cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-8 of 8 results.

A001998 Bending a piece of wire of length n+1; walks of length n+1 on a tetrahedron; also non-branched catafusenes with n+2 condensed hexagons.

Original entry on oeis.org

1, 2, 4, 10, 25, 70, 196, 574, 1681, 5002, 14884, 44530, 133225, 399310, 1196836, 3589414, 10764961, 32291602, 96864964, 290585050, 871725625, 2615147350, 7845353476, 23535971854, 70607649841, 211822683802, 635467254244, 1906400965570, 5719200505225, 17157599124190
Offset: 0

Views

Author

Keywords

Comments

The wire stays in the plane, there are n bends, each is R,L or O; turning the wire over does not count as a new figure.
Equivalently, walks of n+1 steps on a tetrahedron, visiting n+2 vertices, with n "corners"; the symmetry group is S4, reversing a walk does not count as different. Simply interpret R,L,O as instructions to turn R, turn L, or retrace the last step. Walks are not self-avoiding.
Also, it appears that a(n) gives the number of equivalence classes of n-tuples of 0, 1 and 2, where two n-tuples are equivalent if one can be obtained from the other by a sequence of operations R and C, where R denotes reversal and C denotes taking the 2's complement (C(x)=2-x). This has been verified up to a(19)=290585050. Example: for n=3 there are ten equivalence classes {000, 222}, {001, 100, 122, 221}, {002, 022, 200, 220}, {010, 212}, {011, 110, 112, 211}, {012, 210}, {020, 202}, {021, 102, 120, 201}, {101, 121}, {111}, so a(3)=10. - John W. Layman, Oct 13 2009
There exists a bijection between chains of n+2 hexagons and the above described equivalence classes of n-tuples of 0,1, and 2. Namely, for a given chain of n+2 hexagons we take the sequence of the numbers of vertices of degree 2 (0, 1, or 2) between the consecutive contact vertices on one side of the chain; switching to the other side we obtain the 2's complement of this sequence; reversing the order of the hexagons, we obtain the reverse sequence. The inverse mapping is straightforward. For example, to a linear chain of 7 hexagons there corresponds the 5-tuple 11111. - Emeric Deutsch, Apr 22 2013
If we treat two wire bends (or walks, or tuples) related by turning over (or reversing) as different in any of the above-given interpretations of this sequence, we get A007051 (or A124302). Also, a(n-1) is the sum of first 3 terms in n-th row of A284949, see crossrefs therein. - Andrey Zabolotskiy, Sep 29 2017
a(n-1) is the number of color patterns (set partitions) in an unoriented row of length n using 3 or fewer colors (subsets). - Robert A. Russell, Oct 28 2018
From Allan Bickle, Jun 02 2022: (Start)
a(n) is the number of (unlabeled) 3-paths with n+6 vertices. (A 3-path with order n at least 5 can be constructed from a 4-clique by iteratively adding a new 3-leaf (vertex of degree 3) adjacent to an existing 3-clique containing an existing 3-leaf.)
Recurrences appear in the papers by Bickle, Eckhoff, and Markenzon et al. (End)
a(n) is also the number of distinct planar embeddings of the (n+1)-alkane graph (up to at least n=9, and likely for all n). - Eric W. Weisstein, May 21 2024

Examples

			There are 2 ways to bend a piece of wire of length 2 (bend it or not).
For n=4 and a(n-1)=10, the 6 achiral patterns are AAAA, AABB, ABAB, ABBA, ABCA, and ABBC.  The 4 chiral pairs are AAAB-ABBB, AABA-ABAA, AABC-ABCC, and ABAC-ABCB. - _Robert A. Russell_, Oct 28 2018
		

References

  • A. T. Balaban, Enumeration of Cyclic Graphs, pp. 63-105 of A. T. Balaban, ed., Chemical Applications of Graph Theory, Ac. Press, 1976; see p. 75.
  • S. J. Cyvin, B. N. Cyvin, and J. Brunvoll, Enumeration of tree-like octagonal systems: catapolyoctagons, ACH Models in Chem. 134 (1997), 55-70.
  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2.]
  • R. C. Read, The Enumeration of Acyclic Chemical Compounds, pp. 25-61 of A. T. Balaban, ed., Chemical Applications of Graph Theory, Ac. Press, 1976. [I think this reference does not mention this sequence. - N. J. A. Sloane, Aug 10 2006]
  • 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

Column 3 of A320750, offset by one. Column k = 0 of A323942, offset by two.
Cf. A124302 (oriented), A107767 (chiral), A182522 (achiral), with varying offsets.
Column 3 of A320750.
The numbers of unlabeled k-paths for k = 2..7 are given in A005418, A001998, A056323, A056324, A056325, and A345207, respectively.
The sequences above converge to A103293(n+1).

Programs

  • GAP
    a:=[];; for n in [2..45] do if n mod 2 =0 then Add(a,((3^((n-2)/2)+1)/2)^2); else Add(a,  3^((n-3)/2)+(1/4)*(3^(n-2)+1)); fi; od; a; # Muniru A Asiru, Oct 28 2018
  • Maple
    A001998 := proc(n) if n = 0 then 1 elif n mod 2 = 1 then (1/4)*(3^n+4*3^((n-1)/2)+1) else (1/4)*(3^n+2*3^(n/2)+1); fi; end;
    A001998:=(-1+3*z+2*z**2-8*z**3+3*z**4)/(z-1)/(3*z-1)/(3*z**2-1); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence with an extra leading 1
  • Mathematica
    a[n_?OddQ] := (1/4)*(3^n + 4*3^((n - 1)/2) + 1); a[n_?EvenQ] := (1/4)*(3^n + 2*3^(n/2) + 1); Table[a[n], {n, 0, 27}] (* Jean-François Alcover, Jan 25 2013, from formula *)
    LinearRecurrence[{4,0,-12,9},{1,2,4,10},30] (* Harvey P. Dale, Apr 10 2013 *)
    Ach[n_, k_] := Ach[n, k] = If[n<2, Boole[n==k && n>=0], k Ach[n-2,k] + Ach[n-2,k-1] + Ach[n-2,k-2]] (* A304972 *)
    k=3; Table[Sum[StirlingS2[n,j]+Ach[n,j],{j,k}]/2,{n,40}] (* Robert A. Russell, Oct 28 2018 *)
  • PARI
    Vec((1-2*x-4*x^2+6*x^3)/((1-x)*(1-3*x)*(1-3*x^2)) + O(x^50)) \\ Colin Barker, May 15 2016
    

Formula

a(n) = if n mod 2 = 0 then ((3^((n-2)/2)+1)/2)^2 else 3^((n-3)/2)+(1/4)*(3^(n-2)+1).
G.f.: (1-2*x-4*x^2+6*x^3) / ((1-x)*(1-3*x)*(1-3*x^2)). - Corrected by Colin Barker, May 15 2016
a(n) = 4*a(n-1)-12*a(n-3)+9*a(n-4), with a(0)=1, a(1)=2, a(2)=4, a(3)=10. - Harvey P. Dale, Apr 10 2013
a(n) = (1+3^n+3^(1/2*(-1+n))*(2-2*(-1)^n+sqrt(3)+(-1)^n*sqrt(3)))/4. - Colin Barker, May 15 2016
E.g.f.: (2*sqrt(3)*sinh(sqrt(3)*x) + 3*exp(2*x)*cosh(x) + 3*cosh(sqrt(3)*x))/6. - Ilya Gutkovskiy, May 15 2016
From Robert A. Russell, Oct 28 2018: (Start)
a(n-1) = (A124302(n) + A182522(n)) / 2 = A124302(n) - A107767(n-1) = A107767(n-1) + A182522(n).
a(n-1) = Sum_{j=1..k} (S2(n,j) + Ach(n,j)) / 2, where k=3 is the maximum number of colors, S2 is the Stirling subset number A008277, and Ach(n,k) = [n>=0 & n<2 & n==k] + [n>1]*(k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2)).
a(n-1) = A057427(n) + A056326(n) + A056327(n). (End)
a(2*n) = A007051(n)^2; a(2*n+1) = A007051(n)*A007051(n+1). - Todd Simpson, Mar 25 2024

Extensions

Offset and Maple code corrected by Colin Mallows, Nov 12 1999
Term added by Robert A. Russell, Oct 30 2018

A052551 Expansion of 1/((1 - x)*(1 - 2*x^2)).

Original entry on oeis.org

1, 1, 3, 3, 7, 7, 15, 15, 31, 31, 63, 63, 127, 127, 255, 255, 511, 511, 1023, 1023, 2047, 2047, 4095, 4095, 8191, 8191, 16383, 16383, 32767, 32767, 65535, 65535, 131071, 131071, 262143, 262143, 524287, 524287, 1048575, 1048575, 2097151, 2097151
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Equals row sums of triangle A137865. - Gary W. Adamson, Feb 18 2008
Also, the decimal representation of the diagonal from the corner to the origin of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 566", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Jul 05 2017
Number of nonempty subsets of {1,2,...,n+1} that contain only odd numbers. a(0) = a(1) = 1: {1}; a(6) = a(7) = 15: {1}, {3}, {5}, {7}, {1,3}, {1,5}, {1,7}, {3,5}, {3,7}, {5,7}, {1,3,5}, {1,3,7}, {1,5,7}, {3,5,7}, {1,3,5,7}. - Enrique Navarrete, Mar 16 2018
Number of nonempty subsets of {1,2,...,n+2} that contain only even numbers. a(0) = a(1) = 1: {2}; a(4) = a(5) = 7: {2}, {4}, {6}, {2,4}, {2,6}, {4,6}, {2,4,6}. - Enrique Navarrete, Mar 26 2018
Doubling of A000225(n+1), n >= 0 entries. First differences give A077957. - Wolfdieter Lang, Apr 08 2018
a(n-2) is the number of achiral rows or cycles of length n partitioned into two sets or the number of color patterns using exactly 2 colors. An achiral row or cycle is equivalent to its reverse. Two color patterns are equivalent if the colors are permuted. For n = 4, the a(n-2) = 3 row patterns are AABB, ABAB, and ABBA; the cycle patterns are AAAB, AABB, and ABAB. For n = 5, the a(n-2) = 3 patterns for both rows and cycles are AABAA, ABABA, and ABBBA. For n = 6, the a(n-2) = 7 patterns for rows are AAABBB, AABABB, AABBAA, ABAABA, ABABAB, ABBAAB, and ABBBBA; the cycle patterns are AAAAAB, AAAABB, AAABAB, AAABBB, AABAAB, AABABB, and ABABAB. - Robert A. Russell, Oct 15 2018
For integers m > 1, the expansion of 1/((1 - x)*(1 - m*x^2)) generates a(n) = (sqrt(m)^(n + 1)*((-1)^n*(sqrt(m) - 1) + sqrt(m) + 1) - 2)/(2*(m - 1)). It appears, for integer values of n >= 0 and m > 1, that it could be simplified in the integral domain a(n) = (m^(1 + floor(n/2)) - 1)/(m - 1). - Federico Provvedi, Nov 23 2018
From Werner Schulte, Mar 04 2019: (Start)
More generally: For some fixed integers q and r > 0 the expansion of A(q,r; x) = 1/((1-x)*(1-q*x^r)) generates coefficients a(q,r; n) = (q^(1+floor(n/r))-1)/(q-1) for n >= 0; the special case q = 1 leads to a(1,r; n) = 1 + floor(n/r).
The a(q,r; n) satisfy for n > r a linear recurrence equation with constant coefficients. The signature vector is given by the sum of two vectors v and w where v has terms 1 followed by r zeros, i.e., (1,0,0,...,0), and w has r-1 leading zeros followed by q and -q, i.e., (0,0,...,0,q,-q).
Let a_i(q,r; n) be the convolution inverse of a(q,r; n). The terms are given by the sum a_i(q,r; n) = b(n) + c(n) for n >= 0 where b(n) has terms 1 and -1 followed by infinitely zeros, i.e., (1,-1,0,0,0,...), and c(n) has r leading zeros followed by -q, q and infinitely zeros, i.e., (0,0,...,0,-q,q,0,0,0,...).
Here is an example for q = 3 and r = 5: The expansion of A(3,5; x) = 1/((1-x)*(1-3*x^5)) = Sum_{n>=0} a(3,5; n)*x^n generates the sequence of coefficients (a(3,5; n)) = (1,1,1,1,1,4,4,4,4,4,13,13,13,13,13,40,...) where r = 5 controls the repetition and q = 3 the different values.
The a(3,5; n) satisfy for n > 5 the linear recurrence equation with constant coefficients and signature (1,0,0,0,0,0) + (0,0,0,0,3,-3) = (1,0,0,0,3,-3).
The convolution inverse a_i(3,5; n) has terms (1,-1,0,0,0,0,0,0,0,...) + (0,0,0,0,0,-3,3,0,0,...) = (1,-1,0,0,0,-3,3,0,0,...).
For further examples and informations see A014983 (q,r = -3,1), A077925 (q,r = -2,1), A000035 (q,r = -1,1), A000012 (q,r = 0,1), A000027 (q,r = 1,1), A000225 (q,r = 2,1), A003462 (q,r = 3,1), A077953 (q,r = -2,2), A133872 (q,r = -1,2), A004526 (q,r = 1,2), A052551 (this sequence with q,r = 2,2), A077886 (q,r = -2,3), A088911 (q,r = -1,3), A002264 (q,r = 1,3) and A077885 (q,r = 2,3). The offsets might be different.
(End)
a(n) is the number of palindromes of length n over the alphabet {1,2} containing the letter 1. More generally, the number of palindromes of length n over the alphabet {1,2,...,k} containing the letter 1 is given by k^ceiling(n/2)-(k-1)^ceiling(n/2). - Sela Fried, Dec 10 2024

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Crossrefs

Column 2 (offset by two) of A304972.
Cf. A000225 (oriented), A056326 (unoriented), and A122746(n-2) (chiral) for rows.
Cf. A056295 (oriented), A056357 (unoriented), and A059053 (chiral) for cycles.

Programs

  • GAP
    Flat(List([1..21],n->[2^n-1,2^n-1])); # Muniru A Asiru, Oct 16 2018
    
  • Magma
    [2^Floor(n/2)-1: n in [2..50]]; // Vincenzo Librandi, Aug 16 2011
    
  • Maple
    spec := [S,{S=Prod(Sequence(Prod(Z,Union(Z,Z))),Sequence(Z))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    Table[StirlingS2[Floor[n/2] + 2, 2], {n, 0, 50}] (* Robert A. Russell, Dec 20 2017 *)
    Drop[LinearRecurrence[{1, 2, -2}, {0, 1, 1}, 50], 1] (* Robert A. Russell, Oct 14 2018 *)
    CoefficientList[Series[1/((1-x)*(1-2*x^2)), {x, 0, 50}], x] (* Stefano Spezia, Oct 16 2018 *)
    2^(1+Floor[(Range[0,50])/2])-1 (* Federico Provvedi, Nov 22 2018 *)
    ((-1)^#(Sqrt[2]-1)+Sqrt[2]+1)2^((#-1)/2)-1&@Range[0, 50] (* Federico Provvedi, Nov 23 2018 *)
  • PARI
    x='x+O('x^50); Vec(1/((1-x)*(1-2*x^2))) \\ Altug Alkan, Mar 19 2018
    
  • Sage
    [2^(floor(n/2)) -1 for n in (2..50)] # G. C. Greubel, Mar 04 2019

Formula

G.f.: 1/((1 - x)*(1 - 2*x^2)).
Recurrence: a(1) = 1, a(0) = 1, -2*a(n) - 1 + a(n+2) = 0.
a(n) = -1 + Sum((1/2)*(1 + 2*alpha)*alpha^(-1 - n)) where the sum is over alpha = the two roots of -1 + 2*x^2.
a(n) = A016116(n+2) - 1. - R. J. Mathar, Jun 15 2009
a(n) = A060546(n+1) - 1. - Filip Zaludek, Dec 10 2016
From Robert A. Russell, Oct 15 2018: (Start)
a(n-2) = S2(floor(n/2)+1,2), where S2 is the Stirling subset number A008277.
a(n-2) = 2*A056326(n) - A000225(n) = A000225(n) - 2*A122746(n-2) = A056326(n) - A122746(n-2).
a(n-2) = 2*A056357(n) - A056295(n) = A056295(n) - 2*A059053(n) = A056357(n) - A059053(n). (End)
From Federico Provvedi, Nov 22 2018: (Start)
a(n) = 2^( 1 + floor(n/2) ) - 1.
a(n) = ( (-1)^n*(sqrt(2)-1) + sqrt(2) + 1 ) * 2^( (n - 1)/2 ) - 1. (End)
E.g.f.: 2*cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x) - cosh(x) - sinh(x). - Franck Maminirina Ramaharo, Nov 23 2018

Extensions

More terms from James Sellers, Jun 06 2000

A284949 Triangle read by rows: T(n,k) = number of reversible string structures of length n using exactly k different symbols.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 5, 4, 1, 1, 9, 15, 6, 1, 1, 19, 50, 37, 9, 1, 1, 35, 160, 183, 76, 12, 1, 1, 71, 502, 877, 542, 142, 16, 1, 1, 135, 1545, 3930, 3523, 1346, 242, 20, 1, 1, 271, 4730, 17185, 21393, 11511, 2980, 390, 25, 1
Offset: 1

Views

Author

Andrew Howroyd, Apr 06 2017

Keywords

Comments

A string and its reverse are considered to be equivalent. Permuting the colors will not change the structure.
Number of k-block partitions of an n-set up to reflection.
T(n,k) = pi_k(P_n) which is the number of non-equivalent partitions of the path on n vertices, with exactly k parts. Two partitions P1 and P2 of a graph G are said to be equivalent if there is a nontrivial automorphism of G which maps P1 onto P2. - Mohammad Hadi Shekarriz, Aug 21 2019

Examples

			Triangle begins:
1;
1,   1;
1,   2,    1;
1,   5,    4,     1;
1,   9,   15,     6,     1;
1,  19,   50,    37,     9,     1;
1,  35,  160,   183,    76,    12,    1;
1,  71,  502,   877,   542,   142,   16,   1;
1, 135, 1545,  3930,  3523,  1346,  242,  20,  1;
1, 271, 4730, 17185, 21393, 11511, 2980, 390, 25, 1;
		

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Crossrefs

Columns 2..6 are A056326, A056327, A056328, A056329, A056330.
Row sums are A103293.
Partial row sums include A005418, A001998(n-1), A056323, A056324, A056325.
Cf. A277504, A008277 (set partitions), A152175 (up to rotation), A152176 (up to rotation and reflection), A304972 (achiral patterns).

Programs

  • Mathematica
    (* achiral color patterns for row of n colors containing k different colors *)
    Ach[n_, k_] := Ach[n, k] = Switch[k, 0, If[0==n, 1, 0], 1, If[n>0, 1, 0],
       (* else *) _, If[OddQ[n],
       Sum[Binomial[(n-1)/2, i] Ach[n-1-2i, k-1], {i, 0, (n-1)/2}],
       Sum[Binomial[n/2-1, i] (Ach[n-2-2i, k-1] + 2^i Ach[n-2-2i, k-2]),
       {i, 0, n/2-1}]]]
    Table[(StirlingS2[n, k] + Ach[n, k])/2, {n, 1, 15}, {k, 1, n}] // Flatten
    (* Robert A. Russell, Feb 10 2018 *)
  • PARI
    \\ see A056391 for Polya enumeration functions
    T(n,k) = NonequivalentStructsExactly(ReversiblePerms(n), k); \\ Andrew Howroyd, Oct 14 2017
    
  • PARI
    \\ Ach is A304972 as square matrix.
    Ach(n)={my(M=matrix(n,n,i,k,i>=k)); for(i=3, n, for(k=2, n, M[i,k]=k*M[i-2,k] + M[i-2,k-1] + if(k>2, M[i-2,k-2]))); M}
    T(n)={(matrix(n, n, i, k, stirling(i, k, 2)) + Ach(n))/2}
    { my(A=T(10)); for(n=1, #A, print(A[n,1..n])) } \\ Andrew Howroyd, Sep 18 2019

A056324 Number of reversible string structures with n beads using a maximum of five different colors.

Original entry on oeis.org

1, 1, 2, 4, 11, 32, 116, 455, 1993, 9134, 43580, 211659, 1041441, 5156642, 25640456, 127773475, 637624313, 3184387574, 15910947980, 79521737939, 397510726681, 1987259550002, 9935420646296, 49674470817195, 248364482308833, 1241798790172214
Offset: 0

Views

Author

Keywords

Comments

A string and its reverse are considered to be equivalent. Permuting the colors will not change the structure. Thus aabc, cbaa and bbac are all considered to be identical.
Number of set partitions of an unoriented row of n elements with five or fewer nonempty subsets. - Robert A. Russell, Oct 28 2018
There are nonrecursive formulas, generating functions, and computer programs for A056272 and A305751, which can be used in conjunction with the formula. - Robert A. Russell, Oct 28 2018
From Allan Bickle, Jun 02 2022: (Start)
a(n) is the number of (unlabeled) 5-paths with n+7 vertices. (A 5-path with order n at least 7 can be constructed from a 5-clique by iteratively adding a new 5-leaf (vertex of degree 5) adjacent to an existing 5-clique containing an existing 5-leaf.)
Recurrences appear in the papers by Bickle, Eckhoff, and Markenzon et al. (End)

Examples

			For a(4)=11, the 7 achiral patterns are AAAA, AABB, ABAB, ABBA, ABCA, ABBC, and ABCD.  The 4 chiral pairs are AAAB-ABBB, AABA-ABAA, AABC-ABCC, and ABAC-ABCB.
		

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Crossrefs

Cf. A032122.
Column 5 of A320750.
Cf. A056272 (oriented), A320935 (chiral), A305751 (achiral).
The numbers of unlabeled k-paths for k = 2..7 are given in A005418, A001998, A056323, A056324, A056325, and A345207, respectively.
The sequences above converge to A103293(n+1).

Programs

  • Mathematica
    Ach[n_, k_] := Ach[n, k] = If[n<2, Boole[n==k && n>=0], k Ach[n-2,k] + Ach[n-2,k-1] + Ach[n-2,k-2]] (* A304972 *)
    k=5; Table[Sum[StirlingS2[n,j]+Ach[n,j],{j,0,k}]/2,{n,0,40}]  (* Robert A. Russell, Oct 28 2018 *)
    LinearRecurrence[{11, -34, -16, 247, -317, -200, 610, -300}, {1, 1, 2, 4, 11, 32, 116, 455, 1993}, 40] (* Robert A. Russell, Oct 28 2018 *)

Formula

Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.
G.f.: (1-10x+25x^2+32x^3-196x^4+149x^5+225x^6-321x^7+85x^8)/((1-x)*(1-2x)*(1-3x)*(1-5x)*(1-2x^2)*(1-5x^2)). - Colin Barker, Nov 24 2012 [Adapted to offset 0 by Robert A. Russell, Nov 07 2018]
From Robert A. Russell, Oct 28 2018: (Start)
a(n) = (A056272(n) + A305751(n)) / 2.
a(n) = A056272(n) - A320935(n) = A320935(n) + A305751(n).
a(n) = Sum_{j=0..k} (S2(n,j) + Ach(n,j)) / 2, where k=5 is the maximum number of colors, S2 is the Stirling subset number A008277, and Ach(n,k) = [n>=0 & n<2 & n==k] + [n>1]*(k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2)).
a(n) = A000007(n) + A057427(n) + A056326(n) + A056327(n) + A056328(n) + A056329(n). (End)
For n>8, a(n) = 11*a(n-1) - 34*a(n-2) - 16*a(n-3) + 247*a(n-4) - 317*a(n-5) - 200*a(n-6) + 610*a(n-7) - 300*a(n-8). - Muniru A Asiru, Oct 30 2018
From Allan Bickle, Jun 04 2022: (Start)
a(n) = 5^n/240 + 3^n/24 + 2^n/12 + 13*5^(n/2)/120 + 2^(n/2)/6 + 5/16 for n>0 even;
a(n) = 5^n/240 + 3^n/24 + 2^n/12 + 5^((n+1)/2)/24 + 2^((n+1)/2)/12 + 5/16 for n>0 odd. (End)

Extensions

Terms added by Robert A. Russell, Oct 30 2018
a(0)=1 prepended by Robert A. Russell, Nov 07 2018

A056323 Number of reversible string structures with n beads using a maximum of four different colors.

Original entry on oeis.org

1, 1, 2, 4, 11, 31, 107, 379, 1451, 5611, 22187, 87979, 350891, 1400491, 5597867, 22379179, 89500331, 357952171, 1431743147, 5726775979, 22906841771, 91626580651, 366505274027, 1466017950379, 5864067607211
Offset: 0

Views

Author

Keywords

Comments

A string and its reverse are considered to be equivalent. Permuting the colors will not change the structure. Thus aabc, cbaa and bbac are all considered to be identical.
Number of set partitions of an unoriented row of n elements with four or fewer nonempty subsets. - Robert A. Russell, Oct 28 2018
There are nonrecursive formulas, generating functions, and computer programs for A124303 and A305750, which can be used in conjunction with the formula. - Robert A. Russell, Oct 28 2018
From Allan Bickle, Jun 02 2022: (Start)
a(n) is the number of (unlabeled) 4-paths with n+6 vertices. (A 4-path with order n at least 6 can be constructed from a 5-clique by iteratively adding a new 4-leaf (vertex of degree 4) adjacent to an existing 4-clique containing an existing 4-leaf.)
Recurrences appear in the papers by Bickle, Eckhoff, and Markenzon et al. (End)

Examples

			For a(4)=11, the 7 achiral patterns are AAAA, AABB, ABAB, ABBA, ABCA, ABBC, and ABCD. The 4 chiral pairs are AAAB-ABBB, AABA-ABAA, AABC-ABCC, and ABAC-ABCB.
		

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Crossrefs

Cf. A032121.
Column 4 of A320750.
Cf. A124303 (oriented), A320934 (chiral), A305750 (achiral).
The numbers of unlabeled k-paths for k = 2..7 are given in A005418, A001998, A056323, A056324, A056325, and A345207, respectively.
The sequences above converge to A103293(n+1).

Programs

  • Mathematica
    Ach[n_, k_] := Ach[n, k] = If[n<2, Boole[n==k && n>=0], k Ach[n-2,k] + Ach[n-2,k-1] + Ach[n-2,k-2]] (* A304972 *)
    k=4; Table[Sum[StirlingS2[n,j]+Ach[n,j],{j,0,k}]/2,{n,0,40}] (* Robert A. Russell, Oct 28 2018 *)
    LinearRecurrence[{5, 0, -20, 16}, {1, 1, 2, 4, 11}, 40] (* Robert A. Russell, Oct 28 2018 *)

Formula

Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.
For n > 0, a(n) = (16 + (-2)^n + 15*2^n + 4^n)/48. - Colin Barker, Nov 24 2012
G.f.: (1 - 4x - 3x^2 + 14x^3 - 5x^4) / ((1-x)*(1-4x)*(1-4x^2)). - Colin Barker, Nov 24 2012 [Adapted to offset 0 by Robert A. Russell, Nov 09 2018]
From Robert A. Russell, Oct 28 2018: (Start)
a(n) = (A124303(n) + A305750(n)) / 2.
a(n) = A124303(n) - A320934(n) = A320934(n) + A305750(n).
a(n) = Sum_{j=0..k} (S2(n,j) + Ach(n,j)) / 2, where k=4 is the maximum number of colors, S2 is the Stirling subset number A008277, and Ach(n,k) = [n>=0 & n<2 & n==k] + [n>1]*(k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2)).
a(n) = A000007(n) + A057427(n) + A056326(n) + A056327(n) + A056328(n). (End)

Extensions

a(0)=1 prepended by Robert A. Russell, Nov 09 2018

A056325 Number of reversible string structures with n beads using a maximum of six different colors.

Original entry on oeis.org

1, 1, 2, 4, 11, 32, 117, 467, 2135, 10480, 55091, 301633, 1704115, 9819216, 57365191, 338134521, 2005134639, 11937364184, 71254895955, 426063226937, 2550552314219, 15280103807200, 91588104196415, 549159428968825
Offset: 0

Views

Author

Keywords

Comments

A string and its reverse are considered to be equivalent. Permuting the colors will not change the structure. Thus aabc, cbaa and bbac are all considered to be identical.
Number of set partitions of an unoriented row of n elements with six or fewer nonempty subsets. - Robert A. Russell, Oct 28 2018
There are nonrecursive formulas, generating functions, and computer programs for A056273 and A305752, which can be used in conjunction with the first formula. - Robert A. Russell, Oct 28 2018
From Allan Bickle, Jun 23 2022: (Start)
a(n) is the number of (unlabeled) 6-paths with n+8 vertices. (A 6-path with order n at least 8 can be constructed from a 6-clique by iteratively adding a new 6-leaf (vertex of degree 6) adjacent to an existing 6-clique containing an existing 6-leaf.)
Recurrences appear in the papers by Bickle, Eckhoff, and Markenzon et al. (End)

Examples

			For a(4)=11, the 7 achiral patterns are AAAA, AABB, ABAB, ABBA, ABCA, ABBC, and ABCD.  The 4 chiral pairs are AAAB-ABBB, AABA-ABAA, AABC-ABCC, and ABAC-ABCB.
		

References

  • M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]

Crossrefs

Cf. A056308.
Column 6 of A320750.
Cf. A056273 (oriented), A320936 (chiral), A305752 (achiral).
The numbers of unlabeled k-paths for k = 2..7 are given in A005418, A001998, A056323, A056324, A056325, and A345207, respectively.
The sequences above converge to A103293(n+1).

Programs

  • Mathematica
    Ach[n_, k_] := Ach[n, k] = If[n<2, Boole[n==k && n>=0], k Ach[n-2,k] + Ach[n-2,k-1] + Ach[n-2,k-2]] (* A304972 *)
    k=6; Table[Sum[StirlingS2[n,j]+Ach[n,j],{j,0,k}]/2,{n,0,40}] (* Robert A. Russell, Oct 28 2018 *)
    LinearRecurrence[{16, -84, 84, 685, -2140, 180, 7200, -8244, -4176, 11664, -5184}, {1, 1, 2, 4, 11, 32, 117, 467, 2135, 10480, 55091, 301633}, 40] (* Robert A. Russell, Oct 28 2018 *)
  • PARI
    Vec((1 - 15*x + 70*x^2 - 28*x^3 - 654*x^4 + 1479*x^5 + 783*x^6 - 5481*x^7 + 3512*x^8 + 4640*x^9 - 5922*x^10 + 1530*x^11) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 4*x)*(1 - 6*x)*(1 - 2*x^2)*(1 - 3*x^2)*(1 - 6*x^2)) + O(x^30)) \\ Colin Barker, Apr 15 2020

Formula

Use de Bruijn's generalization of Polya's enumeration theorem as discussed in reference.
From Robert A. Russell, Oct 28 2018: (Start)
a(n) = (A056273(n) + A305752(n)) / 2.
a(n) = A056273(n) - A320936(n) = A320936(n) + A305752(n).
a(n) = Sum_{j=0..k} (S2(n,j) + Ach(n,j)) / 2, where k=6 is the maximum number of colors, S2 is the Stirling subset number A008277, and Ach(n,k) = [n>=0 & n<2 & n==k] + [n>1]*(k*Ach(n-2,k) + Ach(n-2,k-1) + Ach(n-2,k-2)).
a(n) = A000007(n) + A057427(n) + A056326(n) + A056327(n) + A056328(n) + A056329(n) + A056330(n).
(End)
From Colin Barker, Mar 24 2020: (Start)
G.f.: (1 - 15*x + 70*x^2 - 28*x^3 - 654*x^4 + 1479*x^5 + 783*x^6 - 5481*x^7 + 3512*x^8 + 4640*x^9 - 5922*x^10 + 1530*x^11) / ((1 - x)*(1 - 2*x)*(1 - 3*x)*(1 - 4*x)*(1 - 6*x)*(1 - 2*x^2)*(1 - 3*x^2)*(1 - 6*x^2)).
a(n) = 16*a(n-1) - 84*a(n-2) + 84*a(n-3) + 685*a(n-4) - 2140*a(n-5) + 180*a(n-6) + 7200*a(n-7) - 8244*a(n-8) - 4176*a(n-9) + 11664*a(n-10) - 5184*a(n-11) for n>11.
(End)
From Allan Bickle, Jun 23 2022: (Start)
a(n) = (1/1440)*6^n + (1/96)*4^n + (1/36)*3^n + (3/32)*2^n + (19/360)*6^(n/2) + (1/9)*3^(n/2) + (1/8)*2^(n/2) + 17/60 for n > 0 even;
a(n) = (1/1440)*6^n + (1/96)*4^n + (1/36)*3^n + (3/32)*2^n + (13/720)*6^((n+1)/2) + (1/18)*3^((n+1)/2) + (1/16)*2^((n+1)/2) + 17/60 for n > 0 odd. (End)

Extensions

Another term from Robert A. Russell, Oct 29 2018
a(0)=1 prepended by Robert A. Russell, Nov 09 2018

A291883 Number T(n,k) of symmetrically unique Dyck paths of semilength n and height k; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 5, 3, 1, 0, 1, 9, 11, 4, 1, 0, 1, 19, 31, 19, 5, 1, 0, 1, 35, 91, 69, 29, 6, 1, 0, 1, 71, 250, 252, 127, 41, 7, 1, 0, 1, 135, 690, 855, 540, 209, 55, 8, 1, 0, 1, 271, 1863, 2867, 2117, 1005, 319, 71, 9, 1, 0, 1, 527, 5017, 9339, 8063, 4411, 1705, 461, 89, 10, 1
Offset: 0

Views

Author

Alois P. Heinz, Sep 05 2017

Keywords

Examples

			: T(4,2) = 5:       /\      /\        /\/\    /\  /\    /\/\/\
:              /\/\/  \  /\/  \/\  /\/    \  /  \/  \  /      \
:
Triangle T(n,k) begins:
  1;
  0, 1;
  0, 1,   1;
  0, 1,   2,   1;
  0, 1,   5,   3,   1;
  0, 1,   9,  11,   4,   1;
  0, 1,  19,  31,  19,   5,   1;
  0, 1,  35,  91,  69,  29,   6,  1;
  0, 1,  71, 250, 252, 127,  41,  7, 1;
  0, 1, 135, 690, 855, 540, 209, 55, 8, 1;
  ...
		

Crossrefs

Main and first two lower diagonals give A000012, A001477, A028387(n-1) for n>0.
Row sums give A007123(n+1).
T(2n,n) give A291885.

Programs

  • Maple
    b:= proc(x, y, k) option remember; `if`(x=0, z^k, `if`(y0, b(x-1, y-1, k), 0))
        end:
    g:= proc(x, y, k) option remember; `if`(x=0, z^k, `if`(y>0,
          g(x-2, y-1, k), 0)+ g(x-2, y+1, max(y+1, k)))
        end:
    T:= n-> (p-> seq(coeff(p, z, i)/2, i=0..n))(b(2*n, 0$2)+g(2*n, 0$2)):
    seq(T(n), n=0..14);
  • Mathematica
    b[x_, y_, k_] := b[x, y, k] = If[x == 0, z^k, If[y < x - 1, b[x - 1, y + 1, Max[y + 1, k]], 0] + If[y > 0, b[x - 1, y - 1, k], 0]];
    g[x_, y_, k_] := g[x, y, k] = If[x == 0, z^k, If[y > 0, g[x - 2, y - 1, k], 0] + g[x - 2, y + 1, Max[y + 1, k]]];
    T[n_] := Function[p, Table[Coefficient[p, z, i]/2, {i, 0, n}]][b[2*n, 0, 0] + g[2*n, 0, 0]];
    Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Jun 03 2018, from Maple *)
  • Python
    from sympy.core.cache import cacheit
    from sympy import Poly, Symbol, flatten
    z=Symbol('z')
    @cacheit
    def b(x, y, k): return z**k if x==0 else (b(x - 1, y + 1, max(y + 1, k)) if y0 else 0)
    @cacheit
    def g(x, y, k): return z**k if x==0 else (g(x - 2, y - 1, k) if y>0 else 0) + g(x - 2, y + 1, max(y + 1, k))
    def T(n): return 1 if n==0 else [i//2 for i in Poly(b(2*n, 0, 0) + g(2*n, 0, 0)).all_coeffs()[::-1]]
    print(flatten(map(T, range(15)))) # Indranil Ghosh, Sep 06 2017

Formula

T(n,k) = (A080936(n,k) + A132890(n,k))/2.
Sum_{k=1..n} k * T(n,k) = A291886(n).

A092431 Numbers having in binary representation a leading 1 followed by n zeros and n-1 ones.

Original entry on oeis.org

2, 9, 35, 135, 527, 2079, 8255, 32895, 131327, 524799, 2098175, 8390655, 33558527, 134225919, 536887295, 2147516415, 8590000127, 34359869439, 137439215615, 549756338175, 2199024304127, 8796095119359, 35184376283135, 140737496743935, 562949970198527
Offset: 1

Views

Author

Reinhard Zumkeller, Mar 23 2004

Keywords

Comments

Smallest numbers having in binary representation n 0's and n 1's: a(n) = Min{m: A023416(m)=A000120(m)=n}.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{7, -14, 8}, {2, 9, 35}, 40] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)
    Table[FromDigits[Join[PadRight[{1},n,0],PadRight[{},n-2,1]],2],{n,2,30}]//Sort (* or *) Rest[CoefficientList[Series[x (-2+5x)/((x-1)(2x-1)(4x-1)),{x,0,30}],x]] (* Harvey P. Dale, Jul 30 2021 *)

Formula

a(n+1) = 2*a(n) + 4^n + 1.
a(n) = 2^(2*n-1) + 2^(n-1) - 1.
a(n) = A007582(n)-1 = A056326(2n+1) = A005367(n-1)/2 = A063376(n)/2-1 = A032125(n+1)/3-1 = A056309(2n+1)/2 = A028403(n+1)/4-1 = (A001576(n)-3)/2 = (A028400(n+1)-9)/8 = Sum_{k=2..n+1} A049775(k). - Ralf Stephan, Mar 24 2004
G.f.: x*(-2+5*x) / ( (x-1)*(2*x-1)*(4*x-1) ). - R. J. Mathar, Jun 01 2011
E.g.f.: exp(x)*(exp(3*x) + exp(x) - 2)/2. - Stefano Spezia, Sep 27 2023
Showing 1-8 of 8 results.