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 1-5 of 5 results.

A034851 Rows of Losanitsch's triangle T(n, k), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 4, 2, 1, 1, 3, 6, 6, 3, 1, 1, 3, 9, 10, 9, 3, 1, 1, 4, 12, 19, 19, 12, 4, 1, 1, 4, 16, 28, 38, 28, 16, 4, 1, 1, 5, 20, 44, 66, 66, 44, 20, 5, 1, 1, 5, 25, 60, 110, 126, 110, 60, 25, 5, 1, 1, 6, 30, 85, 170, 236, 236, 170, 85, 30, 6, 1, 1, 6, 36, 110, 255
Offset: 0

Views

Author

Keywords

Comments

Sometimes erroneously called "Lossnitsch's triangle". But the author's name is Losanitsch (I have seen the original paper in Chem. Ber.). This is a German version of the Serbian name Lozanic. - N. J. A. Sloane, Jun 29 2008
For n >= 3, a(n-3,k) is the number of series-reduced (or homeomorphically irreducible) trees which become a path P(k+1) on k+1 nodes, k >= 0, when all leaves are omitted (see illustration). Proof by Pólya's enumeration theorem. - Wolfdieter Lang, Jun 08 2001
The number of ways to put beads of two colors in a line, but take symmetry into consideration, so that 011 and 110 are considered the same. - Yong Kong (ykong(AT)nus.edu.sg), Jan 04 2005
Alternating row sums are 1,0,1,0,2,0,4,0,8,0,16,0,... - Gerald McGarvey, Oct 20 2008
The triangle sums, see A180662 for their definitions, link Losanitsch's triangle A034851 with several sequences, see the crossrefs. We observe that the Ze3 and Ze4 sums link Losanitsch's triangle with A005683, i.e., R. K. Guy's Twopins game. - Johannes W. Meijer, Jul 14 2011
T(n-(L-1)k, k) is the number of ways to cover an n-length line by exactly k L-length segments excluding symmetric covers. For L=2 it is corresponds to A102541, for L=3 to A228570 and for L=4 to A228572. - Philipp O. Tsvetkov, Nov 08 2013
Also the number of equivalence classes of ways of placing k 1 X 1 tiles in an n X 1 rectangle under all symmetry operations of the rectangle. - Christopher Hunt Gribble, Feb 16 2014
T(n, k) is the number of non-isomorphic outer planar graphs of order n+3, size n+3+k, and maximum degree k+2. - Christian Barrientos, Oct 18 2018
From Álvar Ibeas, Jun 01 2020: (Start)
T(n, k) is the sum of even-degree coefficients of the Gaussian polynomial [n, k]_q. The area below a NE lattice path between (0,0) and (k, n-k) is even for T(n, k) paths and odd for A034852(n, k) of them.
For a (non-reversible) string of k black and n-k white beads, consider the minimum number of bead transpositions needed to place the black ones to the left and the white ones to the right (in other words, the number of inversions of the permutation obtained by labeling the black beads by integers 1,...,k and the white ones by k+1,...,n, in the same order they take on the string). It is even for T(n, k) strings and odd for A034852(n, k) cases.
(End)
Named after the Serbian chemist, politician and diplomat Simeon Milivoje "Sima" Lozanić (1847-1935). - Amiram Eldar, Jun 10 2021
T(n, k) is the number of caterpillars with a perfect matching, with 2n+2 vertices and diameter 2n-1-k. - Christian Barrientos, Sep 12 2023

Examples

			Triangle begins
  1;
  1,  1;
  1,  1,  1;
  1,  2,  2,  1;
  1,  2,  4,  2,  1;
  1,  3,  6,  6,  3,  1;
  1,  3,  9, 10,  9,  3,  1;
  1,  4, 12, 19, 19, 12,  4,  1;
  1,  4, 16, 28, 38, 28, 16,  4,  1;
  1,  5, 20, 44, 66, 66, 44, 20,  5,  1;
		

Crossrefs

Triangle sums (see the comments): A005418 (Row), A011782 (Related to Row2), A102526 (Related to Kn11, Kn12, Kn13, Kn21, Kn22, Kn23), A005207 (Kn3, Kn4), A005418 (Fi1, Fi2), A102543 (Ca1, Ca2), A192928 (Gi1, Gi2), A005683 (Ze3, Ze4).
Sums of squares of terms in rows equal A211208.

Programs

  • Haskell
    a034851 n k = a034851_row n !! k
    a034851_row 0 = [1]
    a034851_row 1 = [1,1]
    a034851_row n = zipWith (-) (zipWith (+) ([0] ++ losa) (losa ++ [0]))
                                ([0] ++ a204293_row (n-2) ++ [0])
       where losa = a034851_row (n-1)
    a034851_tabl = map a034851_row [0..]
    -- Reinhard Zumkeller, Jan 14 2012
  • Maple
    A034851 := proc(n,k) option remember; local t; if k = 0 or k = n then return(1) fi; if n mod 2 = 0 and k mod 2 = 1 then t := binomial(n/2-1,(k-1)/2) else t := 0; fi; A034851(n-1,k-1)+A034851(n-1,k)-t; end: seq(seq(A034851(n, k), k=0..n), n=0..11);
  • Mathematica
    t[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2; t[n_, k_] := (Binomial[n, k] + Binomial[Quotient[n, 2], Quotient[k, 2]])/2; Flatten[Table[t[n, k], {n, 0, 12}, {k, 0, n}]](* Jean-François Alcover, Feb 07 2012, after PARI *)
  • PARI
    {T(n, k) = (1/2) *(binomial(n, k) + binomial(n%2, k%2) * binomial(n\2, k\2))}; /* Michael Somos, Oct 20 1999 */
    

Formula

T(n, k) = (1/2) * (A007318(n, k) + A051159(n, k)).
G.f. for k-th column (if formatted as lower triangular matrix a(n, k)): x^k*Pe(floor((k+1)/2), x^2)/(((1-x)^(k+1))*(1+x)^(floor((k+1)/2))), where Pe(n, x^2) := Sum_{m=0..floor(n/2)} A034839(n, m)*x^(2*m) (row polynomials of Pascal array even numbered columns). - Wolfdieter Lang, May 08 2001
a(n, k) = a(n-1, k-1) + a(n-1, k) - C(n/2-1, (k-1)/2), where the last term is present only if n is even and k is odd (see Sloane link).
T(n, k) = T(n-2, k-2) + T(n-2, k) + C(n-2, k-1), n > 1.
Let P(n, x, y) = Sum_{m=0..n} a(n, m)*x^m*y^(n-m), then for x > 0, y > 0 we have P(n, x, y) = (x+y)*P(n-1, x, y) for n odd and P(n, x, y) = (x+y)*P(n-1, x, y) - x*y*(x^2+y^2)^((n-2)/2) for n even. - Gerald McGarvey, Feb 15 2005
T(n, k) = T(n-1, k-1) + T(n-1, k) - A204293(n-2, k-1), 0 < k <= n and n > 1. - Reinhard Zumkeller, Jan 14 2012
From Christopher Hunt Gribble, Feb 25 2014: (Start)
It appears that:
T(n,k) = C(n,k)/2, n even, k odd;
T(n,k) = (C(n,k) + C(n/2,k/2))/2, n even, k even;
T(n,k) = (C(n,k) + C((n-1)/2,(k-1)/2))/2, n odd, k odd;
T(n,k) = (C(n,k) + C((n-1)/2,k/2))/2, n odd, k even.
(End)

Extensions

More terms from James Sellers, May 04 2000
Name edited by Johannes W. Meijer, Aug 26 2013

A059260 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 1, 2, 4, 3, 1, 0, 3, 6, 7, 4, 1, 1, 3, 9, 13, 11, 5, 1, 0, 4, 12, 22, 24, 16, 6, 1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1
Offset: 0

Views

Author

N. J. A. Sloane, Jan 23 2001

Keywords

Comments

Coefficients of the (left, normalized) shifted cyclotomic polynomial. Or, coefficients of the basic n-th q-series for q=-2. Indeed, let Y_n(x) = Sum_{k=0..n} x^k, having as roots all the n-th roots of unity except for 0; then coefficients in x of (-1)^n Y_n(-x-1) give exactly the n-th row of A059260 and a practical way to compute it. - Olivier Gérard, Jul 30 2002
The maximum in the (2n)-th row is T(n,n), which is A026641; also T(n,n) ~ (2/3)*binomial(2n,n). The maximum in the (2n-1)-th row is T(n-1,n), which is A014300 (but T does not have the same definition as in A026637); also T(n-1,n) ~ (1/3)*binomial(2n,n). Here is a generalization of the formula given in A026641: T(i,j) = Sum_{k=0..j} binomial(i+k-x,j-k)*binomial(j-k+x,k) for all x real (the proof is easy by induction on i+j using T(i,j) = T(i-1,j) + T(i,j-1)). - Claude Morin, May 21 2002
The second greatest term in the (2n)-th row is T(n-1,n+1), which is A014301; the second greatest term in the (2n+1)-th row is T(n+1,n) = 2*T(n-1,n+1), which is 2*A014301. - Claude Morin
Diagonal sums give A008346. - Paul Barry, Sep 23 2004
Riordan array (1/(1-x^2), x/(1-x)). As a product of Riordan arrays, factors into the product of (1/(1+x),x) and (1/(1-x),1/(1-x)) (binomial matrix). - Paul Barry, Oct 25 2004
Signed version is A239473 with relations to partial sums of sequences. - Tom Copeland, Mar 24 2014
From Robert Coquereaux, Oct 01 2014: (Start)
Columns of the triangle (cf. Example below) give alternate partial sums along nw-se diagonals of the Pascal triangle, i.e., sequences A000035, A004526, A002620 (or A087811), A002623 (or A173196), A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808, etc.
The dimension of the space of closed currents (distributional forms) of degree p on Gr(n), the Grassmann algebra with n generators, equivalently, the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence, is V(n,p) = 2^n T(p,n-1) - (-1)^p.
If p is odd V(n,p) is also the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n).
If p is even the dimension of this cohomology group is V(n,p)+1.
Cf. A193844. (End)
From Peter Bala, Feb 07 2024: (Start)
The following remarks assume the row indexing starts at n = 1.
The sequence of row polynomials R(n,x), beginning R(1,x) = 1, R(2,x) = x, R(3,x) = 1 + x + x^2 , ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd( R(n,x), R(m,x)) = R(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the polynomial sequence {R(n,x): n >= 1} is a divisibility sequence; that is, if n divides m then R(n,x) divides R(m,x) in Z[x]. (End)
From Miquel A. Fiol, Oct 04 2024: (Start)
For j>=1, T(i,j) is the independence number of the (i-j)-supertoken graph FF_(i-j)(S_j) of the star graph S_j with j points.
(Given a graph G on n vertices and an integer k>=1, the k-supertoken (or reduced k-th power) FF_k(G) of G has vertices representing configurations of k indistinguishable tokens in the (not necessarily different) vertices of G, with two configurations being adjacent if one can be obtained from the other by moving one token along an edge. See an example below.)
Following the suggestion of Peter Munn, the k-supertoken graph FF_k(S_j) can also be defined as follows: Consider the Lattice graph L(k,j), whose vertices are the k^j j-vectors with elements in the set {0,..,k-1}, two being adjacent if they differ in just one coordinate by one unity. Then, FF_k(S_j) is the subgraph of L(k+1,j) induced by the vertices at distance at most k from (0,..,0). (End)

Examples

			Triangle begins
  1;
  0,  1;
  1,  1,  1;
  0,  2,  2,  1;
  1,  2,  4,  3,  1;
  0,  3,  6,  7,  4,  1;
  1,  3,  9, 13, 11,  5,  1;
  0,  4, 12, 22, 24, 16,  6,  1;
  1,  4, 16, 34, 46, 40, 22,  7,  1;
  0,  5, 20, 50, 80, 86, 62, 29,  8,  1;
Sequences obtained with _Miquel A. Fiol_'s Sep 30 2024 formula of A(n,c1,c2) for other values of (c1,c2). (In the table, rows are indexed by c1=0..6 and columns by c2=0..6):
A000007  A000012  A000027  A025747  A000292* A000332* A000389*
A059841  A008619  A087811* A002623  A001752  A001753  A001769
A193356  A008794* A005993  A005994  -------  -------  -------
-------  -------  -------  A005995  A018210  -------  A052267
-------  -------  -------  -------  A018211  A018212  -------
-------  -------  -------  -------  -------  A018213  A018214
-------  -------  -------  -------  -------  -------  A062136
*requires offset adjustment.
The 2-supertoken FF_2(S_3) of the star graph S_3 with central vertex 1 and peripheral vertices 2,3,4. (The vertex `ij' of FF_2(S_3) represents the configuration of one token in `ì' and the other token in `j'). The T(5,3)=7 independent vertices are 22, 24, 44, 23, 11, 34, and 33.
     22--12---24---14---44
          | \    / |
         23   11   34
            \  |  /
              13
               |
              33
		

Crossrefs

Cf. A059259. Row sums give A001045.
Seen as a square array read by antidiagonals this is the coefficient of x^k in expansion of 1/((1-x^2)*(1-x)^n) with rows A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808 etc. (allowing for signs). A058393 would then effectively provide the table for nonpositive n. - Henry Bottomley, Jun 25 2001

Programs

  • Maple
    read transforms; 1/(1-y-x*y-x^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);
  • Mathematica
    t[n_, k_] := Sum[ (-1)^(n-j)*Binomial[j, k], {j, 0, n}]; Flatten[ Table[t[n, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Oct 20 2011, after Paul Barry *)
  • PARI
    T(n, k) = sum(j=0, n, (-1)^(n - j)*binomial(j, k));
    for(n=0, 12, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 11 2017
    
  • Python
    from sympy import binomial
    def T(n, k): return sum((-1)**(n - j)*binomial(j, k) for j in range(n + 1))
    for n in range(13): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017
  • Sage
    def A059260_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            if k==0: return 0
            return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
        return [(-1)^(n-k+1)*prec(n+1, n-k+1) for k in (1..n)]
    for n in (1..9): print(A059260_row(n)) # Peter Luschny, Mar 16 2016
    

Formula

G.f.: 1/(1-y-x*y-x^2) = 1 + y + x^2 + xy + y^2 + 2x^2y + 2xy^2 + y^3 + ...
E.g.f: (exp(-t)+(x+1)*exp((x+1)*t))/(x+2). - Tom Copeland, Mar 19 2014
O.g.f. (n-th row): ((-1)^n+(x+1)^(n+1))/(x+2). - Tom Copeland, Mar 19 2014
T(i, 0) = 1 if i is even or 0 if i is odd, T(0, i) = 1 and otherwise T(i, j) = T(i-1, j) + T(i, j-1); also T(i, j) = Sum_{m=j..i+j} (-1)^(i+j+m)*binomial(m, j). - Robert FERREOL, May 17 2002
T(i, j) ~ (i+j)/(2*i+j)*binomial(i+j, j); more precisely, abs(T(i, j)/binomial(i+j, j) - (i+j)/(2*i+j) )<=1/(4*(i+j)-2); the proof is by induction on i+j using the formula 2*T(i, j) = binomial(i+j, j)+T(i, j-1). - Claude Morin, May 21 2002
T(n, k) = Sum_{j=0..n} (-1)^(n-j)binomial(j, k). - Paul Barry, Aug 25 2004
T(n, k) = Sum_{j=0..n-k} binomial(n-j, j)*binomial(j, n-k-j). - Paul Barry, Jul 25 2005
Equals A097807 * A007318. - Gary W. Adamson, Feb 21 2007
Equals A128173 * A007318 as infinite lower triangular matrices. - Gary W. Adamson, Feb 17 2007
Equals A130595*A097805*A007318 = (inverse Pascal matrix)*(padded Pascal matrix)*(Pascal matrix) = A130595*A200139. Inverse is A097808 = A130595*(padded A130595)*A007318. - Tom Copeland, Nov 14 2016
T(i, j) = binomial(i+j, j)-T(i-1, j). - Laszlo Major, Apr 11 2017
Recurrence for row polynomials (with row indexing starting at n = 1): R(n,x) = x*R(n-1,x) + (x + 1)*R(n-2,x) with R(1,x) = 1 and R(2,x) = x. - Peter Bala, Feb 07 2024
From Miquel A. Fiol, Sep 30 2024: (Start)
The triangle can be seen as a slice of a 3-dimensional table that links it to well-known sequences as follows.
The j-th column of the triangle, T(i,j) for i >= j, equals A(n,c1,c2) = Sum_{k=0..floor(n/2)} binomial(c1+2*k-1,2*k)*binomial(c2+n-2*k-1,n-2*k) when c1=1, c2=j, and n=i-j.
This gives T(i,j) = Sum_{k=0..floor((i-j)/2)} binomial(i-2*k-1, j-1). For other values of (c1,c2), see the example below. (End)

Extensions

Formula corrected by Philippe Deléham, Jan 11 2014

A005995 Alkane (or paraffin) numbers l(8,n).

Original entry on oeis.org

1, 3, 12, 28, 66, 126, 236, 396, 651, 1001, 1512, 2184, 3108, 4284, 5832, 7752, 10197, 13167, 16852, 21252, 26598, 32890, 40404, 49140, 59423, 71253, 85008, 100688, 118728, 139128, 162384, 188496, 218025, 250971, 287964, 329004, 374794, 425334, 481404, 543004
Offset: 0

Views

Author

N. J. A. Sloane, Winston C. Yang (yang(AT)math.wisc.edu)

Keywords

Comments

From M. F. Hasler, May 01 2009: (Start)
Also, number of 5-element subsets of {1,...,n+5} whose elements sum to an odd integer, i.e. column 5 of A159916.
A linear recurrent sequence with constant coefficients and characteristic polynomial x^9 - 3*x^8 + 8*x^6 - 6*x^5 - 6*x^4 + 8*x^3 - 3*x + 1. (End)
Equals (1/2)*((1, 6, 21, 56, 126, 252, ...) + (1, 0, 3, 0, 6, 0, 10, ...)), see A000389 and A000217.
Equals row sums of triangle A160770.
F(1,5,n) is the number of bracelets with 1 blue, 5 identical red and n identical black beads. If F(1,5,1) = 3 and F(1,5,2) = 12 taken as a base, F(1,5,n) = n(n+1)(n+2)(n+3)/24 + F(1,3,n) + F(1,5,n-2). [F(1,3,n) is the number of bracelets with 1 blue, 3 identical red and n identical black beads. If F(1,3,1) = 2 and F(1,3,2) = 6 taken as a base F(1,3,n) = n(n+1)/2 + [|n/2|] + 1 + F(1,3,n-2)], where [|x|]: if a is an integer and a<=x

References

  • S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Winston C. Yang (paper in preparation).

Crossrefs

Cf. A160770, A053132 (bisection), A271870 (bisection), A018210 (partial sums).

Programs

  • Maple
    a:= n-> (Matrix([[1, 0$6, -3, -9]]). Matrix(9, (i,j)-> if (i=j-1) then 1 elif j=1 then [3, 0, -8, 6, 6, -8, 0, 3, -1][i] else 0 fi)^n)[1, 1]: seq(a(n), n=0..40); # Alois P. Heinz, Jul 31 2008
  • Mathematica
    a[n_?OddQ] := 1/240*(n+1)*(n+2)*(n+3)*(n+4)*(n+5); a[n_?EvenQ] := 1/240*(n+2)*(n+4)*(n+6)*(n^2+3*n+5); Table[a[n], {n, 0, 32}] (* Jean-François Alcover, Mar 17 2014, after M. F. Hasler *)
    LinearRecurrence[{3, 0, -8, 6, 6, -8, 0, 3, -1},{1, 3, 12, 28, 66, 126, 236, 396, 651},40] (* Ray Chandler, Sep 23 2015 *)
  • PARI
    a(k)= if(k%2,(k+1)*(k+3)*(k+5),(k+6)*(k^2+3*k+5))*(k+2)*(k+4)/240 \\ M. F. Hasler, May 01 2009

Formula

G.f.: (1+3*x^2)/((1-x)^3*(1-x^2)^3).
a(2n-1) = n(n+1)(n+2)(2n+1)(2n+3)/30, a(2n) = (n+1)(n+2)(n+3)(4n^2+6n+5)/ 30. [M. F. Hasler, May 01 2009]
a(n) = (1/240)*(n+1)*(n+2)*(n+3)*(n+4)*(n+5) + (1/16)*(n+2)*(n+4)*(1/2)*(1+(-1)^n). [Yosu Yurramendi, Jun 20 2013]
a(n) = A038163(n)+3*A038163(n-2). - R. J. Mathar, Mar 29 2018

A032094 Number of reversible strings with n-1 beads of 2 colors. 7 beads are black. String is not palindromic.

Original entry on oeis.org

4, 16, 60, 160, 396, 848, 1716, 3200, 5720, 9696, 15912, 25152, 38760, 58080, 85272, 122496, 173052, 240240, 328900, 443872, 592020, 780208, 1017900, 1314560, 1682928, 2135744, 2689808, 3361920, 4173840, 5147328, 6310128, 7689984, 9321780, 11240400
Offset: 9

Keywords

Comments

If the offset is changed to 3, this is the 2nd Witt transform of A000292 [Moree]. - R. J. Mathar, Nov 08 2008
Also 7th column of A159916, i.e., number of 7-element subsets of {1,...,n-1} whose elements add up to an odd integer. - M. F. Hasler, May 02 2009
From Petros Hadjicostas, May 19 2018: (Start)
Let k be an integer >= 2. The g.f. of the BHK[k] transform of the sequence (c(n): n >= 1), with g.f. C(x) = Sum_{n>=1} c(n)*x^n, is A_k(x) = (C(x)^k - C(x^2)^(k/2))/2 if k is even, and A_k(x) = (C(x)/2)*(C(x)^{k-1} - C(x^2)^{(k-1)/2}) if k is odd. This follows easily from the formulae in C. G. Bower's web link below about transforms.
When k is even and c(n) = 1 for all n >= 1, we get C(x) = x/(1-x) and A_k(x) = (1/2)*((x/(1-x))^k - (x^2/(1-x^2))^{k/2}). If (a_k(n): n >= 1) is the output sequence (with g.f. A_k(x)), then it can be proved (using Taylor expansions) that a_k(n) = (1/2)*(binomial(n-1, n-k) - binomial((n/2)-1, (n-k)/2)) for even n >= k+1 and a_k(n) = (1/2)*binomial(n-1, n-k) for odd n >= k+1. (Clearly, a_k(1) = ... = a_k(k) = 0.)
In this sequence, k = 8, and (according to C. G. Bower) a(n) = a_{k=8}(n) is the number of reversible non-palindromic compositions of n with 8 positive parts. If n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7 + b_8 is such a composition of n (with b_i >= 1), then it is equivalent to the composition n = b_8 + b_7 + b_6 + b_5 + b_4 + b_3 + b_2 + b_1, and each equivalent class has two elements because here linear palindromes are not allowed as compositions of n.
The fact that we are finding the BHK[8] transform of 1, 1, 1, ... means that each part of each composition of n can have exactly one color (see Bower's link below about transforms).
In each such composition replace each b_i with one black (B) ball followed by b_i - 1 white (W) balls. Then drop the first black (B) ball. We then get a reversible non-palindromic string of length n-1 that has k-1 = 7 black balls and n-k = n-8 white balls. This process, applied to the equivalent compositions n = b_1 + b_2 + b_3 + b_4 + b_5 + b_6 + b_7 + b_8 = b_8 + b_7 + b_6 + b_5 + b_4 + b_3 + b_2 + b_1, gives two strings of length n-1 with 7 black balls and n-8 white balls that are mirror images of each other.
Hence, for n >= 2, a(n) = a_{k=8}(n) is also the number of reversible non-palindromic strings of length n-1 that have k-1 = 7 black balls and n-k = n-8 white balls. (Clearly, a(n) = a_{k=8}(n) > 0 only for n >= 9.)
(End)

Crossrefs

Cf. A005995, A018210, A032091, A032092, A032093, A159916. - M. F. Hasler, May 02 2009 and Petros Hadjicostas, May 19 2018

Programs

  • Mathematica
    f[n_] := Binomial[n - 1, n - 8]/2 - If[ OddQ@ n, 0, Binomial[(n/2) - 1, (n - 8)/2]/2]; Array[f, 36, 9] (* or *)
    CoefficientList[ Series[ 4x^9 (x^2 + 1)/((x - 1)^8 (x + 1)^4), {x, 0, 40}], x] (* or *)LinearRecurrence[{4, -2, -12, 17, 8, -28, 8, 17, -12, -2, 4, -1}, {4, 16, 60, 160, 396, 848, 1716, 3200, 5720, 9696, 15912, 25152}, 34] (* Robert G. Wilson v, May 20 2018 *)
  • PARI
    A032094(n)=(binomial(n--,7)-if(n%2,binomial(n\2,3)))\2 \\ M. F. Hasler, May 02 2009
    
  • PARI
    Vec(4*x^9*(1+x^2)/((1-x)^8*(1+x)^4) + O(x^100)) \\ Colin Barker, Mar 07 2015

Formula

"BHK[ 8 ]" (reversible, identity, unlabeled, 8 parts) transform of 1, 1, 1, 1, ...
From R. J. Mathar, Nov 08 2008: (Start)
G.f.: 4*x^9*(1+x^2)/((1-x)^8*(1+x)^4).
a(n) = 4*A031164(n-9). (End)
From Colin Barker, Mar 07 2015: (Start)
a(n) = (n^7-28*n^6+322*n^5-1960*n^4+6664*n^3-11872*n^2+8448*n)/10080 if n is even.
a(n) = (n^7-28*n^6+322*n^5-1960*n^4+6769*n^3-13132*n^2+13068*n-5040)/10080 if n is odd.
(End)
From Petros Hadjicostas, May 19 2018: (Start)
a(n) = (1/2)*(binomial(n-1, n-8) - binomial((n/2)-1, (n-8)/2)) if n is even.
a(n) = (1/2)*binomial(n-1, n-8) if n is odd.
G.f.: (1/2)*((x/(1-x))^8 - (x^2/(1-x^2))^4).
These formulae agree with the above formulae by R. J. Mathar and Colin Barker. Clearly, the first two formulae (those about a(n)) can be combined into the one given by M. F. Hasler below in the PROG section.
(End)

A141783 Number of bracelets (turn over necklaces) with n beads: 1 blue, 12 green, and r = n - 13 red.

Original entry on oeis.org

1, 7, 49, 231, 924, 3108, 9324, 25236, 63090, 147070, 323554, 676270, 1352540, 2600612, 4829708, 8692788, 15212379, 25949469, 43249115, 70562765, 112900424, 177412664, 274183208, 417232088, 625848132, 926250780, 1353751140
Offset: 13

Author

Washington Bomfim, Aug 17 2008

Keywords

Programs

  • Maple
    A141783:=n->(1/2)*(binomial(n - 1, 12) + binomial((n - 2 + (n mod 2))/2, 6)); seq(A141783(n), n=13..50); # Wesley Ivan Hurt, Jan 30 2014
  • Mathematica
    Table[(1/2) (Binomial[n - 1, 12] + Binomial[(n - 2 + Mod[n, 2])/2, 6]), {n, 13, 50}] (* Wesley Ivan Hurt, Jan 30 2014 *)

Formula

a(n) = 1/2*(binomial(n-1,12) + binomial((n-2+n mod 2)/2, 6)).
a(n) = (1/(2*12!))*(n+2)*(n+4)*(n+6)*(n+8)*(n+10)*(n+12)*((n+1)*(n+3)*(n+5)*(n+7)*(n+9)*(n+11) + 1*3*5*7*9*11) - (1/15)*(1/2^10)*(n^5+(65/2)*n^4+400*n^3+(4615/2)*n^2+6154*n+(11895/2))*(1/2)*(1-(-1)^n) [Yosu Yurramendi, Jun 24 2013]

Extensions

Revised by Washington Bomfim, Jul 24 2012
Showing 1-5 of 5 results.