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 1-10 of 46 results. Next

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

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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

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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

A152176 Triangle read by rows: T(n,k) is the number of k-block partitions of an n-set up to rotations and reflections.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 3, 5, 2, 1, 1, 7, 14, 11, 3, 1, 1, 8, 31, 33, 16, 3, 1, 1, 17, 82, 137, 85, 27, 4, 1, 1, 22, 202, 478, 434, 171, 37, 4, 1, 1, 43, 538, 1851, 2271, 1249, 338, 54, 5, 1, 1, 62, 1401, 6845, 11530, 8389, 3056, 590, 70, 5, 1, 1, 121, 3838, 26148
Offset: 1

Views

Author

Vladeta Jovovic, Nov 27 2008

Keywords

Comments

Number of bracelet structures of length n using exactly k different colored beads. Turning over will not create a new bracelet. Permuting the colors of the beads will not change the structure. - Andrew Howroyd, Apr 06 2017
The number of achiral structures (A) is given in A140735 (odd n) and A293181 (even n). The number of achiral structures plus twice the number of chiral pairs (A+2C) is given in A152175. These can be used to determine A+C by taking half their average, as is done in the Mathematica program. - Robert A. Russell, Feb 24 2018
T(n,k)=pi_k(C_n) which is the number of non-equivalent partitions of the cycle 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,  1,   1;
  1,  3,   2,    1;
  1,  3,   5,    2,    1;
  1,  7,  14,   11,    3,    1;
  1,  8,  31,   33,   16,    3,   1;
  1, 17,  82,  137,   85,   27,   4,  1;
  1, 22, 202,  478,  434,  171,  37,  4, 1;
  1, 43, 538, 1851, 2271, 1249, 338, 54, 5, 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 A056357, A056358, A056359, A056360, A056361.
Row sums are A084708.
Partial row sums include A000011, A056353, A056354, A056355, A056356.
Cf. A081720, A273891, A008277 (set partitions), A284949 (up to reflection), A152175 (up to rotation).

Programs

  • Mathematica
    Adn[d_, n_] := Adn[d, n] = Which[0==n, 1, 1==n, DivisorSum[d, x^# &],
      1==d, Sum[StirlingS2[n, k] x^k, {k, 0, n}],
      True, Expand[Adn[d, 1] Adn[d, n-1] + D[Adn[d, n - 1], x] x]];
    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}]]] (* achiral loops of length n, k colors *)
    Table[(CoefficientList[DivisorSum[n, EulerPhi[#] Adn[#, n/#] &]/(x n), x]
    + Table[Ach[n, k],{k,1,n}])/2, {n, 1, 20}] // Flatten (* Robert A. Russell, Feb 24 2018 *)
  • PARI
    \\ see A056391 for Polya enumeration functions
    T(n,k) = NonequivalentStructsExactly(DihedralPerms(n), k); \\ Andrew Howroyd, Oct 14 2017
    
  • PARI
    \\ Ach is A304972 and R is A152175 as square matrices.
    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}
    R(n)={Mat(Col([Vecrev(p/y, n) | p<-Vec(intformal(sum(m=1, n, eulerphi(m) * subst(serlaplace(-1 + exp(sumdiv(m, d, y^d*(exp(d*x + O(x*x^(n\m)))-1)/d))), x, x^m))/x))]))}
    T(n)={(R(n) + Ach(n))/2}
    { my(A=T(12)); for(n=1, #A, print(A[n, 1..n])) } \\ Andrew Howroyd, Sep 20 2019

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

A276543 Triangle read by rows: T(n,k) = number of primitive (period n) n-bead bracelet structures using exactly k different colored beads.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 3, 5, 2, 1, 0, 5, 13, 11, 3, 1, 0, 8, 31, 33, 16, 3, 1, 0, 14, 80, 136, 85, 27, 4, 1, 0, 21, 201, 478, 434, 171, 37, 4, 1, 0, 39, 533, 1849, 2270, 1249, 338, 54, 5, 1, 0, 62, 1401, 6845, 11530, 8389, 3056, 590, 70, 5, 1
Offset: 1

Views

Author

Andrew Howroyd, Apr 09 2017

Keywords

Comments

Turning over will not create a new bracelet. Permuting the colors of the beads will not change the structure.

Examples

			Triangle starts:
  1
  0  1
  0  1   1
  0  2   2    1
  0  3   5    2    1
  0  5  13   11    3    1
  0  8  31   33   16    3   1
  0 14  80  136   85   27   4  1
  0 21 201  478  434  171  37  4 1
  0 39 533 1849 2270 1249 338 54 5 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

Partial row sums include A000046, A056362, A056363, A056364, A056365.
Row sums are A276548.

Programs

  • PARI
    \\ Ach is A304972 and R is A152175 as square matrices.
    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}
    R(n)={Mat(Col([Vecrev(p/y, n) | p<-Vec(intformal(sum(m=1, n, eulerphi(m) * subst(serlaplace(-1 + exp(sumdiv(m, d, y^d*(exp(d*x + O(x*x^(n\m)))-1)/d))), x, x^m))/x))]))}
    T(n)={my(M=(R(n)+Ach(n))/2); Mat(vectorv(n,n,sumdiv(n, d, moebius(d)*M[n/d,])))}
    { my(A=T(12)); for(n=1, #A, print(A[n, 1..n])) } \\ Andrew Howroyd, Sep 20 2019

Formula

T(n, k) = Sum_{d|n} mu(n/d) * A152176(d, k).

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

A285012 Irregular triangle read by rows: T(n,k) is the number of periodic palindromic structures of length n using exactly k different symbols, 1 <= k <= n/2 + 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 6, 5, 1, 1, 7, 6, 1, 1, 13, 19, 7, 1, 1, 15, 25, 10, 1, 1, 25, 64, 43, 10, 1, 1, 31, 90, 65, 15, 1, 1, 50, 208, 220, 85, 13, 1, 1, 63, 301, 350, 140, 21, 1, 1, 99, 656, 1059, 618, 154, 17, 1, 1, 127, 966, 1701, 1050, 266, 28, 1
Offset: 1

Views

Author

Andrew Howroyd, Apr 07 2017

Keywords

Comments

For example, aaabbb is not a (finite) palindrome but it is a periodic palindrome. Permuting the symbols will not change the structure.
Equivalently, the number of necklaces, up to permutation of the symbols, which when turned over are unchanged. When comparing with the turned over necklace a rotation is allowed but a permutation of the symbols is not.

Examples

			Triangle starts:
1
1   1
1   1
1   3    1
1   3    1
1   6    5     1
1   7    6     1
1  13   19     7     1
1  15   25    10     1
1  25   64    43    10     1
1  31   90    65    15     1
1  50  208   220    85    13    1
1  63  301   350   140    21    1
1  99  656  1059   618   154   17   1
1 127  966  1701  1050   266   28   1
1 197 2035  4803  4065  1488  258  21  1
1 255 3025  7770  6951  2646  462  36  1
1 391 6250 21105 24915 12857 3222 410 26 1
1 511 9330 34105 42525 22827 5880 750 45 1
...
Example for n=6, k=2:
Periodic symmetry means results are either in the form abccba or abcdcb.
There are 3 binary words in the form abccba that start with 0 and contain a 1 which are 001100, 010010, 011110. Of these, 011110 is equivalent to 001100 after rotation.
There are 7 binary words in the form abcdcb that start with 0 and contain a 1 which are 000100, 001010, 001110, 010001, 010101, 011011, 011111. Of these, 011111 is equivalent to 000100, 010001 is equivalent to 001010 and 011011 is equivalent to 010010 from the first set.
There are therefore a total of 7 + 3 - 4 = 6 equivalence classes so T(6,2) = 6.
		

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 A056508, A056509, A056510, A056511, A056512.
Partial row sums include A056503, A056504, A056505, A056506, A056507.
Row sums are A285013.

Programs

  • PARI
    \\ Ach is A304972, Prim is A285037.
    Ach(n,k=n) = {my(M=matrix(n, k, n, k, n>=k)); for(n=3, n, for(k=2, k, M[n, k]=k*M[n-2, k] + M[n-2, k-1] + if(k>2, M[n-2, k-2]))); M}
    Prim(n,k=n\2+1) = {my(A=Ach(n\2+1,k), S=matrix(n\2+1, k, n, k, stirling(n,k,2))); Mat(vectorv(n, n, sumdiv(n, d, moebius(d)*(S[(n/d+1)\2, ] + S[n/d\2+1, ] + if((n-d)%2, A[(n/d+1)\2, ] + A[n/d\2+1, ]))/if(d%2, 2, 1) )))}
    T(n,k=n\2+1) = {my(A=Prim(n,k)); Mat(vectorv(n, n, sumdiv(n, d, A[d, ])))}
    { my(A=T(20)); for(n=1, matsize(A)[1], print(A[n,1..n\2+1])) } \\ Andrew Howroyd, Oct 02 2019
    
  • PARI
    \\ column sequence using above code.
    ColSeq(n, k=2) = { Vec(T(n,k)[,k]) } \\ Andrew Howroyd, Oct 02 2019

Formula

Column k is inverse Moebius transform of column k of A285037. - Andrew Howroyd, Oct 02 2019

A285037 Irregular triangle read by rows: T(n,k) is the number of primitive (period n) periodic palindromic structures using exactly k different symbols, 1 <= k <= n/2 + 1.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 2, 1, 0, 3, 1, 0, 4, 5, 1, 0, 7, 6, 1, 0, 10, 18, 7, 1, 0, 14, 25, 10, 1, 0, 21, 63, 43, 10, 1, 0, 31, 90, 65, 15, 1, 0, 42, 202, 219, 85, 13, 1, 0, 63, 301, 350, 140, 21, 1, 0, 91, 650, 1058, 618, 154, 17, 1, 0, 123, 965, 1701, 1050, 266, 28, 1
Offset: 1

Views

Author

Andrew Howroyd, Apr 08 2017

Keywords

Comments

Permuting the symbols will not change the structure.
Equivalently, the number of n-bead aperiodic necklaces (Lyndon words) with exactly k symbols, up to permutation of the symbols, which when turned over are unchanged. When comparing with the turned over necklace a rotation is allowed but a permutation of the symbols is not.

Examples

			Triangle starts:
1
0   1
0   1
0   2    1
0   3    1
0   4    5     1
0   7    6     1
0  10   18     7     1
0  14   25    10     1
0  21   63    43    10     1
0  31   90    65    15     1
0  42  202   219    85    13    1
0  63  301   350   140    21    1
0  91  650  1058   618   154   17   1
0 123  965  1701  1050   266   28   1
0 184 2016  4796  4064  1488  258  21  1
0 255 3025  7770  6951  2646  462  36  1
0 371 6220 21094 24914 12857 3222 410 26 1
0 511 9330 34105 42525 22827 5880 750 45 1
...
Example for n=6, k=2:
There are 6 inequivalent solutions to A285012(6,2) which are 001100, 010010, 000100, 001010, 001110, 010101. Of these, 010010 and 010101 have a period less than 6, so T(6,2) = 6-2 = 4.
		

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 1..6 are: A063524, A056518, A056519, A056521, A056522, A056523.
Partial row sums include A056513, A056514, A056515, A056516, A056517.
Row sums are A285042.

Programs

  • PARI
    \\ Ach is A304972
    Ach(n,k=n) = {my(M=matrix(n, k, n, k, n>=k)); for(n=3, n, for(k=2, k, M[n, k]=k*M[n-2, k] + M[n-2, k-1] + if(k>2, M[n-2, k-2]))); M}
    T(n,k=n\2+1) = {my(A=Ach(n\2+1,k), S=matrix(n\2+1, k, n, k, stirling(n,k,2))); Mat(vectorv(n, n, sumdiv(n, d, moebius(d)*(S[(n/d+1)\2, ] + S[n/d\2+1, ] + if((n-d)%2, A[(n/d+1)\2, ] + A[n/d\2+1, ]))/if(d%2, 2, 1) )))}
    { my(A=T(20)); for(n=1, matsize(A)[1], print(A[n,1..n\2+1])) } \\ Andrew Howroyd, Oct 01 2019
    
  • PARI
    \\ column sequence using above code.
    ColSeq(n, k=2) = { Vec(T(n,k)[,k]) } \\ Andrew Howroyd, Oct 01 2019

Formula

T(n, k) = Sum_{d | n} mu(n/d) * A285012(d, k).

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
Showing 1-10 of 46 results. Next