A102526 -id:A102526 - cp's OEIS frontend

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

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

N. J. A. Sloane

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

			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;

Reinhard Zumkeller, Rows n = 0..100 of triangle, flattened
F. Al-Kharousi, R. Kehinde, and A. Umar, Combinatorial results for certain semigroups of partial isometries of a finite chain, The Australasian Journal of Combinatorics, Vol. 58, No. 3 (2014), pp. 363-375.
Tewodros Amdeberhan, Mahir Bilen Can and Victor H. Moll, Broken bracelets, Molien series, paraffin wax and the elliptic curve of conductor 48, SIAM Journal of Discrete Math., Vol. 25, No. 4 (2011), p. 1843-1859; arXiv preprint, arXiv:1106.4693 [math.CO], 2011. See Theorem 2.8.
Johann Cigler, Some remarks on Rogers-Szegö polynomials and Losanitsch's triangle, arXiv:1711.03340 [math.CO], 2017.
Sahir Gill, Bounds for Region Containing All Zeros of a Complex Polynomial, International Journal of Mathematical Analysis Vol. 12, No. 7 (2018), pp. 325-333.
Stephen G. Hartke and A. J. Radcliffe, Signatures of Strings, Annals of Combinatorics, Vol. 17, No. 1 (March, 2013), pp. 131-150.
Rethinasamy K. Kittappa, Combinatorial enumeration of rectangular kolam designs of the Tamil land, Abstracts Amer. Math. Soc., Vol. 29, No. 1 (2008), p. 24 (Abstract 1035-05-543).
Wolfdieter Lang, Illustration of initial rows of triangle.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. Vol. 30 (1897), pp. 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926. (Annotated scanned copy)
Ministry of Foreign Affairs of Serbia, List of the Ministers for Foreign Affairs Since the Forming of the First Government in 1811-Sima Lozanic.
Jesse Pajwani, Herman Rohrbach, and Anna M. Viergever, Compactly supported A^1-Euler characteristics of symmetric powers of cellular varieties, arXiv:2404.08486 [math.AG], 2024. See p. 15.
N. J. A. Sloane, Classic Sequences.
Eric Weisstein's World of Mathematics, Losanitsch's Triangle.
Wikipedia, Sima Lozanic.
Index entries for sequences related to trees

Cf. A007318, A034852, A051159, A055138, A102541, A228570, A228572.
Columns: A008619, A087811, A005993 - A005995, A018210 - A018214, A062136, A141783.
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.

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

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);

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 *)

{T(n, k) = (1/2) *(binomial(n, k) + binomial(n%2, k%2) * binomial(n\2, k\2))}; /* Michael Somos, Oct 20 1999 */

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)

More terms from James Sellers, May 04 2000

Name edited by Johannes W. Meijer, Aug 26 2013

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

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

A060312 Number of distinct ways to tile a 2 X n rectangle with dominoes (solutions are identified if they are rotations or reflections of each other).

Original entry on oeis.org

1, 1, 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

Thomas Ward, Mar 27 2001

Same as A001224 except that there a(2)=2 not 1. - N. J. A. Sloane, Mar 30 2015

			a(3)=2 because of the configurations |= and |||.

G. C. Greubel, Table of n, a(n) for n = 1..1001
A. R. Ashrafi, J. Azarija, K. Fathalikhani, S. Klavzar, et al., Orbits of Fibonacci and Lucas cubes, dihedral transformations, and asymmetric strings, 2014.
R. J. Mathar, Paving rectangular regions with rectangular tiles, ..., arXiv:1311.6135 [math.CO], Table 9.
W. E. Patten (proposer) and S. W. Golomb (solver), Problem E1470, "Covering a 2Xn rectangle with dominoes", Amer. Math. Monthly, 69 (1962), 61-62.
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 A001224, which is the main entry for this sequence. Other versions of the sequence can be found in A068928 and A102526.

[n eq 1 select 1 else (1/2)*(Fibonacci(n+2)+Fibonacci(Floor((n-(-1)^n)/2)+2)): n in [0..40]]; // Vincenzo Librandi, Nov 22 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)];

CoefficientList[Series[-(x^7 + x^6 + x^5 + 2 x^4 - x^3 + x^2 - 1) / ((x^2 + x - 1) (x^4 + x^2 - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 22 2014 *)

If F(n) is the n-th Fibonacci number, then a(2n) = (F(2n+1) + F(n+2))/2 for n > 1 and a(2n-1) = (F(2n) + F(n))/2. [Corrected by Manfred Boergens, Aug 25 2025]
a(n) = (F(n+1)+F(floor((n+3+(-1)^n)/2)))/2 for n!=2. - Manfred Boergens, Aug 25 2025
G.f.: -x*(x^7 + x^6 + x^5 + 2*x^4 - x^3 + x^2 - 1) / ((x^2 + x - 1)*(x^4 + x^2 - 1)). - Colin Barker, Dec 15 2012

Edited by N. J. A. Sloane, Mar 30 2015

cp's OEIS Frontend

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

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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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A060312 Number of distinct ways to tile a 2 X n rectangle with dominoes (solutions are identified if they are rotations or reflections of each other).

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A101366 Perfect Abs: Imaginary part of complex z such that Abs[(Total[Divisors[z]]-z)]=Abs[z].

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