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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A001399 a(n) is the number of partitions of n into at most 3 parts; also partitions of n+3 in which the greatest part is 3; also number of unlabeled multigraphs with 3 nodes and n edges.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14, 16, 19, 21, 24, 27, 30, 33, 37, 40, 44, 48, 52, 56, 61, 65, 70, 75, 80, 85, 91, 96, 102, 108, 114, 120, 127, 133, 140, 147, 154, 161, 169, 176, 184, 192, 200, 208, 217, 225, 234, 243, 252, 261, 271, 280, 290, 300, 310, 320, 331, 341
Offset: 0

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Keywords

Comments

Also number of tripods (trees with exactly 3 leaves) on n vertices. - Eric W. Weisstein, Mar 05 2011
Also number of partitions of n+3 into exactly 3 parts; number of partitions of n in which the greatest part is less than or equal to 3; and the number of nonnegative solutions to b + 2c + 3d = n.
Also a(n) gives number of partitions of n+6 into 3 distinct parts and number of partitions of 2n+9 into 3 distinct and odd parts, e.g., 15 = 11 + 3 + 1 = 9 + 5 + 1 = 7 + 5 + 3. - Jon Perry, Jan 07 2004
Also bracelets with n+3 beads 3 of which are red (so there are 2 possibilities with 5 beads).
More generally, the number of partitions of n into at most k parts is also the number of partitions of n+k into k positive parts, the number of partitions of n+k in which the greatest part is k, the number of partitions of n in which the greatest part is less than or equal to k, the number of partitions of n+k(k+1)/2 into exactly k distinct positive parts, the number of nonnegative solutions to b + 2c + 3d + ... + kz = n and the number of nonnegative solutions to 2c + 3d + ... + kz <= n. - Henry Bottomley, Apr 17 2001
Also coefficient of q^n in the expansion of (m choose 3)_q as m goes to infinity. - Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
From Winston C. Yang (winston(AT)cs.wisc.edu), Apr 30 2002: (Start)
Write 1,2,3,4,... in a hexagonal spiral around 0, then a(n) for n > 0 is formed by the folding points (including the initial 1). The spiral begins:
.
85--84--83--82--81--80
/ \
86 56--55--54--53--52 79
/ / \ \
87 57 33--32--31--30 51 78
/ / / \ \ \
88 58 34 16--15--14 29 50 77
/ / / / \ \ \ \
89 59 35 17 5---4 13 28 49 76
/ / / / / \ \ \ \ \
90 60 36 18 6 0 3 12 27 48 75
/ / / / / / / / / / /
91 61 37 19 7 1---2 11 26 47 74
\ \ \ \ / / / /
62 38 20 8---9--10 25 46 73
\ \ \ / / /
63 39 21--22--23--24 45 72
\ \ / /
64 40--41--42--43--44 71
\ /
65--66--67--68--69--70
.
a(p) is maximal number of hexagons in a polyhex with perimeter at most 2p + 6. (End)
a(n-3) is the number of partitions of n into 3 distinct parts, where 0 is allowed as a part. E.g., at n=9, we can write 8+1+0, 7+2+0, 6+3+0, 4+5+0, 1+2+6, 1+3+5 and 2+3+4, which is a(6)=7. - Jon Perry, Jul 08 2003
a(n) gives number of partitions of n+6 into parts <=3 where each part is used at least once (subtract 6=1+2+3 from n). - Jon Perry, Jul 03 2004
This is also the number of partitions of n+3 into exactly 3 parts (there is a 1-to-1 correspondence between the number of partitions of n+3 in which the greatest part is 3 and the number of partitions of n+3 into exactly three parts). - Graeme McRae, Feb 07 2005
Apply the Riordan array (1/(1-x^3),x) to floor((n+2)/2). - Paul Barry, Apr 16 2005
Also, number of triangles that can be created with odd perimeter 3,5,7,9,11,... with all sides whole numbers. Note that triangles with even perimeter can be generated from the odd ones by increasing each side by 1. E.g., a(1) = 1 because perimeter 3 can make {1,1,1} 1 triangle. a(4) = 3 because perimeter 9 can make {1,4,4} {2,3,4} {3,3,3} 3 possible triangles. - Bruce Love (bruce_love(AT)ofs.edu.sg), Nov 20 2006
Also number of nonnegative solutions of the Diophantine equation x+2*y+3*z=n, cf. Pólya/Szegő reference.
From Vladimir Shevelev, Apr 23 2011: (Start)
Also a(n-3), n >= 3, is the number of non-equivalent necklaces of 3 beads each of them painted by one of n colors.
The sequence {a(n-3), n >= 3} solves the so-called Reis problem about convex k-gons in case k=3 (see our comment to A032279).
a(n-3) (n >= 3) is an essentially unimprovable upper estimate for the number of distinct values of the permanent in (0,1)-circulants of order n with three 1's in every row. (End)
A001399(n) is the number of 3-tuples (w,x,y) having all terms in {0,...,n} and w = 2*x+3*y. - Clark Kimberling, Jun 04 2012
Also, for n >= 3, a(n-3) is the number of the distinct triangles in an n-gon, see the Ngaokrajang links. - Kival Ngaokrajang, Mar 16 2013
Also, a(n) is the total number of 5-curve coin patterns (5C4S type: 5 curves covering full 4 coins and symmetry) packing into fountain of coins base (n+3). See illustration in links. - Kival Ngaokrajang, Oct 16 2013
Also a(n) = half the number of minimal zero sequences for Z_n of length 3 [Ponomarenko]. - N. J. A. Sloane, Feb 25 2014
Also, a(n) equals the number of linearly-independent terms at 2n-th order in the power series expansion of an Octahedral Rotational Energy Surface (cf. Harter & Patterson). - Bradley Klee, Jul 31 2015
Also Molien series for invariants of finite Coxeter groups D_3 and A_3. - N. J. A. Sloane, Jan 10 2016
Number of different distributions of n+6 identical balls in 3 boxes as x,y,z where 0 < x < y < z. - Ece Uslu and Esin Becenen, Jan 11 2016
a(n) is also the number of partitions of 2*n with <= n parts and no part >= 4. The bijection to partitions of n with no part >= 4 is: 1 <-> 2, 2 <-> 1 + 3, 3 <-> 3 + 3 (observing the order of these rules). The <- direction uses the following fact for partitions of 2*n with <= n parts and no part >=4: for each part 1 there is a part 3, and an even number (including 0) of remaining parts 3. - Wolfdieter Lang, May 21 2019
List of the terms in A000567(n>=1), A049450(n>=1), A033428(n>=1), A049451(n>=1), A045944(n>=1), and A003215(n) in nondecreasing order. List of the numbers A056105(n)-1, A056106(n)-1, A056107(n)-1, A056108(n)-1, A056109(n)-1, and A003215(m) with n >= 1 and m >= 0 in nondecreasing order. Numbers of the forms 3n*(n-1)+1, n*(3n-2), n*(3n-1), 3n^2, n*(3n+1), n*(3n+2) with n >= 1 listed in nondecreasing order. Integers m such that lattice points from 1 through m on a hexagonal spiral starting at 1 forms a convex polygon. - Ya-Ping Lu, Jan 24 2024

Examples

			G.f. = 1 + x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 7*x^6 + 8*x^7 + 10*x^8 + 12*x^9 + ...
Recall that in a necklace the adjacent beads have distinct colors. Suppose we have n colors with labels 1,...,n. Two colorings of the beads are equivalent if the cyclic sequences of the distances modulo n between labels of adjacent colors have the same period. If n=4, all colorings are equivalent. E.g., for the colorings {1,2,3} and {1,2,4} we have the same period {1,1,2} of distances modulo 4. So, a(n-3)=a(1)=1. If n=5, then we have two such periods {1,1,3} and {1,2,2} modulo 5. Thus a(2)=2. - _Vladimir Shevelev_, Apr 23 2011
a(0) = 1, i.e., {1,2,3} Number of different distributions of 6 identical balls to 3 boxes as x,y and z where 0 < x < y < z. - _Ece Uslu_, Esin Becenen, Jan 11 2016
a(3) = 3, i.e., {1,2,6}, {1,3,5}, {2,3,4} Number of different distributions of 9 identical balls in 3 boxes as x,y and z where 0 < x < y < z. - _Ece Uslu_, Esin Becenen, Jan 11 2016
From _Gus Wiseman_, Apr 15 2019: (Start)
The a(0) = 1 through a(8) = 10 integer partitions of n with at most three parts are the following. The Heinz numbers of these partitions are given by A037144.
  ()  (1)  (2)   (3)    (4)    (5)    (6)    (7)    (8)
           (11)  (21)   (22)   (32)   (33)   (43)   (44)
                 (111)  (31)   (41)   (42)   (52)   (53)
                        (211)  (221)  (51)   (61)   (62)
                               (311)  (222)  (322)  (71)
                                      (321)  (331)  (332)
                                      (411)  (421)  (422)
                                             (511)  (431)
                                                    (521)
                                                    (611)
The a(0) = 1 through a(7) = 8 integer partitions of n + 3 whose greatest part is 3 are the following. The Heinz numbers of these partitions are given by A080193.
  (3)  (31)  (32)   (33)    (322)    (332)     (333)      (3322)
             (311)  (321)   (331)    (3221)    (3222)     (3331)
                    (3111)  (3211)   (3311)    (3321)     (32221)
                            (31111)  (32111)   (32211)    (33211)
                                     (311111)  (33111)    (322111)
                                               (321111)   (331111)
                                               (3111111)  (3211111)
                                                          (31111111)
Non-isomorphic representatives of the a(0) = 1 through a(5) = 5 unlabeled multigraphs with 3 vertices and n edges are the following.
  {}  {12}  {12,12}  {12,12,12}  {12,12,12,12}  {12,12,12,12,12}
            {13,23}  {12,13,23}  {12,13,23,23}  {12,13,13,23,23}
                     {13,23,23}  {13,13,23,23}  {12,13,23,23,23}
                                 {13,23,23,23}  {13,13,23,23,23}
                                                {13,23,23,23,23}
The a(0) = 1 through a(8) = 10 strict integer partitions of n - 6 with three parts are the following (A = 10, B = 11). The Heinz numbers of these partitions are given by A007304.
  (321)  (421)  (431)  (432)  (532)  (542)  (543)  (643)   (653)
                (521)  (531)  (541)  (632)  (642)  (652)   (743)
                       (621)  (631)  (641)  (651)  (742)   (752)
                              (721)  (731)  (732)  (751)   (761)
                                     (821)  (741)  (832)   (842)
                                            (831)  (841)   (851)
                                            (921)  (931)   (932)
                                                   (A21)   (941)
                                                           (A31)
                                                           (B21)
The a(0) = 1 through a(8) = 10 integer partitions of n + 3 with three parts are the following. The Heinz numbers of these partitions are given by A014612.
  (111)  (211)  (221)  (222)  (322)  (332)  (333)  (433)  (443)
                (311)  (321)  (331)  (422)  (432)  (442)  (533)
                       (411)  (421)  (431)  (441)  (532)  (542)
                              (511)  (521)  (522)  (541)  (551)
                                     (611)  (531)  (622)  (632)
                                            (621)  (631)  (641)
                                            (711)  (721)  (722)
                                                   (811)  (731)
                                                          (821)
                                                          (911)
The a(0) = 1 through a(8) = 10 integer partitions of n whose greatest part is <= 3 are the following. The Heinz numbers of these partitions are given by A051037.
  ()  (1)  (2)   (3)    (22)    (32)     (33)      (322)      (332)
           (11)  (21)   (31)    (221)    (222)     (331)      (2222)
                 (111)  (211)   (311)    (321)     (2221)     (3221)
                        (1111)  (2111)   (2211)    (3211)     (3311)
                                (11111)  (3111)    (22111)    (22211)
                                         (21111)   (31111)    (32111)
                                         (111111)  (211111)   (221111)
                                                   (1111111)  (311111)
                                                              (2111111)
                                                              (11111111)
The a(0) = 1 through a(6) = 7 strict integer partitions of 2n+9 with 3 parts, all of which are odd, are the following. The Heinz numbers of these partitions are given by A307534.
  (5,3,1)  (7,3,1)  (7,5,1)  (7,5,3)   (9,5,3)   (9,7,3)   (9,7,5)
                    (9,3,1)  (9,5,1)   (9,7,1)   (11,5,3)  (11,7,3)
                             (11,3,1)  (11,5,1)  (11,7,1)  (11,9,1)
                                       (13,3,1)  (13,5,1)  (13,5,3)
                                                 (15,3,1)  (13,7,1)
                                                           (15,5,1)
                                                           (17,3,1)
The a(0) = 1 through a(8) = 10 strict integer partitions of n + 3 with 3 parts where 0 is allowed as a part (A = 10):
  (210)  (310)  (320)  (420)  (430)  (530)  (540)  (640)  (650)
                (410)  (510)  (520)  (620)  (630)  (730)  (740)
                       (321)  (610)  (710)  (720)  (820)  (830)
                              (421)  (431)  (810)  (910)  (920)
                                     (521)  (432)  (532)  (A10)
                                            (531)  (541)  (542)
                                            (621)  (631)  (632)
                                                   (721)  (641)
                                                          (731)
                                                          (821)
The a(0) = 1 through a(7) = 7 integer partitions of n + 6 whose distinct parts are 1, 2, and 3 are the following. The Heinz numbers of these partitions are given by A143207.
  (321)  (3211)  (3221)   (3321)    (32221)    (33221)     (33321)
                 (32111)  (32211)   (33211)    (322211)    (322221)
                          (321111)  (322111)   (332111)    (332211)
                                    (3211111)  (3221111)   (3222111)
                                               (32111111)  (3321111)
                                                           (32211111)
                                                           (321111111)
(End)
Partitions of 2*n with <= n parts and no part >= 4: a(3) = 3 from (2^3), (1,2,3), (3^2) mapping to (1^3), (1,2), (3), the partitions of 3 with no part >= 4, respectively. - _Wolfdieter Lang_, May 21 2019
		

References

  • R. Ayoub, An Introduction to the Analytic Theory of Numbers, Amer. Math. Soc., 1963; Chapter III, Problem 33.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 110, D(n); page 263, #18, P_n^{3}.
  • J. L. Gross and J. Yellen, eds., Handbook of Graph Theory, CRC Press, 2004; p. 517.
  • H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 2.
  • F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 88, (4.1.18).
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 275.
  • R. Honsberger, Mathematical Gems III, Math. Assoc. Amer., 1985, p. 39.
  • J. H. van Lint, Combinatorial Seminar Eindhoven, Lecture Notes Math., 382 (1974), see pp. 33-34.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis I (Springer 1924, reprinted 1972), Part One, Chap. 1, Sect. 1, Problem 25.
  • 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

Programs

  • Haskell
    a001399 = p [1,2,3] where
       p _      0 = 1
       p []     _ = 0
       p ks'@(k:ks) m = if m < k then 0 else p ks' (m - k) + p ks m
    -- Reinhard Zumkeller, Feb 28 2013
    
  • Magma
    I:=[1,1,2,3,4,5]; [n le 6 select I[n] else Self(n-1)+Self(n-2)-Self(n-4)-Self(n-5)+Self(n-6): n in [1..80]]; // Vincenzo Librandi, Feb 14 2015
    
  • Magma
    [#RestrictedPartitions(n,{1,2,3}): n in [0..62]]; // Marius A. Burtea, Jan 06 2019
    
  • Magma
    [Round((n+3)^2/12): n in [0..70]]; // Marius A. Burtea, Jan 06 2019
    
  • Maple
    A001399 := proc(n)
        round( (n+3)^2/12) ;
    end proc:
    seq(A001399(n),n=0..40) ;
    with(combstruct):ZL4:=[S,{S=Set(Cycle(Z,card<4))}, unlabeled]:seq(count(ZL4,size=n),n=0..61); # Zerinvary Lajos, Sep 24 2007
    B:=[S,{S = Set(Sequence(Z,1 <= card),card <=3)},unlabelled]: seq(combstruct[count](B, size=n), n=0..61); # Zerinvary Lajos, Mar 21 2009
  • Mathematica
    CoefficientList[ Series[ 1/((1 - x)*(1 - x^2)*(1 - x^3)), {x, 0, 65} ], x ]
    Table[ Length[ IntegerPartitions[n, 3]], {n, 0, 61} ] (* corrected by Jean-François Alcover, Aug 08 2012 *)
    k = 3; 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 *)
    LinearRecurrence[{1,1,0,-1,-1,1},{1,1,2,3,4,5},70] (* Harvey P. Dale, Jun 21 2012 *)
    a[ n_] := With[{m = Abs[n + 3] - 3}, Length[ IntegerPartitions[ m, 3]]]; (* Michael Somos, Dec 25 2014 *)
    k=3 (* Number of red beads in bracelet problem *);CoefficientList[Series[(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 *)
    Table[Length[Select[IntegerPartitions[n,{3}],UnsameQ@@#&]],{n,0,30}] (* Gus Wiseman, Apr 15 2019 *)
  • PARI
    {a(n) = round((n + 3)^2 / 12)}; /* Michael Somos, Sep 04 2006 */
    
  • Python
    [round((n+3)**2 / 12) for n in range(0,62)] # Ya-Ping Lu, Jan 24 2024

Formula

G.f.: 1/((1 - x) * (1 - x^2) * (1 - x^3)) = -1/((x+1)*(x^2+x+1)*(x-1)^3); Simon Plouffe in his 1992 dissertation
a(n) = round((n + 3)^2/12). Note that this cannot be of the form (2*i + 1)/2, so ties never arise.
a(n) = A008284(n+3, 3), n >= 0.
a(n) = 1 + a(n-2) + a(n-3) - a(n-5) for all n in Z. - Michael Somos, Sep 04 2006
a(n) = a(-6 - n) for all n in Z. - Michael Somos, Sep 04 2006
a(6*n) = A003215(n), a(6*n + 1) = A000567(n + 1), a(6*n + 2) = A049450(n + 1), a(6*n + 3) = A033428(n + 1), a(6*n + 4) = A049451(n + 1), a(6*n + 5) = A045944(n + 1).
a(n) = a(n-1) + A008615(n+2) = a(n-2) + A008620(n) = a(n-3) + A008619(n) = A001840(n+1) - a(n-1) = A002620(n+2) - A001840(n) = A000601(n) - A000601(n-1). - Henry Bottomley, Apr 17 2001
P(n, 3) = (1/72) * (6*n^2 - 7 - 9*pcr{1, -1}(2, n) + 8*pcr{2, -1, -1}(3, n)) (see Comtet). [Here "pcr" stands for "prime circulator" and it is defined on p. 109 of Comtet, while the formula appears on p. 110. - Petros Hadjicostas, Oct 03 2019]
Let m > 0 and -3 <= p <= 2 be defined by n = 6*m+p-3; then for n > -3, a(n) = 3*m^2 + p*m, and for n = -3, a(n) = 3*m^2 + p*m + 1. - Floor van Lamoen, Jul 23 2001
72*a(n) = 17 + 6*(n+1)*(n+5) + 9*(-1)^n - 8*A061347(n). - Benoit Cloitre, Feb 09 2003
From Jon Perry, Jun 17 2003: (Start)
a(n) = 6*t(floor(n/6)) + (n%6) * (floor(n/6) + 1) + (n mod 6 == 0?1:0), where t(n) = n*(n+1)/2.
a(n) = ceiling(1/12*n^2 + 1/2*n) + (n mod 6 == 0?1:0).
[Here "n%6" means "n mod 6" while "(n mod 6 == 0?1:0)" means "if n mod 6 == 0 then 1, else 0" (as in C).]
(End)
a(n) = Sum_{i=0..floor(n/3)} 1 + floor((n - 3*i)/2). - Jon Perry, Jun 27 2003
a(n) = Sum_{k=0..n} floor((k + 2)/2) * (cos(2*Pi*(n - k)/3 + Pi/3)/3 + sqrt(3) * sin(2*Pi*(n-k)/3 + Pi/3)/3 + 1/3). - Paul Barry, Apr 16 2005
(m choose 3)_q = (q^m-1) * (q^(m-1) - 1) * (q^(m-2) - 1)/((q^3 - 1) * (q^2 - 1) * (q - 1)).
a(n) = Sum_{k=0..floor(n/2)} floor((3 + n - 2*k)/3). - Paul Barry, Nov 11 2003
A117220(n) = a(A003586(n)). - Reinhard Zumkeller, Mar 04 2006
a(n) = 3 * Sum_{i=2..n+1} floor(i/2) - floor(i/3). - Thomas Wieder, Feb 11 2007
Identical to the number of points inside or on the boundary of the integer grid of {I, J}, bounded by the three straight lines I = 0, I - J = 0 and I + 2J = n. - Jonathan Vos Post, Jul 03 2007
a(n) = A026820(n,3) for n > 2. - Reinhard Zumkeller, Jan 21 2010
Euler transform of length 3 sequence [ 1, 1, 1]. - Michael Somos, Feb 25 2012
a(n) = A005044(2*n + 3) = A005044(2*n + 6). - Michael Somos, Feb 25 2012
a(n) = A000212(n+3) - A002620(n+3). - Richard R. Forberg, Dec 08 2013
a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6). - David Neil McGrath, Feb 14 2015
a(n) = floor((n^2+3)/12) + floor((n+2)/2). - Giacomo Guglieri, Apr 02 2019
From Devansh Singh, May 28 2020: (Start)
Let p(n, 3) be the number of 3-part integer partitions in which every part is > 0.
Then for n >= 3, p(n, 3) is equal to:
(n^2 - 1)/12 when n is odd and 3 does not divide n.
(n^2 + 3)/12 when n is odd and 3 divides n.
(n^2 - 4)/12 when n is even and 3 does not divide n.
(n^2)/12 when n is even and 3 divides n.
For n >= 3, p(n, 3) = a(n-3). (End)
a(n) = floor(((n+3)^2 + 4)/12). - Vladimír Modrák, Zuzana Soltysova, Dec 08 2020
Sum_{n>=0} 1/a(n) = 15/4 - Pi/(2*sqrt(3)) + Pi^2/18 + tanh(Pi/(2*sqrt(3)))*Pi/sqrt(3). - Amiram Eldar, Sep 29 2022
E.g.f.: exp(-x)*(9 + exp(2*x)*(47 + 42*x + 6*x^2) + 16*exp(x/2)*cos(sqrt(3)*x/2))/72. - Stefano Spezia, Mar 05 2023
a(6n) = 1+6*A000217(n); Sum_{i=1..n} a(6*i) = A000578(n+1). - David García Herrero, May 05 2024

Extensions

Name edited by Gus Wiseman, Apr 15 2019

A008805 Triangular numbers repeated.

Original entry on oeis.org

1, 1, 3, 3, 6, 6, 10, 10, 15, 15, 21, 21, 28, 28, 36, 36, 45, 45, 55, 55, 66, 66, 78, 78, 91, 91, 105, 105, 120, 120, 136, 136, 153, 153, 171, 171, 190, 190, 210, 210, 231, 231, 253, 253, 276, 276, 300, 300, 325, 325, 351, 351, 378, 378, 406, 406, 435, 435
Offset: 0

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Author

Keywords

Comments

Number of choices for nonnegative integers x,y,z such that x and y are even and x + y + z = n.
Diagonal sums of A002260, when arranged as a number triangle. - Paul Barry, Feb 28 2003
a(n) = number of partitions of n+4 such that the differences between greatest and smallest parts are 2: a(n-4) = A097364(n,2) for n>3. - Reinhard Zumkeller, Aug 09 2004
For n >= i, i=4,5, a(n-i) is the number of incongruent two-color bracelets of n beads, i from them are black (cf. A005232, A032279), having a diameter of symmetry. - Vladimir Shevelev, May 03 2011
Prefixing A008805 by 0,0,0,0 gives the sequence c(0), c(1), ... defined by c(n)=number of (w,x,y) such that w = 2x+2y, where w,x,y are all in {1,...,n}; see A211422. - Clark Kimberling, Apr 15 2012
Partial sums of positive terms of A142150. - Reinhard Zumkeller, Jul 07 2012
The sum of the first parts of the nondecreasing partitions of n+2 into exactly two parts, n >= 0. - Wesley Ivan Hurt, Jun 08 2013
Number of the distinct symmetric pentagons in a regular n-gon, see illustration for some small n in links. - Kival Ngaokrajang, Jun 25 2013
a(n) is the number of nonnegative integer solutions to the equation x + y + z = n such that x + y <= z. For example, a(4) = 6 because we have 0+0+4 = 0+1+3 = 0+2+2 = 1+0+3 = 1+1+2 = 2+0+2. - Geoffrey Critzer, Jul 09 2013
a(n) is the number of distinct opening moves in n X n tic-tac-toe. - I. J. Kennedy, Sep 04 2013
a(n) is the number of symmetry-allowed, linearly-independent terms at n-th order in the series expansion of the T2 X t2 vibronic perturbation matrix, H(Q) (cf. Opalka & Domcke). - Bradley Klee, Jul 20 2015
a(n-1) also gives the number of D_4 (dihedral group of order 4) orbits of an n X n square grid with squares coming in either of two colors and only one square has one of the colors. - Wolfdieter Lang, Oct 03 2016
Also, this sequence is the third column in the triangle of the coefficients of the sum of two consecutive Fibonacci polynomials F(n+1, x) and F(n, x) (n>=0) in ascending powers of x. - Mohammad K. Azarian, Jul 18 2018
In an n-person symmetric matching pennies game (a zero-sum normal-form game) with n > 2 symmetric and indistinguishable players, each with two strategies (viz. heads or tails), a(n-3) is the number of distinct subsets of players that must play the same strategy to avoid incurring losses (single pure Nash equilibrium in the reduced game). The total number of distinct partitions is A000217(n-1). - Ambrosio Valencia-Romero, Apr 17 2022
a(n) is the number of connected bipartite graphs with n+1 edges and a stable set of cardinality 2. - Christian Barrientos, Jun 15 2022
a(n) is the number of 132-avoiding odd Grassmannian permutations of size n+2. - Juan B. Gil, Mar 10 2023
Consider a regular n-gon with all diagonals drawn. Define a "layer" to be the set of all regions sharing an edge with the exterior. Removing a layer creates another layer. Count the layers, removing them until none remain. The number of layers is a(n-2). See illustration. - Christopher Scussel, Nov 07 2023

Examples

			a(5) = 6, since (5) + 2 = 7 has three nondecreasing partitions with exactly 2 parts: (1,6),(2,5),(3,4). The sum of the first parts of these partitions = 1 + 2 + 3 = 6. - _Wesley Ivan Hurt_, Jun 08 2013
		

References

  • H. D. Brunk, An Introduction to Mathematical Statistics, Ginn, Boston, 1960; p. 360.

Crossrefs

Cf. A000217, A002260, A002620, A006918 (partial sums), A054252, A135276, A142150, A158920 (binomial trans.).

Programs

  • GAP
    List([0..60], n-> (2*n +3 +(-1)^n)*(2*n +7 +(-1)^n)/32); # G. C. Greubel, Sep 12 2019
    
  • Haskell
    import Data.List (transpose)
    a008805 = a000217 . (`div` 2) . (+ 1)
    a008805_list = drop 2 $ concat $ transpose [a000217_list, a000217_list]
    -- Reinhard Zumkeller, Feb 01 2013
    
  • Magma
    [(2*n+3+(-1)^n)*(2*n+7+(-1)^n)/32 : n in [0..50]]; // Wesley Ivan Hurt, Apr 22 2015
    
  • Maple
    A008805:=n->(2*n+3+(-1)^n)*(2*n+7+(-1)^n)/32: seq(A008805(n), n=0..50); # Wesley Ivan Hurt, Apr 22 2015
  • Mathematica
    CoefficientList[Series[1/(1-x^2)^2/(1-x), {x, 0, 50}], x]
    Table[Binomial[Floor[n/2] + 2, 2], {n, 0, 57}] (* Michael De Vlieger, Oct 03 2016 *)
  • PARI
    a(n)=(n\2+2)*(n\2+1)/2
    
  • Python
    def A008805(n): return (m:=(n>>1)+1)*(m+1)>>1 # Chai Wah Wu, Oct 20 2023
  • Sage
    [(2*n +3 +(-1)^n)*(2*n +7 +(-1)^n)/32 for n in (0..60)] # G. C. Greubel, Sep 12 2019
    

Formula

G.f.: 1/((1-x)*(1-x^2)^2) = 1/((1+x)^2*(1-x)^3).
E.g.f.: (exp(x)*(2*x^2 +12*x+ 11) - exp(-x)*(2*x -5))/16.
a(-n) = a(-5+n).
a(n) = binomial(floor(n/2)+2, 2). - Vladimir Shevelev, May 03 2011
From Paul Barry, May 31 2003: (Start)
a(n) = ((2*n +5)*(-1)^n + (2*n^2 +10*n +11))/16.
a(n) = Sum_{k=0..n} ((k+2)*(1+(-1)^k))/4. (End)
From Paul Barry, Apr 16 2005: (Start)
a(n) = Sum_{k=0..n} floor((k+2)/2)*(1-(-1)^(n+k-1))/2.
a(n) = Sum_{k=0..floor(n/2)} floor((n-2k+2)/2). (End)
A signed version is given by Sum_{k=0..n} (-1)^k*floor(k^2/4). - Paul Barry, Aug 19 2003
a(n) = A108299(n-2,n)*(-1)^floor((n+1)/2) for n>1. - Reinhard Zumkeller, Jun 01 2005
a(n) = A004125(n+3) - A049798(n+2). - Carl Najafi, Jan 31 2013
a(n) = Sum_{i=1..floor((n+2)/2)} i. - Wesley Ivan Hurt, Jun 08 2013
a(n) = (1/2)*floor((n+2)/2)*(floor((n+2)/2)+1). - Wesley Ivan Hurt, Jun 08 2013
From Wesley Ivan Hurt, Apr 22 2015: (Start)
a(n) = a(n-1) +2*a(n-2) -2*a(n-3) -a(n-4) +a(n-5).
a(n) = (2*n +3 +(-1)^n)*(2*n +7 +(-1)^n)/32. (End)
a(n-1) = A054252(n,1) = A054252(n^2-1), n >= 1. See a Oct 03 2016 comment above. - Wolfdieter Lang, Oct 03 2016
a(n) = A000217(A008619(n)). - Guenther Schrack, Sep 12 2018
From Ambrosio Valencia-Romero, Apr 17 2022: (Start)
a(n) = a(n-1) if n odd, a(n) = a(n-1) + (n+2)/2 if n is even, for n > 0, a(0) = 1.
a(n) = (n+1)*(n+3)/8 if n odd, a(n) = (n+2)*(n+4)/8 if n is even, for n >= 0.
a(n) = A002620(n+2) - a(n-1), for n > 0, a(0) = 1.
a(n) = A142150(n+2) + a(n-1), for n > 0, a(0) = 1.
a(n) = A000217(n+3)/2 - A135276(n+3)/2. (End)

A052307 Triangle read by rows: T(n,k) = number of bracelets (reversible necklaces) with n beads, k of which are black and n - k are white.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 3, 3, 3, 1, 1, 1, 1, 3, 4, 4, 3, 1, 1, 1, 1, 4, 5, 8, 5, 4, 1, 1, 1, 1, 4, 7, 10, 10, 7, 4, 1, 1, 1, 1, 5, 8, 16, 16, 16, 8, 5, 1, 1, 1, 1, 5, 10, 20, 26, 26, 20, 10, 5, 1, 1, 1, 1, 6, 12, 29, 38, 50, 38, 29, 12, 6, 1, 1, 1, 1, 6, 14, 35, 57, 76, 76, 57, 35, 14, 6, 1, 1
Offset: 0

Views

Author

Christian G. Bower, Nov 15 1999

Keywords

Comments

Equivalently, T(n,k) is the number of orbits of k-element subsets of the vertices of a regular n-gon under the usual action of the dihedral group D_n, or under the action of Euclidean plane isometries. Note that each row of the table is symmetric and unimodal. - Austin Shapiro, Apr 20 2009
Also, the number of k-chords in n-tone equal temperament, up to (musical) transposition and inversion. Example: there are 29 tetrachords, 38 pentachords, 50 hexachords in the familiar 12-tone equal temperament. Called "Forte set-classes," after Allen Forte who first cataloged them. - Jon Wild, May 21 2004

Examples

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
   1;
   1,  1;
   1,  1,  1;
   1,  1,  1,  1;
   1,  1,  2,  1,  1;
   1,  1,  2,  2,  1,  1;
   1,  1,  3,  3,  3,  1,  1;
   1,  1,  3,  4,  4,  3,  1,  1;
   1,  1,  4,  5,  8,  5,  4,  1,  1;
   1,  1,  4,  7, 10, 10,  7,  4,  1,  1;
   1,  1,  5,  8, 16, 16, 16,  8,  5,  1,  1;
   1,  1,  5, 10, 20, 26, 26, 20, 10,  5,  1,  1;
   1,  1,  6, 12, 29, 38, 50, 38, 29, 12,  6,  1,  1;
   ...
		

References

  • Martin Gardner, "New Mathematical Diversions from Scientific American" (Simon and Schuster, New York, 1966), pages 245-246.
  • N. Zagaglia Salvi, Ordered partitions and colourings of cycles and necklaces, Bull. Inst. Combin. Appl., 27 (1999), 37-40.

Crossrefs

Programs

  • Maple
    A052307 := proc(n,k)
            local hk,a,d;
            if k = 0 then
                    return 1 ;
            end if;
            hk := k mod 2 ;
            a := 0 ;
            for d in numtheory[divisors](igcd(k,n)) do
                    a := a+ numtheory[phi](d)*binomial(n/d-1,k/d-1) ;
            end do:
            %/k + binomial(floor((n-hk)/2),floor(k/2)) ;
            %/2 ;
    end proc: # R. J. Mathar, Sep 04 2011
  • Mathematica
    Table[If[m*n===0,1,1/2*If[EvenQ[n], If[EvenQ[m], Binomial[n/2, m/2], Binomial[(n-2)/2, (m-1)/2 ]], If[EvenQ[m], Binomial[(n-1)/2, m/2], Binomial[(n-1)/2, (m-1)/2]]] + 1/2*Fold[ #1 +(EulerPhi[ #2]*Binomial[n/#2, m/#2])/n &, 0, Intersection[Divisors[n], Divisors[m]]]], {n,0,12}, {m,0,n}] (* Wouter Meeussen, Aug 05 2002, Jan 19 2009 *)
  • PARI
    B(n,k)={ if(n==0, return(1)); GCD = gcd(n, k); S = 0;
    for(d = 1, GCD, if((k%d==0)&&(n%d==0), S+=eulerphi(d)*binomial(n/d,k/d)));
    return (binomial(floor(n/2)- k%2*(1-n%2), floor(k/2))/2 + S/(2*n)); }
    n=0;k=0; for(L=0, 8645, print(L, " ", B(n,k)); k++; if(k>n, k=0; n++))
    /* Washington Bomfim, Jun 30 2012 */
    
  • Python
    from sympy import binomial as C, totient, divisors, gcd
    def T(n, k): return 1 if n==0 else C((n//2) - k%2 * (1 - n%2), (k//2))/2 + sum(totient(d)*C(n//d, k//d) for d in divisors(gcd(n, k)))/(2*n)
    for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 23 2017

Formula

T(0,0) = 1. If n > 0, T(n,k) = binomial(floor(n/2) - (k mod 2) * (1 - (n mod 2)), floor(k/2)) / 2 + Sum_{d|n, d|k} (phi(d)*binomial(n/d, k/d)) / (2*n). - Washington Bomfim, Jun 30 2012 [edited by Petros Hadjicostas, May 29 2019]
From Freddy Barrera, Apr 21 2019: (Start)
T(n,k) = (1/2) * (A119963(n,k) + A047996(n,k)).
T(n,k) = T(n, n-k) for each k < n (Theorem 2 of H. Gupta). (End)
G.f. for column k >= 1: (x^k/2) * ((1/k) * Sum_{m|k} phi(m)/(1 - x^m)^(k/m) + (1 + x)/(1 - x^2)^floor((k/2) + 1)). (This formula is due to Herbert Kociemba.) - Petros Hadjicostas, May 25 2019
Bivariate o.g.f.: Sum_{n,k >= 0} T(n, k)*x^n*y^k = (1/2) * ((x + 1) * (x*y + 1) / (1 - x^2 * (y^2 + 1)) + 1 - Sum_{d >= 1} (phi(d)/d) * log(1 - x^d * (1 + y^d))). - Petros Hadjicostas, Jun 13 2019

A005232 Expansion of (1-x+x^2)/((1-x)^2*(1-x^2)*(1-x^4)).

Original entry on oeis.org

1, 1, 3, 4, 8, 10, 16, 20, 29, 35, 47, 56, 72, 84, 104, 120, 145, 165, 195, 220, 256, 286, 328, 364, 413, 455, 511, 560, 624, 680, 752, 816, 897, 969, 1059, 1140, 1240, 1330, 1440, 1540, 1661, 1771, 1903, 2024, 2168, 2300, 2456, 2600, 2769, 2925, 3107, 3276
Offset: 0

Views

Author

Keywords

Comments

Also number of n X 2 binary matrices under row and column permutations and column complementations (if offset is 0).
Also Molien series for certain 4-D representation of dihedral group of order 8.
With offset 4, number of bracelets (turnover necklaces) of n-bead of 2 colors with 4 red beads. - Washington Bomfim, Aug 27 2008
From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of 4 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=4 (see our comment to A032279). (End)
Number of 2 X 2 matrices with nonnegative integer values totaling n under row and column permutations. - Gabriel Burns, Nov 08 2016
From Petros Hadjicostas, Jan 12 2019: (Start)
By "necklace", Vladimir Shevelev (above) means "turnover necklace", i.e., a bracelet. Zagaglia Salvi (1999) also uses this terminology: she calls a bracelet "necklace" and a necklace "cycle".
According to Cyvin et al. (1997), the sequence (a(n): n >= 0) consists of "the total numbers of isomers of polycyclic conjugated hydrocarbons with q + 1 rings and q internal carbons in one ring (class Q_q)", where q = 4 and n is the hydrogen content (i.e., we count certain isomers of C_{n+2*q} H_n with q = 4 and n >= 0). (End)

Examples

			G.f. = 1 + x + 3*x^2 + 4*x^3 + 8*x^4 + 10*x^5 + 16*x^6 + 20*x^7 + 29*x^8 + ...
There are 8 4 X 2 matrices up to row and column permutations and column complementations:
  [1 1] [1 0] [1 0] [0 1] [0 1] [0 1] [0 1] [0 0]
  [1 1] [1 1] [1 0] [1 0] [1 0] [1 0] [0 1] [0 1]
  [1 1] [1 1] [1 1] [1 1] [1 0] [1 0] [1 0] [1 0]
  [1 1] [1 1] [1 1] [1 1] [1 1] [1 0] [1 0] [1 1].
There are 8 2 X 2 matrices of nonnegative integers totaling 4 up to row and column permutations:
  [4 0] [3 1] [2 2] [2 1] [2 1] [3 0] [2 0] [1 1]
  [0 0] [0 0] [0 0] [0 1] [1 0] [1 0] [2 0] [1 1].
		

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

Row n=2 of A343875.
Column k=4 of A052307.

Programs

  • Maple
    A005232:=-(-1-z-2*z**3+2*z**2+z**7-2*z**6+2*z**4)/(z**2+1)/(1+z)**2/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence apart from an initial 1
  • Mathematica
    k = 4; 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 + x^2)/((1 - x)^2(1 - x^2)(1 - x^4)), {x, 0, 51}], x] (* Robert G. Wilson v, Mar 29 2006 *)
    LinearRecurrence[{2,0,-2,2,-2,0,2,-1},{1,1,3,4,8,10,16,20},60] (* Harvey P. Dale, Oct 24 2012 *)
    k=4 (* Number of red beads in bracelet problem *); CoefficientList[Series[(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) = (n^3 + 9*n^2 + (32-9*(n%2))*n + [48, 15, 36, 15][n%4+1]) / 48}; \\ Michael Somos, Feb 01 2007
    
  • PARI
    {a(n) = my(s=1); if( n<-5, n = -6-n; s=-1); if( n<0, 0, s * polcoeff( (1 - x + x^2) / ((1 - x)^2 * (1 - x^2) * (1 - x^4)) + x * O(x^n), n))}; \\ Michael Somos, Feb 01 2007
    
  • PARI
    a(n) = round((n^3 +9*n^2 +(32-9*(n%2))*n)/48 +0.6) \\ Washington Bomfim, Jul 17 2008
    
  • PARI
    a(n) = ceil((n+1)*(2*n^2+16*n+39+9*(-1)^n)/96) \\ Tani Akinari, Aug 23 2013
    
  • Python
    a=lambda n: sum((k//2+1)*((n-k)//2+1) for k in range((n-1)//2+1))+(n+1)%2*(((n//4+1)*(n//4+2))//2)  # Gabriel Burns, Nov 08 2016

Formula

G.f.: (1+x^3)/((1-x)*(1-x^2)^2*(1-x^4)).
G.f.: (1/8)*(1/(1-x)^4+3/(1-x^2)^2+2/(1-x)^2/(1-x^2)+2/(1-x^4)). - Vladeta Jovovic, Aug 05 2000
Euler transform of length 6 sequence [ 1, 2, 1, 1, 0, -1 ]. - Michael Somos, Feb 01 2007
a(2n+1) = A006918(2n+2)/2;
a(2n) = (A006918(2n+1) + A008619(n))/2.
a(n) = -a(-6 - n) for all n in Z. - Michael Somos, Feb 05 2011
From Vladimir Shevelev, Apr 22 2011: (Start)
if n == 0 (mod 4), then a(n) = n*(n^2-3*n+8)/48;
if n == 1, 3 (mod 4), then a(n) = (n^2-1)*(n-3)/48;
if n == 2 (mod 4), then a(n) = (n-2)*(n^2-n+6)/48. (End)
a(n) = 2*a(n-1) - 2*a(n-3) + 2*a(n-4) - 2*a(n-5) + 2*a(n-7) - a(n-8) with a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 4, a(4) = 8, a(5) = 10, a(6) = 16, a(7) = 20. - Harvey P. Dale, Oct 24 2012
a(n) = ((n+3)*(2*n^2+12*n+19+9*(-1)^n) + 6*(-1)^((2*n-1+(-1)^n)/4)*(1+(-1)^n))/96. - Luce ETIENNE, Mar 16 2015
a(n) = |A128498(n)| + |A128498(n-3)|. - R. J. Mathar, Jun 11 2019

Extensions

Sequence extended by Christian G. Bower

A005513 Number of n-bead bracelets (turnover necklaces) of two colors with 6 red beads and n-6 black beads.

Original entry on oeis.org

1, 1, 4, 7, 16, 26, 50, 76, 126, 185, 280, 392, 561, 756, 1032, 1353, 1782, 2277, 2920, 3652, 4576, 5626, 6916, 8372, 10133, 12103, 14448, 17063, 20128, 23528, 27474, 31824, 36822, 42315, 48564, 55404, 63133, 71554, 81004
Offset: 6

Views

Author

Keywords

Comments

From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent (turnover) necklaces of 6 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=6 (see our comment to A032279).
(End)
Also number of reverse invariant anonymous and neutral equivalence classes of preference profiles with 3 alternatives and (n-6) agents (IANC model). - Alexander Karpov, Apr 12 2018
Also the number of weighted cubic graphs with weight n derived from one of the 2 cubic graphs on 6 vertices (contributing to A321306). - R. J. Mathar, Nov 05 2018

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

Column k=6 of A052307.

Programs

  • Maple
    A005513 := proc(n) if n mod 6 = 0 then (24*binomial(n-1,5)+3*(n+1)*(n-2)*(n-4)+16*n)/288 elif n mod 6 = 3 then (24*binomial(n-1,5)+3*(n-1)*(n-3)*(n-5)+16*n-48)/288 elif n mod 6 = 2 or n mod 6 = 4 then (8*binomial(n-1,5)+(n+1)*(n-2)*(n-4))/96 elif n mod 6 = 1 or n mod 6 = 5 then (8*binomial(n-1,5)+(n-1)*(n-3)*(n-5))/96 fi: end: seq(A005513(n),n=6..44); # Johannes W. Meijer, Aug 11 2011
  • Mathematica
    k = 6; 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=6; 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 *)
    CoefficientList[Series[(1/12) (1/(1 - x)^6 + 4/(1 - x^2)^3 + 2/(1 - x^3)^2 + 3/((1 - x)^2 (1 - x^2)^2) + 2/(1 - x^6)), {x, 0, 43}], x] (* Vincenzo Librandi, Apr 24 2018 *)

Formula

S. J. Cyvin et al. (1997) give a g.f.
G.f.: (x^6/12)*(1/(1-x)^6+4/(1-x^2)^3+2/(1-x^3)^2+3/((1-x)^2*(1-x^2)^2)+2/(1-x^6)). - Vladeta Jovovic, Feb 28 2007
G.f.: x^6*(1-x+x^2+x^3+2*x^4+2*x^6+x^8-x^5) / ( (x^2-x+1)*(1+x+x^2)^2*(1+x)^3*(x-1)^6 ). - R. J. Mathar, Sep 18 2011
From Vladimir Shevelev, Apr 23 2011: (Start)
if n==0 mod 6, a(n)=(24*C(n-1,5)+3*(n+1)*(n-2)*(n-4)+16*n)/288;
if n==3 mod 6, a(n)=(24*C(n-1,5)+3*(n-1)*(n-3)*(n-5)+16*n-48)/288;
if n==2,4 mod 6, a(n)=(8*C(n-1,5)+(n+1)*(n-2)*(n-4))/96;
if n==1,5 mod 6, a(n)=(8*C(n-1,5)+(n-1)*(n-3)*(n-5))/96.
(End)

Extensions

Sequence extended and description corrected by Christian G. Bower
Name edited by Petros Hadjicostas, Jan 10 2019

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

Views

Author

Keywords

Comments

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

A032280 Number of bracelets (turnover necklaces) of n beads of 2 colors, 7 of them black.

Original entry on oeis.org

1, 1, 4, 8, 20, 38, 76, 133, 232, 375, 600, 912, 1368, 1980, 2829, 3936, 5412, 7293, 9724, 12760, 16588, 21287, 27092, 34112, 42640, 52819, 65008, 79392, 96405, 116280, 139536, 166464, 197676, 233529, 274740, 321741, 375364
Offset: 7

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From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of nonequivalent necklaces of 7 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=7 (see our comment to A032279).
(End)

References

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

Crossrefs

Column k=7 of A052307.

Programs

  • Mathematica
    k = 7; 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[-(4 x^6 - 2 x^5 - 2 x^4 + 4 x^3 + x^2 - 2 x + 1)/((x - 1)^7 (x + 1)^3 (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 19 2013 *)
    k=7; 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.
"DIK[ 7 ]" (necklace, indistinct, unlabeled, 7 parts) transform of 1, 1, 1, 1...
From Vladimir Shevelev, Apr 23 2011: (Start)
Put s(n,k,d) = 1, if n == k(mod d); 0, otherwise. Then
a(n) = (3/7)*s(n,0,7) + (48*C(n-1,6) + 7*(n-2)*(n-4)*(n-6))/672, if n is even;
a(n) = (3/7)*s(n,0,7) + (48*C(n-1,6) + 7*(n-1)*(n-3)*(n-5))/672, if n is odd. (End)
G.f.: -x^7*(4*x^6-2*x^5-2*x^4+4*x^3+x^2-2*x+1) / ((x-1)^7*(x+1)^3*(x^6+x^5+x^4+x^3+x^2+x+1)). - Colin Barker, Feb 06 2013
From Herbert Kociemba, Nov 05 2016: (Start)
G.f.: (1/2)*x^7*((1+x)/(1-x^2)^4 + 1/7*(1/(1-x)^7 + 6/(1-x^7))).
G.f.: k=7, x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^floor((k+2)/2))/2. [edited by Petros Hadjicostas, Jul 18 2018] (End)

A032281 Number of bracelets (turnover necklaces) of n beads of 2 colors, 9 of them black.

Original entry on oeis.org

1, 1, 5, 12, 35, 79, 185, 375, 750, 1387, 2494, 4262, 7105, 11410, 17930, 27407, 41107, 60335, 87154, 123695, 173173, 238957, 325845, 438945, 585265, 772252, 1009868, 1308742, 1682660, 2146420, 2718806, 3419924, 4274905, 5310667, 6560225, 8059021, 9849925
Offset: 9

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Comments

From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of 9 beads each of them painted by one of n colors.
The sequence solves the so-called Reis problem about convex k-gons in the case k=9 (see our comment to A032279). (End)

References

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

Crossrefs

Column k=9 of A052307.

Programs

  • Mathematica
    k = 9; 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=9;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

"DIK[ 9 ]" (necklace, indistinct, unlabeled, 9 parts) transform of 1, 1, 1, 1...
Put s(n,k,d) = 1, if n == k (mod d), and s(n,k,d) = 0, otherwise. Then a(n) =(1/3)*s(n,0,9) + (n-3)*(n-6)*s(n,0,3)/162 + (n-2)(n-4)*(n-6)*(n-8)*(945 + (n-1)*(n-3)*(n-5)*(n-7))/725760, if n is even; a(n) = (1/3)*s(n,0,9) + (n-3)*(n-6)*s(n,0,3)/162 +(n-1)*(n-3)*(n-5)*(n-7)*(945 + (n-2)*(n-4)*(n-6)*(n-8))/725760, if n is odd. - Vladimir Shevelev, Apr 23 2011
From Herbert Kociemba, Nov 05 2016: (Start)
G.f.: (1/2)*x^9*((1+x)/(1-x^2)^5 + 1/9*(1/(1-x)^9 - 2/(-1+x^3)^3 - 6/(-1+x^9))).
G.f.: k=9, x^k*((1/k)*(Sum_{d|k} phi(d)*(1-x^d)^(-k/d)) + (1+x)/(1-x^2)^floor((k+2)/2))/2. [edited by Petros Hadjicostas, Jul 18 2018] (End)

A032282 Number of bracelets (turnover necklaces) of n beads of 2 colors, 11 of them black.

Original entry on oeis.org

1, 1, 6, 16, 56, 147, 392, 912, 2052, 4262, 8524, 16159, 29624, 52234, 89544, 148976, 242086, 384111, 597506, 911456, 1367184, 2017509, 2934559, 4209504, 5963464, 8347612, 11558232, 15837472, 21493712, 28903332
Offset: 11

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From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent necklaces of 11 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=11 (see our comment to A032279).
(End)

References

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

Crossrefs

Column k=11 of A052307.

Programs

  • Mathematica
    k = 11; 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=11;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

"DIK[ 11 ]" (necklace, indistinct, unlabeled, 11 parts) transform of 1, 1, 1, 1...
From Vladimir Shevelev, Apr 23 2011: (Start)
Put s(n,k,d)=1, if n==k(mod d), and s(n,k,d)=0, otherwise. Then
a(n) = 5*s(n,0,11)/11+(3840*C(n-1,10)+11*(n-2)*(n-4)*(n-6)(n-8)*(n-10))/84480, if n is even;
a(n) = 5*s(n,0,11)/11+(3840*C(n-1,10)+11*(n-1)*(n-3)*(n-5)*(n-7)*(n-9))/84480, if n is odd.
(End)
From Herbert Kociemba, Nov 05 2016: (Start)
G.f.: 1/22*x^11*(1/(1-x)^11 + 11/((-1+x)^6*(1+x)^5) - 10/(-1+x^11)).
G.f.: k=11, x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^floor[(k+2)/2])/2. [edited by Petros Hadjicostas, Jul 18 2018]
(End)

A005515 Number of n-bead bracelets (turnover necklaces) of two colors with 10 red beads and n-10 black beads.

Original entry on oeis.org

1, 1, 6, 14, 47, 111, 280, 600, 1282, 2494, 4752, 8524, 14938, 25102, 41272, 65772, 102817, 156871, 235378, 346346, 502303, 716859, 1010256, 1404624, 1931540, 2625658, 3534776, 4711448, 6226148, 8156396, 10603704, 13679696, 17527595, 22304765, 28209566, 35459694
Offset: 10

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Comments

From Vladimir Shevelev, Apr 23 2011: (Start)
Also number of non-equivalent (turnover) necklaces of 10 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=10 (see our comment to A032279). (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

Column k=10 of A052307.

Programs

  • Mathematica
    k = 10; 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=10;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

From Vladimir Shevelev, Apr 23 2011: (Start)
Put s(n,k,d) = 1, if n == k (mod d), and s(n,k,d) = 0, otherwise. Then a(n) = n*s(n,0,5)/25 + (384*C(n-1,9) + (n+1)*(n-2)*(n-4)*(n-6)*(n-8))/7680, if n is even; a(n) = (n-5)*s(n,0,5)/25 + (384*C(n-1,9) + (n-1)*(n-3)*(n-5)*(n-7)*(n-9))/7680, if n is odd. (End)
From Herbert Kociemba, Nov 04 2016: (Start)
G.f.: (1/20)*x^10*(1/(-1+x)^10 + 10/((-1+x)^6*(1+x)^5) + 1/(1-x^2)^5 + 4/(-1+x^5)^2 - 4/(-1+x^10)).
G.f.: k=10, x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^floor((k+2)/2))/2. [edited by Petros Hadjicostas, Jan 10 2019] (End)

Extensions

Sequence extended and description corrected by Christian G. Bower
Name edited by Petros Hadjicostas, Jan 10 2019
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