A060312 -id:A060312 - cp's OEIS frontend

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).

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

Gerald McGarvey, Feb 24 2005

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

			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

Cf. A034851, A005685, A005688, A068930, A102526, A102543, A192928.

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)

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 *)

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

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

A001224 If F(n) is the n-th Fibonacci number, then a(2n) = (F(2n+1) + F(n+2))/2 and a(2n+1) = (F(2n+2) + F(n+1))/2.

Original entry on oeis.org

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: 1

N. J. A. Sloane

Arises from a problem of finding the number of inequivalent ways to pack a 2 X n rectangle with dominoes. The official solution is given in A060312. The present sequence gives the correct answer provided n != 2, when it gives 2 instead of 1. To put it another way, the present sequence gives the number of tilings of a 2 x n rectangle with dominoes when left-to-right mirror images are not regarded as distinct. - N. J. A. Sloane, Mar 30 2015
Also the number of inequivalent ways to tile a 2 X n rectangle with a combination of squares with sides 1 or 2. - John Mason, Nov 30 2022
Slavik V. Jablan observes that this is also the number of generating rational knots and links. See reference.
Also the number of distinct binding configurations on an n-site one-dimensional linear lattice, where the molecules cannot touch each other. This number determines the order of recurrence for the partition function of binding to a two-dimensional n X m lattice.
From Petros Hadjicostas, Jan 08 2018: (Start)
Consider Christian G. Bower's theory of transforms given in a weblink below. For each positive integer k and each input sequence (b(n): n>=1) with g.f. B(x) = Sum_{n>=1} b(n)*x^n, let (a_k(n): n>=1) = BIK[k](b(n): n>=1). (We thus change some of the notation in Bower's weblink.) Here, BIK[k] is the "reversible, indistinct, unlabeled" transform for k boxes.
If BIK[k](x) = Sum_{n>=1} a_k(n)*x^n is the g.f. of the output sequence, then it can be proved that BIK[k](x) = (B(x)^k + B(x^2)^{k/2})/2, when k is even, and = B(x)*BIK[k-1](x), when k is odd. (We assume BIK[0](x) = 1.)
If (a(n): n>=1) = BIK(b(n): n>=1) with g.f. BIK(x) = 1 + Sum_{n>=1} a(n)*x^n, then BIK(x) = 1 + Sum_{k>=1} BIK[k](x). (The addition of the extra 1 in the g.f. seems arbitrary.) We then get BIK(x) = 1 + (1/2)*(B(x)/(1 - B(x)) + (B(x) + B(x^2))/(1 - B(x^2))).
For this sequence, the input sequence satisfies b(1) = b(2) = 1 and b(n) = 0 for n >= 3. Hence, B(x) = x + x^2 and BIK(x) = (1+x)*(1-x-x^3)/((1-x-x^2)*(1-x^2-x^4)), which equals Christian G. Bower's g.f. in the formula section below. (End)

			From _Petros Hadjicostas_, Jan 08 2018: (Start)
We give some examples to explain _Christian G. Bower's theory of transforms given in the weblink above. We have boxes of two sizes here: boxes that can hold one ball and boxes that can hold two balls. (This is because we want the BIK transform of x + x^2. See the comments above.) Two boxes of the same size are considered identical (indistinct and unlabeled). We place the boxes in a line that can be read in either direction. Here, a(n) = total number of ways of placing such boxes in such a line so that the total number of balls in the boxes is n.
When we have 4 balls in total inside the boxes, we have the following configurations of boxes in a line that can be read in either direction: 1111, 121, 211, and 22. (Note that 211 = 112.) Hence, a(4) = 4.
When n = 5, we have the following configurations of boxes: 11111, 2111, 1211, 221, and 212. Hence, a(5) = 5.
When n = 6, we have: 111111, 21111, 12111, 11211, 2211, 2121, 2112, 1221, and 222. Hence, a(6) = 9. (End)

S. Golomb, Polyominoes, Princeton Univ. Press 1994.
S. Jablan S. and R. Sazdanovic, LinKnot: Knot Theory by Computer, World Scientific Press, 2007.
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).

T. D. Noe, Table of n, a(n) for n = 1..500
Ali Reza Ashrafi, Jernej Azarija, Khadijeh Fathalikhani, Sandi Klavžar, and Marko Petkovšek, Vertex and edge orbits of Fibonacci and Lucas cubes, arXiv:1407.4962 [math.CO], 2014.
M. Assis, J. L. Jacobsen, I. Jensen, J.-M. Maillard, and B. M. McCoy, Integrability vs non-integrability: Hard hexagons and hard squares compared, arXiv preprint 1406.5566 [math-ph], 2014.
Christian G. Bower, Transforms (2).
Petros Hadjicostas, Generalized colored circular palindromic compositions, Moscow Journal of Combinatorics and Number Theory, 9(2) (2020), 173-186.
S. V. Jablan, Geometry of Links, XII Yugoslav Geometric Seminar (Novi Sad, 1998), Novi Sad J. Math. 29(3) (1999), 121-139. [Broken link]
S. V. Jablan, Geometry of Links, XII Yugoslav Geometric Seminar (Novi Sad, 1998), Novi Sad J. Math. 29(3) (1999), 121-139.
Y. Kong, General recurrence theory of ligand binding on a three-dimensional lattice, J. Chem. Phys. Vol. 111 (1999), 4790-4799. (Table I)
W. E. Patten (proposer) and S. W. Golomb (solver), Problem E1470, "Covering a 2Xn rectangle with dominoes", Amer. Math. Monthly, 69 (1962), 61-62.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
J. Riordan, Letter, Jul 13 1970.
N. J. A. Sloane, Annotated scan of Monthly problem E1470 with illustration of a(4)=4 (Page 1).
N. J. A. Sloane, Annotated scan of Monthly problem E1470 with illustration of a(4)=4 (Page 2).
Index entries for sequences related to dominoes
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,0,-1,-1).

Essentially the same as A060312, A068928 and A102526.

Cf. A000045.

[(1/2)*((Fibonacci(n+1))+Fibonacci(((n+3+(-1)^n) div 2))): n in [1..40]]; // Vincenzo Librandi, Nov 23 2014

# Maple code for A060312 and A001224 from N. J. A. Sloane, Mar 30 2015
with(combinat); F:=fibonacci;
f:=proc(n) option remember;
if n=2 then 1 # change this to 2 to get A001224
elif (n mod 2) = 0 then (F(n+1)+F(n/2+2))/2;
else (F(n+1)+F((n+1)/2))/2; fi; end;
[seq(f(n),n=1..50)];
A001224:=-(-1-z+2*z**2+z**3+z**4+z**5)/(z**4+z**2-1)/(z**2+z-1); # conjectured by Simon Plouffe in his 1992 dissertation
a:= n-> (Matrix([[5,4,2,2,1,1]]). Matrix(6, (i,j)-> if (i=j-1) then 1 elif j=1 then [1,2,-1,0,-1,-1][i] else 0 fi)^n)[1,6]: seq(a(n), n=1..38); # Alois P. Heinz, Aug 26 2008

a[n_?EvenQ] := (Fibonacci[n + 1] + Fibonacci[n/2 + 2])/2; a[n_?OddQ] := (Fibonacci[n + 1] + Fibonacci[(n + 1)/2])/2; Table[a[n], {n, 38}] (* Jean-François Alcover, Oct 06 2011, after formula *)
LinearRecurrence[{1, 2, -1, 0, -1, -1}, {1, 2, 2, 4, 5, 9}, 38] (* Jean-François Alcover, Sep 21 2017 *)

a(2n+1) = A051450(n+1) and a(2n) = A005207(n+1).
From Christian G. Bower, May 09 2000: (Start)
G.f.: (2-(x+x^2)^2)/(2*(1-x-x^2)) + (1+x+x^2)*(x^2+x^4)/(2*(1-x^2-x^4)).
"BIK" transform of x+x^2. (End)
If F(n) is the n-th Fibonacci number, then a(2n) = (F(2n+1) + F(n+2))/2 and a(2n+1) = (F(2n+2) + F(n+1))/2.
G.f.: (1+x)*(1-x-x^3)/((1-x-x^2)*(1-x^2-x^4)). (See the Comments section above.) - Petros Hadjicostas, Jan 08 2018
From Manfred Boergens, Aug 25 2025: (Start)
a(n) = (Fibonacci(n+1) + Fibonacci(floor((n+3+(-1)^n)/2)))/2.
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-5) - a(n-6). (End)

More terms from Christian G. Bower, May 09 2000
Typo in references corrected by Jernej Azarija, Oct 23 2013
Edited by N. J. A. Sloane, Mar 30 2015

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

Gerald McGarvey, Feb 24 2005

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

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

Johann Cigler, Some remarks on Rogers-Szegö polynomials and Losanitsch's triangle, arXiv:1711.03340 [math.CO], 2017.
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,0,-1,-1).

Cf. A034851.

Essentially the same as A001224, A060312 and A068928.

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

LinearRecurrence[{1, 2, -1, 0, -1, -1}, {1, 1, 2, 2, 4, 5}, 40] (* Jean-François Alcover, Nov 17 2017 *)

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

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

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

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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).

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A001224 If F(n) is the n-th Fibonacci number, then a(2n) = (F(2n+1) + F(n+2))/2 and a(2n+1) = (F(2n+2) + F(n+1))/2.

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

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A232622 Number of incongruent domino tilings of the 3 X (2n) board.

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A329279 Number of distinct tilings of a 2n X 2n square with 1 x n polyominoes.

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