cp's OEIS Frontend

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

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A005418 Number of (n-1)-bead black-white reversible strings; also binary grids; also row sums of Losanitsch's triangle A034851; also number of caterpillar graphs on n+2 vertices.

Original entry on oeis.org

1, 2, 3, 6, 10, 20, 36, 72, 136, 272, 528, 1056, 2080, 4160, 8256, 16512, 32896, 65792, 131328, 262656, 524800, 1049600, 2098176, 4196352, 8390656, 16781312, 33558528, 67117056, 134225920, 268451840, 536887296, 1073774592, 2147516416, 4295032832
Offset: 1

Views

Author

Keywords

Comments

Equivalently, walks on triangle, visiting n+2 vertices, so length n+1, n "corners"; the symmetry group is S3, reversing a walk does not count as different. Walks are not self-avoiding. - Colin Mallows
Slavik V. Jablan observes that this is also the number of rational knots and links with n+2 crossings (cf. A018240). See reference. [Corrected by Andrey Zabolotskiy, Jun 18 2020]
Number of bit strings of length (n-1), not counting strings which are the end-for-end reversal or the 0-for-1 reversal of each other as different. - Carl Witty (cwitty(AT)newtonlabs.com), Oct 27 2001
The formula given in page 1095 of the Balasubramanian reference can be used to derive this sequence. - Parthasarathy Nambi, May 14 2007
Also number of compositions of n up to direction, where a composition is considered equivalent to its reversal, see example. - Franklin T. Adams-Watters, Oct 24 2009
Number of normally non-isomorphic realizations of the associahedron of type I starting with dimension 2 in Ceballos et al. - Tom Copeland, Oct 19 2011
Number of fibonacenes with n+2 hexagons. See the Balaban and the Dobrynin references. - Emeric Deutsch, Apr 21 2013
From the point of view of binary grids, it is a (1,n)-rectangular grid. A225826 to A225834 are the numbers of binary pattern classes in the (m,n)-rectangular grid, 1 < m < 11. - Yosu Yurramendi, May 19 2013
Number of n-vertex difference graphs (bipartite 2K_2-free graphs) [Peled & Sun, Thm. 9]. - Falk Hüffner, Jan 10 2016
The offset should be 0, since the first row of A034851 is row 0. The name would then be: "Number of n bead...". - Daniel Forgues, Jul 26 2018
a(n) is the number of non-isomorphic generalized rigid ladders with n cells. A generalized rigid ladder with n cells is a graph with vertex set is the union of {u_0, u_1, ..., u_n} and {v_0, v_1, ..., v_n}, and for every 0 <= i <= n-1, the edges are of the form {u_i,u_i+1}, {v_i, v_i+1}, {u_i,v_i} and either {u_i,v_i+1} or {u_i+1,v_i}. - Christian Barrientos, Jul 29 2018
Also number of non-isomorphic stairs with n+1 cells. A stair is a snake polyomino allowing only two directions for adjacent cells: east and north. - Christian Barrientos and Sarah Minion, Jul 29 2018
From Robert A. Russell, Oct 28 2018: (Start)
There are two different unoriented row colorings using two colors that give us very similar results here, a difference of one in the offset. In an unoriented row, chiral pairs are counted as one.
a(n) is the number of color patterns (set partitions) of an unoriented row of length n using two or fewer colors (subsets). Two color patterns are equivalent if the colors are permutable.
a(n+1) is the number of ways to color an unoriented row of length n using two noninterchangeable colors (one need not use both colors).
See the examples below of these two different colorings. (End)
Also arises from the enumeration of types of based polyhedra with exactly two triangular faces [Rademacher]. - N. J. A. Sloane, Apr 24 2020
a(n) is the number of (unlabeled) 2-paths with n+4 vertices. (A 2-path with order n at least 4 can be constructed from a 3-clique by iteratively adding a new 2-leaf (vertex of degree 2) adjacent to an existing 2-clique containing an existing 2-leaf.) - Allan Bickle, Apr 05 2022
a(n) is the number of caterpillars with a perfect matching and order 2n+2. - Christian Barrientos, Sep 12 2023
a(n) is also the number of distinct planar embeddings of the (n+2)-centipede graph (up to at least n=8 and likely for all larger n). - Eric W. Weisstein, May 21 2024
a(n) is also the number of distinct planar embeddings of the 2 X (n+2) grid graph i.e., the (n+2)-ladder graph. - Eric W. Weisstein, May 21 2024
Dimension of the homogeneous component of degree n of the free Jordan algebra on two generators (or, in this case, the free special Jordan algebra on two generators). It follows from (Shirshov 1956, Cohn 1959). - Vladimir Dotsenko, Mar 29 2025

Examples

			a(5) = 10 because there are 16 compositions of 5 (shown as <vectors>) but only 10 equivalence classes (shown as {sets}): {<5>}, {<4,1>,<1,4>}, {<3,2>,<2,3>}, {<3,1,1>,<1,1,3>}, {<1,3,1>},{<2,2,1>,<1,2,2>}, {<2,1,2>}, {<2,1,1,1>,<1,1,1,2>}, {<1,2,1,1>,<1,1,2,1>}, {<1,1,1,1,1>}. - _Geoffrey Critzer_, Nov 02 2012
G.f. = x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + 20*x^6 + 36*x^7 + 72*x^8 + ... - _Michael Somos_, Jun 24 2018
From _Robert A. Russell_, Oct 28 2018: (Start)
For a(5)=10, the 4 achiral patterns (set partitions) are AAAAA, AABAA, ABABA, and ABBBA. The 6 chiral pairs are AAAAB-ABBBB, AAABA-ABAAA, AAABB-AABBB, AABAB-ABABB, AABBA-ABBAA, and ABAAB-ABBAB. The colors are permutable.
For n=4 and a(n+1)=10, the 4 achiral colorings are AAAA, ABBA, BAAB, and BBBB. The 6 achiral pairs are AAAB-BAAA, AABA-ABAA, AABB-BBAA, ABAB-BABA, ABBB-BBBA, and BABB-BBAB. The colors are not permutable. (End)
		

References

  • K. Balasubramanian, "Combinatorial Enumeration of Chemical Isomers", Indian J. Chem., (1978) vol. 16B, pp. 1094-1096. See page 1095.
  • Wayne M. Dymacek, Steinhaus graphs. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 399--412, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561065 (81f:05120)
  • Jablan S. and Sazdanovic R., LinKnot: Knot Theory by Computer, World Scientific Press, 2007.
  • Joseph S. Madachy: Madachy's Mathematical Recreations. New York: Dover Publications, Inc., 1979, p. 46 (first publ. by Charles Scribner's Sons, New York, 1966, under the title: Mathematics on Vacation)
  • 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.]
  • C. A. Pickover, Keys to Infinity, Wiley 1995, p. 75.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column 2 of A320750 (set partitions).
Cf. A131577 (oriented), A122746(n-3) (chiral), A016116 (achiral), for set partitions with up to two subsets.
Column 2 of A277504, offset by one (colors not permutable).
Cf. A000079 (oriented), A122746(n-2) (chiral), and A060546 (achiral), for a(n+1).

Programs

  • Haskell
    a005418 n = sum $ a034851_row (n - 1) -- Reinhard Zumkeller, Jan 14 2012
    
  • Maple
    A005418 := n->2^(n-2)+2^(floor(n/2)-1): seq(A005418(n), n=1..34);
  • Mathematica
    LinearRecurrence[{2,2,-4}, {1,2,3}, 40] (* or *) Table[2^(n-2)+2^(Floor[n/2]-1), {n,40}] (* Harvey P. Dale, Jan 18 2012 *)
  • PARI
    A005418(n)= 2^(n-2) + 2^(n\2-1); \\ Joerg Arndt, Sep 16 2013
    
  • Python
    def A005418(n): return 1 if n == 1 else 2**((m:= n//2)-1)*(2**(n-m-1)+1) # Chai Wah Wu, Feb 03 2022

Formula

a(n) = 2^(n-2) + 2^(floor(n/2) - 1).
G.f.: -x*(-1 + 3*x^2) / ( (2*x - 1)*(2*x^2 - 1) ). - Simon Plouffe in his 1992 dissertation
G.f.: x*(1+2*x)*(1-3*x^2)/((1-4*x^2)*(1-2*x^2)), not reduced. - Wolfdieter Lang, May 08 2001
a(n) = 6*a(n - 2) - 8*a(n - 4). a(2*n) = A063376(n - 1) = 2*a(2*n - 1); a(2*n + 1) = A007582(n). - Henry Bottomley, Jul 14 2001
a(n+2) = 2*a(n+1) - A077957(n) with a(1) = 1, a(2) = 2. - Yosu Yurramendi, Oct 24 2008
a(n) = 2*a(n-1) + 2*a(n-2) - 4*a(n-3). - Jaume Oliver Lafont, Dec 05 2008
Union of A007582 and A161168. Union of A007582 and A063376. - Jaroslav Krizek, Aug 14 2009
G.f.: G(0); G(k) = 1 + 2*x/(1 - x*(1+2^(k+1))/(x*(1+2^(k+1)) + (1+2^k)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 12 2011
a(2*n) = 2*a(2*n-1) and a(2*n+1) = a(2*n) + 4^(n-1) with a(1) = 1. - Johannes W. Meijer, Aug 26 2013
From Robert A. Russell, Oct 28 2018: (Start)
a(n) = (A131577(n) + A016116(n)) / 2 = A131577(n) - A122746(n-3) = A122746(n-3) + A016116(n), for set partitions with up to two subsets.
a(n+1) = (A000079(n) + A060546(n)) / 2 = A000079(n) - A122746(n-2) = A122746(n-2) + A060546(n), for two colors that do not permute.
a(n) = Sum_{j=0..k} (S2(n,j) + Ach(n,j)) / 2, where k=2 is the maximum number of colors, S2(n,k) 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) = (k^n + k^ceiling(n/2)) / 2, where k=2 is number of colors we can use. (End)
E.g.f.: (cosh(2*x) + 2*cosh(sqrt(2)*x) + sinh(2*x) + sqrt(2)*sinh(sqrt(2)*x) - 3)/4. - Stefano Spezia, Jun 01 2022

A102541 Triangle read by rows, formed from antidiagonals of Losanitsch's triangle. T(n, k) = A034851(n-k, k), n >= 0 and 0 <= k <= floor(n/2).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 3, 4, 1, 1, 3, 6, 2, 1, 4, 9, 6, 1, 1, 4, 12, 10, 3, 1, 5, 16, 19, 9, 1, 1, 5, 20, 28, 19, 3, 1, 6, 25, 44, 38, 12, 1, 1, 6, 30, 60, 66, 28, 4, 1, 7, 36, 85, 110, 66, 16, 1, 1, 7, 42, 110, 170, 126, 44, 4, 1, 8, 49, 146, 255, 236, 110, 20, 1, 1, 8, 56
Offset: 0

Views

Author

Gerald McGarvey, Feb 24 2005

Keywords

Comments

Row sums A102526 are essentially the same as A001224, A060312 and A068928.
Moving the terms in each column of this triangle, see the example, upwards to row 0 gives Losanitsch's triangle A034851 as a square array. - Johannes W. Meijer, Aug 24 2013
The number of ways to cover n-length line by exactly k 2-length segments excluding symmetric covers. - Philipp O. Tsvetkov, Nov 08 2013
Also the number of equivalence classes of ways of placing k 2 X 2 tiles in an n X 2 rectangle under all symmetry operations of the rectangle. - Christopher Hunt Gribble, Feb 16 2014
T(n, k) is the number of irreducible caterpillars with n+3 edges and diameter k+2. - Christian Barrientos, Apr 05 2020

Examples

			The first few rows of triangle T(n, k) are:
n/k: 0, 1, 2, 3
0:   1
1:   1
2:   1, 1
3:   1, 1
4:   1, 2, 1
5:   1, 2, 2
6:   1, 3, 4, 1
7:   1, 3, 6, 2
		

Crossrefs

Programs

  • Maple
    From Johannes W. Meijer, Aug 24 2013: (Start)
    T := proc(n,k) option remember: if n <0 then return(0) fi: if k < 0 or k > floor(n/2) then return(0) fi: A034851(n-k, k) end: 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(T(n, k), k=0..floor(n/2)), n=0..16);  # End first program
    T := proc(n,k) option remember: if n < 0 then return(0) fi: if k < 0 or k > floor(n/2) then return(0) fi: if n=0 then return(1) fi: if type(n, even) or type(k, even) then procname(n-1, k) + procname(n-2, k-1) else procname(n-1, k) + procname(n-2, k-1) - binomial((n-3)/2-(k-1)/2, (n-3)/2-(k-1)) fi: end: seq(seq(T(n, k), k=0..floor(n/2)), n=0..16); # End second program (End)
  • Mathematica
    t[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2;
    t[n_, k_] := (Binomial[n, k] + Binomial[Quotient[n, 2], Quotient[k, 2]])/2;
    T[n_, k_] := t[n - k, k];
    Table[T[n, k], {n, 0, 16}, {k, 0, Quotient[n, 2]}] // Flatten (* Jean-François Alcover, Jul 21 2022 *)

Formula

T(n, k) = A034851(n-k, k), n >= 0 and 0 <= k <= floor(n/2).
T(n, k) = T(n-1, k) + T(n-2, k-1) - C((n-3)/2-(k-1)/2, (n-3)/2-(k-1)) except when n or k even then T(n, k) = T(n-1, k) + T(n-2, k-1) with T(0, 0) = 1, T(n, 0) = 0 for n<0 and T(n, k) = 0 for k < 0 and k > floor(n/2). - Johannes W. Meijer, Aug 24 2013

Extensions

Definition edited, incorrect formula deleted, keyword corrected and extended by Johannes W. Meijer, Aug 24 2013

A228570 Triangle read by rows, formed from antidiagonals of triangle A102541. T(n, k) = A034851(n-2*k, k), n>= 0 and 0 <= k <= floor(n/3).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 3, 4, 1, 4, 6, 1, 1, 4, 9, 2, 1, 5, 12, 6, 1, 5, 16, 10, 1, 1, 6, 20, 19, 3, 1, 6, 25, 28, 9, 1, 7, 30, 44, 19, 1, 1, 7, 36, 60, 38, 3, 1, 8, 42, 85, 66, 12, 1, 8, 49, 110, 110, 28, 1
Offset: 0

Views

Author

Johannes W. Meijer, Aug 26 2013

Keywords

Comments

The row sums of this triangle are A102543. The antidiagonal sums are given by A192928 and the backwards antidiagonal sums are given by A228571.
Moving the terms in each column of this triangle, see the example, upwards to row 0 gives Losanitsch’s triangle A034851 as a square array.
Also the number of equivalence classes of ways of placing k 3 X 3 tiles in an n X 3 rectangle under all symmetry operations of the rectangle. - Christopher Hunt Gribble, Feb 16 2014

Examples

			The first few rows of triangle T(n, k) are:
   n/k: 0,  1,  2,  3
   0:   1
   1:   1
   2:   1
   3:   1,  1
   4:   1,  1
   5:   1,  2
   6:   1,  2,  1
   7:   1,  3,  2
   8:   1,  3,  4
   9:   1,  4,  6,  1
  10:   1,  4,  9,  2
  11:   1,  5, 12,  6
		

Crossrefs

Programs

  • Maple
    T := proc(n,k) option remember: if n <0 then return(0) fi: if k < 0 or k > floor(n/3) then return(0) fi: A034851(n-2*k, k) end: 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(T(n, k), k=0..floor(n/3)), n=0..18);  # End first program
    T := proc(n,k) option remember: if n=0 and k=0 or n=1 and k=0 or n=2 and k=0 then return(1) fi: if k <0 or k > floor(n/3) then return(0) fi: if type(n, even) and type(k, odd) then procname(n-1, k) + procname(n-3, k-1) - binomial((n-4)/2-2*(k-1)/2, (k-1)/2) else procname(n-1, k) + procname(n-3, k-1) fi: end: seq(seq(T(n,k), k=0..floor(n/3)), n=0..18); # End second program
  • Mathematica
    T[n_, k_] := (Binomial[n - 2k, k] + Boole[EvenQ[k] || OddQ[n]] Binomial[(n - 2k - Mod[n, 2])/2, Quotient[k, 2]])/2; Table[T[n, k], {n, 0, 20}, {k, 0, Quotient[n, 3]}] // Flatten (* Jean-François Alcover, Oct 06 2017, after Andrew Howroyd *)
  • PARI
    T(n,k)={(binomial(n-2*k,k) + (k%2==0||n%2==1)*binomial((n-2*k-n%2)/2,k\2))/2}
    for(n=1,20,for(k=0,(n\3), print1(T(n,k), ", "));print) \\ Andrew Howroyd, May 30 2017

Formula

T(n, k) = A034851(n-2*k, k), n >= 0 and 0 <= k <= floor(n/3).
T(n, k) = T(n-1, k) + T(n-3, k-1) - C((n-4)/2 - 2*(k-1)/2, (k-1)/2) where the last term is present only if n even and k odd; T(0, 0) = 1, T(1, 0) = 1, T(2, 0) = 1, T(n, k) = 0 for n < 0 and T(n, k) = 0 for k < 0 and k > floor(n/3).

A228572 Triangle read by rows, formed from antidiagonals of triangle A228570: T(n,k) = A034851(n-3*k, k) for n >= 0 and 0 <= k <= floor(n/4).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 3, 1, 1, 3, 2, 1, 4, 4, 1, 4, 6, 1, 5, 9, 1, 1, 5, 12, 2, 1, 6, 16, 6, 1, 6, 20, 10, 1, 7, 25, 19, 1, 1, 7, 30, 28, 3, 1, 8, 36, 44, 9, 1, 8, 42, 60, 19, 1, 9, 49, 85, 38, 1, 1, 9, 56, 110, 66, 3
Offset: 0

Views

Author

Johannes W. Meijer, Aug 26 2013

Keywords

Comments

The row sums of this triangle are A192928.
Moving the terms in each column of this triangle, see the example, upwards to row 0 gives Losanitsch’s triangle A034851 as a square array. Observe A102541 and A228570 for the same phenomenom. The number of zeros in the columns for these three triangles are multiples of 2, 3 and 4 respectively.
Also the number of equivalence classes of ways of placing k 4 X 4 tiles in an n X 4 rectangle under all symmetry operations of the rectangle. - Christopher Hunt Gribble, Apr 24 2015

Examples

			The first few rows of triangle T(n, k) are:
   n/k: 0, 1, 2, 3
   0:   1
   1:   1
   2:   1
   3:   1
   4:   1, 1
   5:   1, 1
   6:   1, 2
   7:   1, 2
   8:   1, 3, 1
   9:   1, 3, 2
  10:   1, 4, 4
  11:   1, 4, 6
  12:   1, 5, 9, 1
		

Crossrefs

Programs

  • Maple
    T := proc(n, k) option remember: if n <0 then return(0) fi: if k < 0 or k > floor(n/4) then return(0) fi: A034851(n-3*k, k) end: 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(T(n, k), k=0..floor(n/4)), n=0..21); # End first program
    T := proc(n,k) option remember: if n=0 and k=0 or n=1 and k=0 or n=2 and k=0 or n=3 and k=0 then return(1) fi: if k <0 or k > floor(n/4) then return(0) fi: if type(n, odd) and type(k, odd) then procname(n-1,k) + procname(n-4, k-1) - binomial((n+3)/2 - 3*(k+1)/2 - 1,(k+1)/2-1) else procname(n-1, k) + procname(n-4, k-1)  fi: end: seq(seq(T(n, k), k=0..floor(n/4)), n=0..21); # End second program
  • Mathematica
    T[n_, k_] := (Binomial[n - 3k, k] + Boole[EvenQ[k] || EvenQ[n]]* Binomial[(n - 3k - Mod[k, 2] - Mod[n, 2])/2, Quotient[k, 2]])/2; Table[T[n, k], {n, 0, 20}, {k, 0, Quotient[n, 4]}] // Flatten (* Jean-François Alcover, Oct 06 2017, after Andrew Howroyd *)
  • PARI
    T(n,k)={(binomial(n-3*k,k) + (k%2==0||n%2==0)*binomial((n-3*k-k%2-n%2)/2,k\2))/2}
    for(n=1,20,for(k=0,(n\4), print1(T(n,k), ", "));print) \\ Andrew Howroyd, May 29 2017

Formula

T(n, k) = A034851(n-3*k, k) for n >= 0 and 0 <= k <= floor(n/4).
T(n, k) = T(n-1, k) + T(n-4, k-1) - C((n+3)/2 - 3*(k+1)/2-1, (k+1)/2-1), where the last term is present only if n odd and k odd; T(0, 0) = 1, T(1, 0) = 1, T(2, 0) = 1, T(3, 0) = 1, T(n, k) = 0 for n < 0 and T(n, k) = 0 for k < 0 and k > floor(n/4).

A062136 Twelfth column of Losanitsch's triangle A034851 (formatted as lower triangular matrix).

Original entry on oeis.org

1, 6, 42, 182, 693, 2184, 6216, 15912, 37854, 83980, 176484, 352716, 676270, 1248072, 2229096, 3863080, 6519591, 10737090, 17299646, 27313650, 42337659, 64512240, 96770544, 143048880, 208616044, 300402648, 427500360, 601661144, 838033836, 1155900720, 1579738736
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

Also seventh column (m=6) of triangle A062135.
Number of homeomorphically irreducible (or series-reduced) trees (no vertices of degree 2) with n+9 leaves which become tree P(7) (path on 7 nodes (vertices) or 6 edges (links) when all leaves are omitted. A leave is an edge together with a node of degree 1 at one end). Proof by Polya enumeration. See illustration for A034851.

Crossrefs

Cf. A018213.

Programs

  • Magma
    [(1/(2*Factorial(11)))*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6)*(n + 7)*(n + 8)*(n + 9)*(n + 10)*(n + 11) + (1/15)*(1/2^9)*(n + 2)*(n + 4)*(n + 6)*(n + 8)*(n + 10)*(1/2)*(1 + (-1)^n): n in [0..30]]; // G. C. Greubel, Nov 24 2017
  • Mathematica
    Table[(1/(2*11!))*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6)*(n + 7)*(n + 8)*(n + 9)*(n + 10)*(n + 11) + (1/15)*(1/2^9)*(n + 2)*(n + 4)*(n + 6)*(n + 8)*(n + 10)*(1/2)*(1 + (-1)^n), {n, 0, 50}] (* G. C. Greubel, Nov 24 2017 *)
  • PARI
    for(n=0,50, print1((1/(2*11!))*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6)*(n + 7)*(n + 8)*(n + 9)*(n + 10)*(n + 11) + (1/15)*(1/2^9)*(n + 2)*(n + 4)*(n + 6)*(n + 8)*(n + 10)*(1/2)*(1 + (-1)^n), ", ")) \\ G. C. Greubel, Nov 24 2017
    

Formula

G.f.: Pe(6, x^2)/((1-x)^(2*6)*(1+x)^6), with Pe(6, x^2) := Sum_{m=0..3} A034839(6, m)*x^(2*m) = 1+15*x^2+15*x^4+x^6.
a(n) = A034851(n+11,11).
a(2n+1) = A001288(2n+12)/2; a(2n) = (A001288(2n+11)+A000389(n+5))/2. - Gary W. Adamson, Dec 15 2010
a(n) = (1/(2*11!))*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(n+6)*(n+7)*(n+8)*(n+9)*(n+10)*(n+11) + (1/15)*(1/2^9)*(n+2)*(n+4)*(n+6)*(n+8)*(n+10)*(1/2)*(1+(-1)^n). - Yosu Yurramendi, Jun 24 2013

A102526 Antidiagonal sums of Losanitsch's triangle (A034851).

Original entry on oeis.org

1, 1, 2, 2, 4, 5, 9, 12, 21, 30, 51, 76, 127, 195, 322, 504, 826, 1309, 2135, 3410, 5545, 8900, 14445, 23256, 37701, 60813, 98514, 159094, 257608, 416325, 673933, 1089648, 1763581, 2852242, 4615823, 7466468, 12082291, 19546175, 31628466
Offset: 0

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Author

Gerald McGarvey, Feb 24 2005

Keywords

Comments

This is an interleaving of A005207 and A051450. Thus a(2*m) = A005207(m) = (F(2*m-1) + F(m+1)) / 2, a(2*m - 1) = A051450(m) = (F(2*m) + F(m)) / 2 where F() are Fibonacci numbers (A000045). - Max Alekseyev, Jun 28 2006
The Kn11(n) and Kn21(n) sums, see A180662 for their definitions, of Losanitsch's triangle A034851 equal a(n), while the Kn12(n) and Kn22(n) sums equal (a(n+2)-A000012(n)) and the Kn13(n) and Kn23(n) sums equal (a(n+4)-A008619(n+4)). - Johannes W. Meijer, Jul 14 2011
a(n) is the number of homeomorphically irreducible caterpillars with n + 3 edges. - Christian Barrientos, Sep 12 2020

References

  • Jablan S. and Sazdanovic R., LinKnot: Knot Theory by Computer, World Scientific Press, 2007.

Crossrefs

Cf. A034851.
Essentially the same as A001224, A060312 and A068928.

Programs

  • Maple
    with(combinat): A102526 :=proc(n): if type(n, even) then (fibonacci(n+1)+fibonacci(n/2+2))/2 else (fibonacci(n+1)+fibonacci((n+1)/2))/2 fi: end: seq(A102526(n), n=0..38); # Johannes W. Meijer, Jul 14 2011
  • Mathematica
    LinearRecurrence[{1, 2, -1, 0, -1, -1}, {1, 1, 2, 2, 4, 5}, 40] (* Jean-François Alcover, Nov 17 2017 *)
  • PARI
    Vec((1+x)*(1-x-x^3)/(x^2+x-1)/(x^4+x^2-1)+O(x^99)) \\ Charles R Greathouse IV, Nov 17 2017
    
  • PARI
    a(n)=([0,1,0,0,0,0; 0,0,1,0,0,0; 0,0,0,1,0,0; 0,0,0,0,1,0; 0,0,0,0,0,1; -1,-1,0,-1,2,1]^n*[1;1;2;2;4;5])[1,1] \\ Charles R Greathouse IV, Nov 17 2017

Formula

G.f.: -(1+x)*(x^3+x-1) / ( (x^2+x-1)*(x^4+x^2-1) ). - R. J. Mathar, Nov 09 2013
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-5) - a(n-6). - Wesley Ivan Hurt, Sep 17 2020

A192928 The Gi1 and Gi2 sums of Losanitsch's triangle A034851.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 3, 5, 6, 9, 11, 16, 20, 29, 37, 53, 69, 98, 130, 183, 245, 343, 463, 646, 877, 1220, 1664, 2310, 3161, 4381, 6009, 8319, 11430, 15811, 21751, 30070, 41405, 57216, 78836, 108906, 150130, 207346
Offset: 0

Views

Author

Johannes W. Meijer, Jul 14 2011

Keywords

Comments

The Gi1 and Gi2 sums, see A180662 for their definitions, of Losanitsch's triangle A034851 equal this sequence.
From Johannes W. Meijer, Aug 26 2013: (Start)
The a(n) are also the Ca1 and Ca2 sums of McGarvey’s triangle A102541.
Furthermore they are the Kn11 and Kn12 sums of triangle A228570.
And finally the terms of this sequence are the row sums of triangle A228572. (End)

Crossrefs

Programs

  • Maple
    A192928 := proc(n): (A003269(n+1)+x(n)+x(n-1)+x(n-4))/2 end: A003269 := proc(n): sum(binomial(n-1-3*j, j), j=0..(n-1)/3) end: x:=proc(n): if type(n,even) then A003269(n/2+1) else 0 fi: end: seq(A192928(n),n=0..42);
  • Mathematica
    LinearRecurrence[{1, 1, -1, 1, 0, -1, 0, 1, -1, 0, 0, -1}, {1, 1, 1, 1, 2, 2, 3, 3, 5, 6, 9, 11}, 43] (* Jean-François Alcover, Nov 16 2017 *)

Formula

G.f.: (-1/2)*(1/(x^4+x-1) + (1+x+x^4)/(x^8+x^2-1))= -(1+x)*(x^7-x^6+x^5+x-1) / ( (x^4+x-1)*(x^8+x^2-1) ).
a(n) = (A003269(n+1)+x(n)+x(n-1)+x(n-4))/2 with x(2*n) = A003269(n+1) and x(2*n+1) = 0.
From Johannes W. Meijer, Aug 26 2013: (Start)
a(n) = sum(A228572(n, k), k=0..n)
a(n) = sum(A228570(n-k, k), k=0..floor(n/2))
a(n) = sum(A102541(n-2*k, k), k=0..floor(n/3))
a(n) = sum(A034851(n-3*k, k), k=0..floor(n/4)) (End)

A005654 Number of bracelets (turn over necklaces) with n red, 1 pink and n-1 blue beads; also reversible strings with n red and n-1 blue beads; also next-to-central column in Losanitsch's triangle A034851.

Original entry on oeis.org

1, 2, 6, 19, 66, 236, 868, 3235, 12190, 46252, 176484, 676270, 2600612, 10030008, 38781096, 150273315, 583407990, 2268795980, 8836340260, 34461678394, 134564560988, 526024917288, 2058358034616, 8061901596814, 31602652961516, 123979635837176, 486734861612328
Offset: 1

Views

Author

Keywords

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A034851.

Programs

  • Magma
    [((Binomial(2*n-1, n)+Binomial(n-1, Floor(n/2)))/2): n in [1..30]]; // Vincenzo Librandi, May 24 2012
  • Maple
    A005654:=n->(1/2)*(binomial(2*n-1,n)+binomial(n-1,floor(n/2))): seq(A005654(n), n=1..40); # Wesley Ivan Hurt, Jan 29 2017
  • Mathematica
    Table[(Binomial[2n-1,n]+Binomial[n-1,Floor[n/2]])/2,{n,30}] (* Harvey P. Dale, May 17 2012 *)
  • PARI
    C(n,k)=binomial(n,k)
    a(n)=(1/2)*(C(2*n-1,n)+C(n-1,n\2))
    

Formula

a(n) = (1/2) * (binomial(2*n-1, n) + binomial(n-1, floor(n/2))). - Michael Somos
a(n) = A034851(2*n-1, n-1).
Conjecture: n*(n-2)*a(n) - (5*n-3)*(n-2)*a(n-1) + 4*(n-2)*a(n-2) + 4*(5*n^2-27*n+37)*a(n-3) - 8*(2*n-7)*(n-4)*a(n-4) = 0. - R. J. Mathar, Nov 09 2013

Extensions

Sequence extended and description corrected by Christian G. Bower

A055138 Matrix inverse of Losanitsch's triangle A034851.

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 1, 0, -2, 1, -1, 2, 0, -2, 1, -1, -3, 6, 0, -3, 1, 4, -3, -7, 8, 0, -3, 1, -1, 18, -18, -13, 17, 0, -4, 1, -19, -4, 56, -28, -22, 20, 0, -4, 1, 31, -127, 60, 136, -98, -34, 36, 0, -5, 1, 120, 163, -511, 80, 288, -126, -50, 40, 0, -5, 1
Offset: 0

Views

Author

Christian G. Bower, May 09 2000

Keywords

Comments

Also the adjoint matrix of Losanitsch's triangle, since the determinants of n X n subarrays, rooted at (0,0), of Losanitsch's triangle are 1 and since for a matrix A, A^(-1) = 1/det(A) * (adjoint of A). - Gerald McGarvey, Oct 30 2007

Examples

			Triangle begins:
   1;
  -1, 1;
   0,-1, 1;
   1, 0,-2, 1;
  -1, 2, 0,-2, 1;
  ...
		

Crossrefs

Cf. A034851.

Programs

  • Maple
    b:= proc(n, k) (binomial(irem(n, 2, 'i'), irem(k, 2, 'j'))*
                    binomial(i, j)+binomial(n, k))/2 end:
    T:= n-> (M-> seq(M[n+1, j], j=1..n+1))(Matrix(n+1,
                 (i, j)-> b(i-1, j-1))^(-1)):
    seq(T(n), n=0..10);  # Alois P. Heinz, Mar 01 2022
  • Mathematica
    nmax = 10;
    b[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2;
    b[n_, k_] := (Binomial[n, k] + Binomial[Quotient[n, 2], Quotient[k, 2]])/2;
    M = Table[b[n, k], {n, 0, nmax}, {k, 0, nmax}] // Inverse;
    T[n_, k_] := M[[n+1, k+1]];
    Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 13 2022 *)
  • PARI
    T(n, k) = (1/2)*(binomial(n, k) + binomial(n%2, k%2) * binomial(n\2, k\2)); \\ A034851
    row(n) = my(m=matrix(n+1, n+1, i, j, i--; j--; T(i, j))); vector(n+1, i, (1/m)[n+1,i]); \\ Michel Marcus, Mar 01 2022

Extensions

Typo in definition corrected by Georg Fischer, Mar 01 2022

A062135 Odd-numbered columns of Losanitsch triangle A034851 formatted as triangle with an additional first column.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 2, 6, 3, 1, 0, 3, 10, 12, 4, 1, 0, 3, 19, 28, 20, 5, 1, 0, 4, 28, 66, 60, 30, 6, 1, 0, 4, 44, 126, 170, 110, 42, 7, 1, 0, 5, 60, 236, 396, 365, 182, 56, 8, 1, 0, 5, 85, 396, 868, 1001, 693, 280, 72, 9, 1
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

Because the sequence of column m=2*k, k >= 1, of A034851 is the partial sum sequence of the one of column m=2*k-1 the present triangle is essentially Losanitsch's triangle A034851.
Row sums give A051450 with A051450(0) := 1. Column sequences (without leading zeros) are for m=0..6: A000007, A008619, A005993, A005995, A018211, A018213, A062136.

Examples

			Triangle begins:
  {1};
  {0,1};
  {0,1,1};
  {0,2,2,1};
  ...
Pe(4,x^2)=1+6*x^2+x^4.
		

Crossrefs

Programs

  • 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 - 1 + m, n - m], {n, 0, 12}, {m, 0, n}]] (* Michael De Vlieger, Sep 28 2024, after Jean-François Alcover at A034851  *)

Formula

T(n, m) = A034851(n-1+m, n-m), n >= m >= 0; A034851(n-1, n) := 0, n >= 1, A034851(-1, 0) := 1.
T(n, m) = 0 if n= 1; T(n, m) = T(n-1, m)+sum(T(k, m-1), k=m-1..n-1) if n+m even and T(n, m) = T(n-1, m)+sum(T(k, m-1), k=m-1..n-1)-binomial((n+m-3)/2, m-1) if n+m odd, n >= m >= 1.
G.f. for column m: x^m*Pe(m, x^2)/(((1-x)^(2*m))*(1+x)^m), m >= 0, with Pe(m, x^2)= sum(A034839(m, k)*x^(2*k), k=0..floor(n/2)), the row polynomial of array A034839 (even-indexed entries of the rows of Pascal's triangle).

Extensions

More terms from Michael De Vlieger, Sep 28 2024
Showing 1-10 of 67 results. Next