = 3 from two representative color multinomials c[1]^3*c[2] and c[1]^2*c[2]^2 (from the two m=2 partitions of n=4: [3,1] and [2,2]). The first one has one representative bracelet, namely cyclic(1112), the second one leads to the two representative bracelets: cyclic(1122) and cyclic(1212). Together these are the 3 bracelets counted by a(4,2). The first color class c[.]^3*c[.] consists of 4*3=12 bracelets, when all 4 colors are used. The second one consists of 2*6=12 bracelets. Together they sum up to the 24 bracelets counted by A213941(4,2). In this example the necklace case does not differ from the bracelet one.
		

Crossrefs

Columns k=2..5 are A213942, A214307, A214309, A214311.
Cf. A213934 (cyclic symmetry).

Programs

  • PARI
    Cyc(v)={my(s=vecsum(v)); sumdiv(gcd(v), d, eulerphi(d)*(s/d)!/prod(i=1, #v, (v[i]/d)!))/s}
    CPal(v)={my(odds=#select(t->t%2,v), s=vecsum(v));  if(odds>2, 0, ((s-odds)/2)!/prod(i=1, #v, (v[i]\2)!))}
    T(n,k)={my(t=0); forpart(p=n, t+=Cyc(Vec(p))+CPal(Vec(p)), [1,n], [k,k]); t/2}
    for(n=1, 10, for(k=1,n, print1(T(n,k), ", ")); print); \\ Andrew Howroyd, Sep 26 2017
    
  • PARI
    \\ faster version; here U is A226874 as vector of polynomials.
    U(n)={Vec(serlaplace(prod(k=1, n, 1/(1-y*x^k/k!) + O(x*x^n))))}
    T(n)={my(t=U(n)); vector(n, n, vector(n, k, ((1/n)*sumdiv(n, d, eulerphi(n/d) * polcoeff(t[d+1], k)) + if(n%2, sum(d=0, (n-1)/2, binomial((n-1)/2, d)*polcoeff(t[d+1], (k-1))), polcoeff(t[n/2+1], k) + sum(d=0, n/2-1, binomial(n/2-1, d)*(2^d + if(d%2, 0, binomial(d, d/2)))*polcoeff(t[n/2-d], k-2))/2))/2))}
    { my(t=T(10)); for(n=1, #t, print(t[n])) } \\ Andrew Howroyd, Dec 22 2017

Formula

a(n,m) = Sum_{j=1..p(n,m)}A213939(n,k(n,m,1)+j-1), with k(n,m,1) the position where in the list of partitions of n in A-St order the first with m parts appears, and p(n,m) is the number of partitions of n with m parts shown in the array A008284. E.g., n=5, m=3: k(5,3,1)=4, p(5,3)=2.

A213941 Partition array a(n,k) with the total number of bracelets (D_n symmetry) with n beads, each available in n colors, with color signature given by the k-th partition of n in Abramowitz-Stegun(A-St) order.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 12, 12, 24, 3, 5, 20, 40, 60, 120, 120, 12, 6, 30, 90, 45, 180, 720, 220, 600, 1440, 900, 60, 7, 42, 126, 168, 315, 1890, 1050, 1890, 2100, 12600, 6720, 6300, 18900, 7560, 360, 8, 56, 224, 280, 224, 672, 4032, 6384, 5544, 6384, 5880, 45360
Offset: 1

Views

Author

Wolfdieter Lang, Jul 20 2012

Keywords

Comments

This array is obtained by multiplying the entry of the array A213939(n,k) (number of bracelets (dihedral D_n symmetry) with n beads, each available in n colors, with color representative given by the n-multiset representative obtained from the k-th partition of n in A-St order after 'exponentiation') with the entry of the array A035206(n,k) (number of members in the equivalence class represented by the color multiset considered for A213939(n,k)): a(n,k)=A213939(n,k)*A035206(n,k), k=1..p(n)=A000041(n), n>=1. The row sums then give the total number of bracelets with n beads from n colors, given by A081721(n).
See A212359 for references, the 'exponentiation', and a link. For multiset signatures and representative multisets defining color multinomials see also a link in A213938.
The corresponding triangle with the summed row entries related to partitions of n with fixed number of parts is A214306.

Examples

			n\k 1   2    3    4    5     6     7     8     9    10   11
1   1
2   2   1
3   3   6    1
4   4  12   12   24    3
5   5  20   40   60  120   120    12
6   6  30   90   45  180   720   220   600  1440   900   60
...
Row m=7 is: 7 42 126 168 315 1890 1050 1890 2100 12600 6720 6300 18900 7560 360.
For the rows n=1 to n=15 see the link.
a(3,1) = 3 because the 3 bracelets with 3 beads coming in 3 colors have the color multinomials (here monomials) c[1]^3=c[1]*c[1]*c[1], c[2]^3 and c[3]^3. The partition of 3 is [3], the color representative is c[1]^3, and the equivalence class with color signature from the partition [3] has the three given members. There is no difference between necklace and bracelet numbers in this case.
a(3,2) = 6 from the color signature 2,1 with the representative multinomial c[1]^2 c[2] with coefficient A213939(3,2) = 1, the only 3-bracelet cyclic(112) (taking j for the color c[j]), and A035206(3,2) = 6 members of the whole color equivalence class: cyclic(112), cyclic(113), cyclic(221), cyclic(223), cyclic(331) and cyclic(332). There is no difference between necklaces and bracelets numbers in this case.
a(3,3) = 1, color signature 1^3 = 1,1,1 with representative multinomial c[1]*c[2]*c[3] with coefficient A213939(3,3)=1 from the bracelet cyclic(1,2,3). The necklace (1,3,2) becomes equivalent to this one under D_3 operation. There are no other members in this class (A035206(3,3)=1).
The sum of row No. 3 is 10 = A081721(3). The bracelets are 111, 222, 333, 112, 113, 221, 223, 331, 332 and 123, all taken cyclically.
		

Crossrefs

Formula

a(n,k) = A213939(n,k)*A035206(n,k), k=1, 2, ..., p(n) = A000041(n), n >= 1.

A133799 a(2) = 1, a(3)=3; for n >= 4, a(n) = (n-2)!*Stirling_2(n,n-1)/2 = n!/4.

Original entry on oeis.org

1, 3, 6, 30, 180, 1260, 10080, 90720, 907200, 9979200, 119750400, 1556755200, 21794572800, 326918592000, 5230697472000, 88921857024000, 1600593426432000, 30411275102208000, 608225502044160000, 12772735542927360000, 281000181944401920000
Offset: 2

Views

Author

N. J. A. Sloane, Jan 17 2008

Keywords

Comments

a(n-1), n>=5, gives the number of necklaces with n beads (C_n symmetry) with color signature determined from the partition 2^2,1^(n-4) of n. Only n-2 distinct colors, say c[1], c[2], ..., c[n-2] are used, and the representative necklaces have the color c[1] and c[2] each twice. E.g., n=5, partition 2,2,1, color signature (take the parts as exponents) c[1]c[1]c[2]c[2]c[3], with the a(4)=6 necklaces (write j for color c[j]) 11223, 11232, 11322, 12213, 12123 and 12132, all taken cyclically. See A212359 for the numbers for general partitions or color signatures. - Wolfdieter Lang, Jun 27 2012

Crossrefs

A diagonal of triangle A133800.
Cf. A212359.

Programs

  • Magma
    [n le 3 select 2*n-3 else Factorial(n)/4: n in [2..30]]; // G. C. Greubel, Sep 28 2024
    
  • Mathematica
    Join[{1,3},Range[4,30]!/4] (* Harvey P. Dale, Aug 13 2013 *)
  • PARI
    concat([1,3],vector(66,n,(n+3)!/4)) \\ Joerg Arndt, Aug 14 2013
    
  • SageMath
    def A133799(n): return (factorial(n) +2*int(n==2) +6*int(n==3))//4
    [A133799(n) for n in range(2,31)] # G. C. Greubel, Sep 28 2024

Formula

a(n) = numerator(n!/(2*(n! - 2))) for n > 2. - Stefano Spezia, Dec 06 2023

Extensions

Corrected parameters in definition. - Geoffrey Critzer, Apr 26 2009

A214306 Triangle with entry a(n,m) giving the total number of bracelets of n beads (D_n symmetry) with n colors available for each bead, but only m distinct colors present, with m from {1, 2, ..., n} and n >= 1.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 24, 24, 3, 5, 60, 180, 120, 12, 6, 165, 1120, 2040, 900, 60, 7, 336, 5145, 21420, 25200, 7560, 360, 8, 784, 23016, 183330, 442680, 335160, 70560, 2520, 9, 1584, 91056, 1320480, 5846400, 8890560, 4656960, 725760, 20160, 10
Offset: 1

Views

Author

Wolfdieter Lang, Jul 20 2012

Keywords

Comments

This triangle is obtained from the array A213941 by summing in row n, for n >= 1, all entries related to partitions of n with the same number of parts m.
a(n,m) is the total number of necklaces of n beads (dihedral D_n symmetry) corresponding to all the color multinomials obtained from all p(n,m) = A008284(n,m) partitions of n with m parts, written in nonincreasing form, by 'exponentiation'. Therefore only m from the available n colors are present, and a(n,m) gives the number of bracelets with n beads with only m of the n available colors present, for m from 1,2,...,n, and n >= 1. All of the possible color assignments are counted.
See the comments on A212359 for the Abramowitz-Stegun (A-St) order of partitions, and the 'exponentiation' to obtain multisets, used to encode color multinomials, from partitions. See a link in A213938 for representative multisets for given signature used to define color multinomials.
The row sums of this triangle coincide with the ones of array A213941, and they are given by A081721.

Examples

			n\m 1    2     3       4       5       8       7      8     9
1   1
2   2    1
3   3    6     1
4   4   24    24       3
5   5   60   180     120      12
6   6  165  1120    2040     900     60
7   7  336  5145   21420   25200   7560      360
8   8  784 23016  183330  442680  33516    70560   2520
9   9 1584 91056 1320480 5846400 8890560 4656960 725760 20160
...
Row n=10:  10, 3420, 357480, 8691480, 64420272, 172609920, 177811200, 68040000, 8164800, 181440;
Row n=11:  11, 6820, 1327095, 52727400, 622175400, 2714009760, 4837417200, 3592512000, 1047816000, 99792000, 1814400.
a(2,2) = 1 from the color monomial c[1]^1*c[2]^1 = c[1]*c[2] (from the m=2 partition [1,1] of n=2). The bracelet in question is cyclic(12) (we use j for color c[j] in these examples). The same holds for the necklace case.
a(5,3) = 60 + 120 = 180, from A213941(5,4) + A213941(5,5), because k(5,3,1) = A214314(5,3) = 4 and p(5,3)=2.
a(3,1) = 3 from the color monomials c[1]^3, c[2]^3 and c[3]^3. The three bracelets are cyclic(111), cyclic(222) and cyclic(333). The same holds for the necklace case.
In general a(n,1)=n from the partition [n] providing the color signature (exponent), and the n color choices.
a(3,2) = 6 from the color signature c[.]^2 c[.]^1, (from the m=2 partition [2,1] of n=3), and there are 6 choices for the color indices. The 6 bracelets are cyclic(112), cyclic(113), cyclic(221), cyclic(223), cyclic(331) and cyclic(332). The same holds for the necklace case.
a(3,3) = 1. The color multinomial is c[1]*c[2]*c[3] (from the m=3 partition [1,1,1]). All three available colors are used. There is only one bracelet: cyclic(1,2,3). The necklace cyclic(1,3,2) becomes equivalent under D_3 operation.
a(4,2) = 24 from two color signatures c[.]^3 c[.] and c[.]^2 c[.]^2 (from the two m=2 partitions of n=4: [3,1] and [2,2]). The first one produces 4*3=12 bracelets, namely 1112, 1113, 1114, 2221, 2223, 2224, 3331, 3332, 3334, 4441, 4442 and 4443, all taken cyclically. The second color signature leads to another 6*2=12 bracelets: 1122, 1133, 1144, 2233, 2244, 3344, 1212, 1313, 1414, 2323, 2424 and 3434, all taken cyclically. Together they provide the 24 bracelets counted by a(4,2). The same holds for the necklace case.
a(4,3) = 24 from the color signature c[.]^2 c[.]c[.]. There are 4*3 =12 color choices each with two bracelets: 1123, 1213, 1124, 1214, 1134, 1314, 2213, 2123, 2214, 2124, 2234, 2324, 3312, 3132, 3314, 3134, 3324, 3234, 4412, 4142, 4413, 4143, 4423 and 4243, each taken cyclically.
		

Crossrefs

Cf. A081721, A212360 (necklaces), A213941, A273891.

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_] := Binomial[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 08 2017, after Andrew Howroyd *)

Formula

a(n,m) = Sum_{j=1..p(n,m)} A213941(n, k(n,m,1)+j-1), with k(n,m,1) = A214314(n,m) the position where in the list of partitions of n in A-St order the first with m parts appears, and p(n,m) is the number of partitions of n with m parts shown in the array A008284. E.g., n=5, m=3: k(5,3,1) = A214314(5,3) = 4, p(5,3) = 2.
a(n,m) = binomial(n,m) * A273891(n,m). - Andrew Howroyd, Mar 25 2017

A213936 Number triangle with entry a(n,k), n>=1, m=1, 2, ..., n, giving the number of representative necklaces with n beads (C_n symmetry) corresponding to the color multinomial c[1]^k*c[2]*...*c[n+1-k].

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 6, 3, 1, 1, 24, 12, 4, 1, 1, 120, 60, 20, 5, 1, 1, 720, 360, 120, 30, 6, 1, 1, 5040, 2520, 840, 210, 42, 7, 1, 1, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 1, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 1
Offset: 1

Views

Author

Wolfdieter Lang, Jul 10 2012

Keywords

Comments

This table coincides with A173333 but has an extra main diagonal with entries 1.
a(n,k) is the number of necklaces of n beads (C_N symmetry), with colors from the repertoire {c[1],c[2],...,c[n]}, corresponding to the representative color multinomials obtained from the partition [k,1^(n-k)] of n with m=n-k+1 parts by 'exponentiation' (taking the parts in the given order as exponents of the colors), hence only m from the available n colors are present. As representative necklaces one takes the ones where the color c[1] appears k times. In particular, for k=1 the partition is [1^n] and all n colors are used, and there are (n-1)! necklaces from permuting the n colors.
a(n,k) appears in the representative necklace partition array A212359 in row n at the position l(n,n+1-k,1), with l(n,m,1) the position of the first partition with m parts in the list of partitions of n in A-St order. E.g., n=5, k=4: l(5,5-3,1) =2 with the partition [4,1] (used in reverse order compared to A-St).
See the comments on A212359 for the Abramowitz-Stegun (A-St) reference, and the 'exponentiation' to obtain multisets, used to encode color multinomials, from partitions.
The row sums of this triangle are given by A213937.

Examples

			n\k      1       2      3      4     5    6   7  8  9 10 ...
1        1
2        1       1
3        2       1      1
4        6       3      1      1
5       24      12      4      1     1
6      120      60     20      5     1    1
7      720     360    120     30     6    1   1
8     5040    2520    840    210    42    7   1  1
9    40320   20160   6720   1680   336   56   8  1  1
10  362880  181440  60480  15120  3024  504  72  9  1  1 ...
a(4,3) = 1 because  the partition is [3,1], the color signature (exponentiation) c[.]^3 c[.]^1, and the one representative necklace (we use j for color c[j] here) is: cyclic(1112).
a(4,2) = 3  because  the partition is [2,1^2], the color signature c[.]^2 c[.] c[.], and the three representative necklaces are: cyclic(1123), cyclic(1132) and cyclic(1213).
a(5,3) = 4  because the color signature is  c[.]^3 c[.] c[.]  (from the partition [3,1^2]). and the four representative necklaces are 11123, 11132, 11213 and 11312, all taken cyclically.
		

Crossrefs

Cf. A212359, A213937 (row sums). For columns and diagonals see the links under A173333 (after an additional 1 has been supplied for each columns).

Formula

a(n,n)=1, a(n,k) = (n-1)!/k! if 1 <= k < n, else 0.
See also A212359 with a link for the formula for general partitions.
a(n,k) = A173333(n-1,k), 1 <= k < n.

A212360 Partition array a(n,k) with the total number of necklaces (C_n symmetry) with n beads, each available in n colors, with color signature given by the k-th partition of n in Abramowitz-Stegun(A-St) order.

Original entry on oeis.org

1, 2, 1, 3, 6, 2, 4, 12, 12, 36, 6, 5, 20, 40, 120, 180, 240, 24, 6, 30, 90, 60, 300, 1200, 320, 1200, 2700, 1800, 120, 7, 42, 126, 210, 630, 3150, 2100, 3150, 4200, 25200, 12600, 12600, 37800, 15120, 720, 8, 56, 224, 392, 280, 1176, 7056, 11760, 9072, 11760, 11760, 88200, 58800, 176400, 22260, 58800, 470400, 352800, 141120, 529200, 141120, 5040
Offset: 1

Views

Author

Wolfdieter Lang, Jun 25 2012

Keywords

Comments

This array is obtained by multiplying the entry of the array A212359(n,k) (number of necklaces (C_n symmetry) with n beads, each available in n colors, with color representative given by the n-multiset representative obtained from the k-th partition of n in A-St order after 'exponentiation') with the entry of the array A035206(n,k) (number of members in the equivalence class represented by the color multiset considered for A212359(n,k)): a(n,k)=A212359(n,k)* A035206(n,k), k=1..p(n)= A000041(n), n>=1. The row sums then give the total number of necklaces with beads from n colors, given by A056665(n).
See A212359 for references, the 'exponentiation', and a link.
The corresponding triangle with the summed row entries which belong to partitions of n with fixed number of parts is A213935. [From Wolfdieter Lang, Jul 12 2012]

Examples

			n\k  1   2   3   4    5     6    7     8     9    10   11
1    1
2    2   1
3    3   6   2
4    4  12  12  36    6
5    5  20  40 120  180   240   24
6    6  30  90  60  300  1200  320  1200  2700  1800  120
...
See the link for the rows n=1..15.
a(3,1)=3 because the 3 necklaces with 3 beads coming in 3 colors have the color multinomials (here monomials)  c[1]^3=c[1]*c[1]*c[1], c[2]^3 and c[3]^3. The partition of 3 is 3, the color representative is c[1]^3, and the equivalence class with color signature from the partition 3 has the three given members.
a(3,2)=6 from the color signature 2,1 with the representative multinomial c[1]^2 c[2] with coefficient A212359(3,2)=1, the only 3-necklace cyclic(112) (taking j for the color  c[j]), and  A035206(3,2)=6 members of the whole color equivalence class: cyclic(112), cyclic(113),  cyclic(221), cyclic(223), cyclic(331) and cyclic(332).
a(3,3)=2, color signature 1^3=1,1,1 with representative multinomial  c[1]*c[2]*c[3] with coefficient A212359(3,3)=2 from the two necklaces cyclic(1,2,3) and cyclic (1,3,2). There are no other members in this class (A035206(3,3)=1).
The sum of row nr. 3 is 11=A056665(3). See the example given there with c[1]=R, c[2]=G and c[3]=B.
		

Crossrefs

Formula

a(n,k) = A212359(n,k)*A035206(n,k), k=1,2,...,p(n)= A000041(n), n>=1.

A213934 Triangle with entry a(n,m) giving the number of necklaces of n beads (C_N symmetry) with n colors available for each bead, but only m distinct fixed colors, say c[1],...,c[m], are present, with m from {1,...,n} and n>=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 3, 3, 6, 1, 3, 10, 12, 24, 1, 8, 31, 50, 60, 120, 1, 9, 71, 180, 300, 360, 720, 1, 22, 187, 815, 1260, 2100, 2520, 5040, 1, 29, 574, 2324, 6496, 10080, 16800, 20160, 40320, 1, 66, 1373, 9570, 32268, 58464, 90720, 151200, 181440, 362880
Offset: 1

Views

Author

Wolfdieter Lang, Jun 27 2012

Keywords

Comments

This triangle is obtained from the partition array A212359 by summing in the row number n, for n>=1, all entries related to partitions of n with the same number of parts m.
a(n,m) is the number of necklaces of n beads (C_N symmetry) corresponding to the representative color multinomials obtained from all partitions of n with m parts by 'exponentiation', hence only m from the available n colors are present. As a representative multinomial of each of the p(n,m)=A008284(n,m) such m-color classes we take the one where the considered m part partition of n, [p[1],...,p[m]], written in a nonincreasing way, is distributed as exponents over c[1]^p[1]*...*c[m]^p[m]. That is only the first m colors from the n available ones are involved.
See the comments on A212359 for the Abramowitz-Stegun (A-St) order of partitions, and the 'exponentiation' to obtain multisets, used to encode color multinomials, from partitions.
The row sums of this triangle coincide with the ones of array A212359, and they are given by A072605.
Number of necklaces with n beads w over a k-ary alphabet {a1,a2,...,ak} such that #(w,a1) >= #(w,a2) >= ... >= #(w,ak) >= 1, where #(w,x) counts the letters x in word w (necklace analog of A226874). - Andrew Howroyd, Dec 20 2017

Examples

			n\m  1  2    3    4     5     6     7      8      9     10 ...
1    1
2    1  1
3    1  1    2
4    1  3    3    6
5    1  3   10   12    24
6    1  8   31   50    60   120
7    1  9   71  180   300   360   720
8    1 22  187  815  1260  2100  2520   5040
9    1 29  574 2324  6496 10080 16800  20160  40320
10   1 66 1373 9570 32268 58464 90720 151200 181440 362880
...
a(5,3) = 4 + 6 = 10, from A212359(5,4) + A212359(5,5), because k(5,3,1) = 4 and p(5,3) = 2.
a(2,1) = 1 because the partition [2] of n=2 with part number m=1 corresponds to the representative color multinomial (here monomial) c[1]^2=c[1]*c[1], and there is one such representative necklace. There is another necklace color monomial in this class of n=2 colors where only m=1 color is active: c[2]*c[2]. See the triangle entry A213935(2,1)=2.
a(3,1) = 1 from the color monomial representative c[1]^3. This class has 2 other members: c[2]^3 and c[3]^3. See A213935(3,1)=3.
In general a(n,1)=1 and A213935(n,1)=n from the partition [n] providing the color signature and a representative c[1]^n.
a(3,2)=1 from the representative color multinomial c[1]^2*c[2] (from the m=2 partition [2,1] of n=3) leading to just one representative necklace cyclic(112) (when one uses j for color c[j]). The whole class consists of A213935(3,2)=6 necklaces: cyclic(112), cyclic(113), cyclic(221), cyclic(223), cyclic(331) and  cyclic(332).
a(3,3)=2. The representative color multinomial is c[1]*c[2]*c[3] (from the m=3 partition [1,1,1]). There are the two non-equivalent representative necklaces cyclic(1,2,3) and cyclic(1,3,2) which constitute already the whole class (A213935(3,3)=2).
a(4,2) = 3 from two representative color multinomials c[1]^3*c[2] and c[1]^2*c[2]^2 (from the two m=2 partitions of n=4: [3,1] and [2,2]). The first one has one representative necklace, namely cyclic(1112), the second one originates from two representative necklaces: cyclic(1122) and cyclic(1212). Together these are the 3 necklaces counted by a(4,2). The class with the first representative consists of 4*3=12 necklaces, when all 4 colors are used. The class of the second representative consists of 2*6=12 necklaces. Together they sum up to the 24 necklaces counted by A213935(4,2).
		

Crossrefs

Cf. A008284, A072605 (row sums), A212359, A213935.
Cf. A213940 (bracelets), A226874 (words).

Programs

  • Mathematica
    b[n_, i_, t_] := b[n, i, t] = If[t == 1, 1/n!, Sum[b[n - j, j, t - 1]/j!, {j, i, n/t}]];
    a226874[n_, k_] := If[n k == 0, If[n == k, 1, 0], n! b[n, 1, k]];
    T[n_, k_] := (1/n) Sum[EulerPhi[n/d] a226874[d, k], {d, Divisors[n]}];
    Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 06 2018, after Alois P. Heinz and Andrew Howroyd *)
  • PARI
    \\ here U is A226874 as vector of polynomials.
    U(n)={Vec(serlaplace(prod(k=1, n, 1/(1-y*x^k/k!) + O(x*x^n))))}
    C(n)={my(t=U(n)); vector(n, n, vector(n, k, (1/n)*sumdiv(n, d, eulerphi(n/d) * polcoeff(t[d+1], k))))}
    { my(t=C(10)); for(n=1, #t, print(t[n])) } \\ Andrew Howroyd, Dec 20 2017

Formula

a(n,m) = Sum_{j=1..p(n,m)}A212359(n,k(n,m,1)+j-1), with k(n,m,1) the position where in the list of partitions of n in A-St order the first with m parts appears, and p(n,m) the number of partitions of n with m parts shown in the array A008284. E.g., n=5, m=3: k(5,3,1)=4, p(5,3)=2.
T(n,k) = (1/n)*Sum_{d|n} phi(n/d)*A226874(d, k). - Andrew Howroyd, Dec 20 2017
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