A102541 -id:A102541 - cp's OEIS frontend

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).

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

Johannes W. Meijer, Aug 26 2013

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

			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

Andrew Howroyd, Table of n, a(n) for n = 0..1000

Cf. A034851, A102541, A228572, A102543, A192928, A228571, A005691.

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

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 *)

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

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).

A011973 Irregular triangle read by rows: T(n,k) = binomial(n-k, k), n >= 0, 0 <= k <= floor(n/2); or, coefficients of (one version of) Fibonacci polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 4, 3, 1, 5, 6, 1, 1, 6, 10, 4, 1, 7, 15, 10, 1, 1, 8, 21, 20, 5, 1, 9, 28, 35, 15, 1, 1, 10, 36, 56, 35, 6, 1, 11, 45, 84, 70, 21, 1, 1, 12, 55, 120, 126, 56, 7, 1, 13, 66, 165, 210, 126, 28, 1, 1, 14, 78, 220, 330, 252, 84, 8, 1, 15, 91, 286, 495, 462

Offset: 0

N. J. A. Sloane

T(n,k) is the number of subsets of {1,2,...,n-1} of size k and containing no consecutive integers. Example: T(6,2)=6 because the subsets of size 2 of {1,2,3,4,5} with no consecutive integers are {1,3},{1,4},{1,5},{2,4},{2,5} and {3,5}. Equivalently, T(n,k) is the number of k-matchings of the path graph P_n. - Emeric Deutsch, Dec 10 2003
T(n,k) = number of compositions of n+2 into k+1 parts, all >= 2. Example: T(6,2)=6 because we have (2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3) and (3,3,2). - Emeric Deutsch, Apr 09 2005
Given any recurrence sequence S(k) = x*a(k-1) + a(k-2), starting (1, x, x^2+1, ...); the (k+1)-th term of the series = f(x) in the k-th degree polynomial: (1, (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 + 1), ...). Example: let x = 2, then S(k) = 1, 2, 5, 12, 29, 70, 169, ... such that A000129(7) = 169 = f(x), x^6 + 5x^4 + 6x^2 + 1 = (64 + 80 + 24 + 1). - Gary W. Adamson, Apr 16 2008
Row k gives the nonzero coefficients of U(k,x/2) where U is the Chebyshev polynomial of the second kind. For example, row 6 is 1,5,6,1 and U(6,x/2) = x^6 - 5x^4 + 6x^2 - 1. - David Callan, Jul 22 2008
T(n,k) is the number of nodes at level k in the Fibonacci tree f(k-1). The Fibonacci trees f(k) of order k are defined as follows: 1. f(-1) and f(0) each consist of a single node. 2. For k >= 1, to the root of f(k-1), taken as the root of f(k), we attach with a rightmost edge the tree f(k-2). See the Iyer and Reddy references. These trees are not the same as the Fibonacci trees in A180566. Example: T(3,0)=1 and T(3,1)=2 because in f(2) = /\ we have 1 node at level 0 and 2 nodes at level 1. - Emeric Deutsch, Jun 21 2011
Triangle, with zeros omitted, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
Riordan array (1/(1-x),x^2/(1-x). - Philippe Deléham, Dec 12 2011
This sequence is the elements on the rising diagonals of the Pascal triangle, where the sum of the elements in each rising diagonal represents a Fibonacci number. - Mohammad K. Azarian, Mar 08 2012
If we set F(0;x) = 0, F(1;x) = 1, F(n+1;x) = x*F(n;x) + F(n-1;x), then we obtain the sequence of Vieta-Fibonacci polynomials discussed by Gary W. Adamson above. We note that F(n;x) = (-i)^n * U(n;i*x/2), where U denotes the respective Chebyshev polynomial of the second kind (see David Callan's remark above). Let us fix a,b,f(0),f(1) in C, b is not the zero, and set f(n) = a*f(n-1) + b*f(n-2). Then we deduce the relation: f(n) = b^((n-1)/2) * F(n;a/sqrt(b))*f(1) + b^(n/2) * F(n-1;a/sqrt(b))*f(0), where for a given value of the complex root sqrt(b) we set b^(n/2) = (sqrt(b))^n. Moreover, if b=1 then we get f(n+k) + (-1)^k * f(n-k) = L(k;a)*f(n), for every k=0,1,...,n, and where L(0;a)=2, L(1;a)=a, L(n+1;a)=a*L(n;a) + L(n-1;a) are the Vieta-Lucas polynomials. Let us observe that L(n+2;a) = F(n+2;a) + F(n;a), L(m+n;a) = L(m;a)*F(n;a) + L(m-1;a)*F(n-1;a), which implies also L(n+1;a) = a*F(n;a) + 2*F(n-1;a). Further we have L(n;a) = 2*(-i)^n * T(n;i*x/2), where T(n;x) denotes the n-th Chebyshev polynomial of the first kind. For the proofs, other relations and facts - see Witula-Slota's papers. - Roman Witula, Oct 12 2012
The diagonal sums of this triangle are A000930. - John Molokach, Jul 04 2013
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19). - Tom Copeland, Oct 11 2014
For a mirrored, shifted version showing the relation of these coefficients to the Pascal triangle, Fibonacci, and other number triangles, see A030528. See also A053122 for a relation to Cartan matrices. - Tom Copeland, Nov 04 2014
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al. - Tom Copeland, Nov 29 2014
A reversed, signed and aerated version is given by A049310, related to Chebyshev polynomials. - Tom Copeland, Dec 06 2015
For n >= 3, the n-th row gives the coefficients of the independence polynomial of the (n-2)-path graph P_{n-2}. - Eric W. Weisstein, Apr 07 2017
For n >= 2, the n-th row gives the coefficients of the matching-generating polynomial of the (n-1)-path graph P_{n-1}. - Eric W. Weisstein, Apr 10 2017
Antidiagonals of the Pascal matrix A007318 read bottom to top. These are also the antidiagonals read from top to bottom of the numerical coefficients of the Maurer-Cartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper), which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse is A102426. - Tom Copeland, Jul 02 2018
T(n,k) is the number of Markov equivalence classes with skeleton the path on n+1 nodes having exactly k immoralities. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
T(n, k) = number of compositions of n+1 into n+1-2*k odd parts. For example, T(6,2) = 6 because 7 = 5+1+1 = 3+3+1 = 3+1+3 = 1+1+5 = 1+3+3 = 1+1+5. - Michael Somos, Sep 19 2019
From Gary W. Adamson, Apr 25 2022: (Start)
Alternate rows can be parsed into those with odd integer coefficients to the right of the leftmost 1, and those with even integer coefficients to the right of the leftmost 1. The first set is shown in A054142 and are characteristic polynomials of submatrices of an infinite tridiagonal matrix (A332602) with all -1's in the super and subdiagonals and (1,2,2,2,...) as the main diagonal. For example, the characteristic equation of the 3 X 3 submatrix (1,-1,0; -1,2,-1; 0,-1,2) is x^3 - 5x^2 + 6x - 1. The roots are the Beraha constants B(7,1) = 3.24697...; B(7,2) = 1.55495...; and B(7,3) = 0.198062.... For n X n matrices of this form, the largest eigenvalue is B(2n+1, 1). The 3 X 3 matrix has an eigenvalue of 3.24697... = B(7,1).
Polynomials with even integer coefficients to the right of the leftmost 1 are in A053123 with roots being the even-indexed Beraha constants. The generating Cartan matrices are those with (2,2,2,...) as the main diagonal and -1's as the sub- and superdiagonals. The largest eigenvalue of n X n matrices of this form are B(2n+2,1). For example, the largest eigenvalue of (2,-1,0; -1,2,-1; 0,-1,2) is 3.414... = B(8,1) = a root to x^3 - 6x^2 + 10x - 4. (End)
T(n,k) is the number of edge covers of P_(n+2) with (n-k) edges. For example, T(6,2)=6 because among edges 1, 2, ..., 7 of P_8, we can eliminate any two non-consecutive edges among 2-6. These numbers can be found using the recurrence relation for the edge cover polynomial of P_n, which is E(P_n,x) = xE(P_(n-1),x)+xE(P_(n-2),x) and E(P_1,x)=0, E(P_2,x)=x (ref. Akbari and Oboudi). - Feryal Alayont, Jun 03 2022
T(n,k) is the number of ways to tile an n-board (an n X 1 array of 1 X 1 cells) using k dominoes and n-2*k squares. - Michael A. Allen, Dec 28 2022
T(n,k) is the number of positive integer sequences (s(1),s(2),...,s(n-2k)) such that s(i) < s(i+1), s(1) is odd, s(n-2k) <= n, and s(i) and s(i+1) have opposite parity (ref. Donnelly, Dunkum, and McCoy). Example: T(6,0)=1 corresponds to 123456; T(6,1)=5 corresponds to 1234, 1236, 1256, 1456, 3456; T(6,2)=6 corresponds to 12, 14, 16, 34, 36; and T(6,3)=1 corresponds to the empty sequence () with length 0. - Molly W. Dunkum, Jun 27 2023

			The first few Fibonacci polynomials (defined here by F(0,x) = 0, F(1,x) = 1; F(n+1, x) = F(n, x) + x*F(n-1, x)) are:
0: 0
1: 1
2: 1
3: 1 + x
4: 1 + 2*x
5: 1 + 3*x + x^2
6: (1 + x)*(1 + 3*x)
7: 1 + 5*x + 6*x^2 + x^3
8: (1 + 2*x)*(1 + 4*x + 2*x^2)
9: (1 + x)*(1 + 6*x + 9*x^2 + x^3)
10: (1 + 3*x + x^2 )*(1 + 5*x + 5*x^2)
11: 1 + 9*x + 28*x^2 + 35*x^3 + 15*x^4 + x^5
From _Roger L. Bagula_, Feb 20 2009: (Start)
  1
  1
  1   1
  1   2
  1   3   1
  1   4   3
  1   5   6   1
  1   6  10   4
  1   7  15  10   1
  1   8  21  20   5
  1   9  28  35  15   1
  1  10  36  56  35   6
  1  11  45  84  70  21   1
  1  12  55 120 126  56   7 (End)
For n=9 and k=4, T(9,4) = C(5,4) = 5 since there are exactly five size-4 subsets of {1,2,...,8} that contain no consecutive integers, namely, {1,3,5,7}, {1,3,5,8}, {1,3,6,8}, {1,4,6,8}, and {2,4,6,8}. - _Dennis P. Walsh_, Mar 31 2011
When the rows of the triangle are displayed as centered text, the falling diagonal sums are A005314. The first few terms are row1 = 1 = 1; row2 = 1+1 = 2; row3 = 2+1 = 3; row4 = 1+3+1 = 5; row5 = 1+3+4+1 = 9; row6 = 4+6+5+1 = 16; row7 = 1+10+10+6+1 = 28; row8 = 1+5+20+15+7+1 = 49; row9 = 6+15+35+21+8+1 = 86; row10 = 1+21+35+56+28+9+1 = 151. - _John Molokach_, Jul 08 2013
In the example, you can see that the n-th row of Pascal's triangle is given by T(n, 0), T(n+1, 1), ..., T(2n-1, n-1), T(2n, n). - _Daniel Forgues_, Jul 07 2018

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Index entries for triangles and arrays related to Pascal's triangle

Row sums = A000045(n+1) (Fibonacci numbers). - Michael Somos, Apr 02 1999
Cf. A000129, A054123, A052553, A000930, A030528, A053122.
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways.
Cf. A007318, A025581, A055087.
Cf. A000108, A049310, A055151, A098474, A124644, A134264.
Cf. A054142, A053123, A332602.

a011973 n k = a011973_tabf !! n !! k
a011973_row n = a011973_tabf !! n
a011973_tabf = zipWith (zipWith a007318) a025581_tabl a055087_tabf
-- Reinhard Zumkeller, Jul 14 2015

a := proc(n) local k; [ seq(binomial(n-k,k),k=0..floor(n/2)) ]; end;
T := proc(n, k): if k<0 or k>floor(n/2) then return(0) fi: binomial(n-k, k) end: seq(seq(T(n,k), k=0..floor(n/2)), n=0..15); # Johannes W. Meijer, Aug 26 2013

(* first: sum method *) Table[CoefficientList[Sum[Binomial[n - m + 1, m]*x^m, {m, 0, Floor[(n + 1)/2]}], x], {n, 0, 12}] (* Roger L. Bagula, Feb 20 2009 *)
(* second: polynomial recursion method *) Clear[L, p, x, n, m]; L[x, 0] = 1; L[x, 1] = 1 + x; L[x_, n_] := L[x, n - 1] + x*L[x, n - 2]; Table[ExpandAll[L[x, n]], {n, 0, 10}]; Table[CoefficientList[ExpandAll[L[x, n]], x], {n, 0, 12}]; Flatten[%] (* Roger L. Bagula, Feb 20 2009 *)
(* Center option shows falling diagonals are A224838 *) Column[Table[Binomial[n - m, m], {n, 0, 25}, {m, 0, Floor[n/2]}], Center] (* John Molokach, Jul 26 2013 *)
Table[ Select[ CoefficientList[ Fibonacci[n, x], x], Positive] // Reverse, {n, 1, 18} ] // Flatten (* Jean-François Alcover, Oct 21 2013 *)
CoefficientList[LinearRecurrence[{1, x}, {1 + x, 1 + 2 x}, {-1, 10}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
CoefficientList[Table[x^((n - 1)/2) Fibonacci[n, 1/Sqrt[x]], {n, 15}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)

{T(n, k) = if( k<0 || 2*k>n, 0, binomial(n-k, k))};

# Prints the table; cf. A145574.
for n in (2..20): [Compositions(n, length=m, min_part=2).cardinality() for m in (1..n//2)]  # Peter Luschny, Oct 18 2012

Let F(n, x) be the n-th Fibonacci polynomial in x; the g.f. for F(n, x) is Sum_{n>=0} F(n, x)*y^n = (1 + x*y)/(1 - y - x*y^2). - Paul D. Hanna
T(m, n) = 0 for n != 0 and m <= 1 T(0, 0) = T(1, 0) = 1 T(m, n) = T(m - 1, n) + T(m-2, n-1) for m >= 2 (i.e., like the recurrence for Pascal's triangle A007318, but going up one row as well as left one column for the second summand). E.g., T(7, 2) = 10 = T(6, 2) + T(5, 1) = 6 + 4. - Rob Arthan, Sep 22 2003
G.f. for k-th column: x^(2*k-1)/(1-x)^(k+1).
Identities for the Fibonacci polynomials F(n, x):
F(m+n+1, x) = F(m+1, x)*F(n+1, x) + x*F(m, x)F(n, x).
F(n, x)^2-F(n-1, x)*F(n+1, x) = (-x)^(n-1).
The degree of F(n, x) is floor((n-1)/2) and F(2p, x) = F(p, x) times a polynomial of equal degree which is 1 mod p.
From Roger L. Bagula, Feb 20 2009: (Start)
p(x,n) = Sum_{m=0..floor((n+1)/2)} binomial(n-m+1, m)*x^m;
p(x,n) = p(x, n - 1) + x*p(x, n - 2). (End)
T(n, k) = A102541(2*n+2, 2*k+1) + A102541(2*n+1, 2*k) - A102541(2*n+3, 2*k+1), n >= 0 and 0 <= k <= floor(n/2). - Johannes W. Meijer, Aug 26 2013
G.f.: 1/(1-x-y*x^2) = R(0)/2, where R(k) = 1 + 1/(1 - (2*k+1+ x*y)*x/((2*k+2+ x*y)*x + 1/R(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
O.g.f. G(x,t) = x/(1-x-tx^2) = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... has the inverse Ginv(x,t) = -[1+x-sqrt[(1+x)^2 + 4tx^2]]/(2tx) = x - x^2 + (1-t) x^3 + (-1+3t) x^4 + ..., an o.g.f. for the signed Motzkin polynomials of A055151, consistent with A134264 with h_0 = 1, h_1 = -1, h_2 = -t, and h_n = 0 otherwise. - Tom Copeland, Jan 21 2016
O.g.f. H(x,t) = x (1+tx)/ [1-x(1+tx)] = x + (1+t) x^2 + (1+2t) x^3 + ... = -L[Cinv(-tx)/t], where L(x) = x/(1+x) with inverse Linv(x) = x/(1-x) and Cinv(x) = x (1-x) is the inverse of C(x) = (1-sqrt(1-4x))/2, the o.g.f. of the shifted Catalan numbers A000108. Then Hinv(x,t) = -C[t Linv(-x)]/t = [-1 + sqrt(1+4tx/(1+x))]/2t = x - (1+t) x^2 + (1+2t+2t^2) x^3 - (1+3t+6t^2+5t^3) x^4 + ..., which is signed A098474, reverse of A124644. - Tom Copeland, Jan 25 2016
T(n, k) = GegenbauerC(k, (n+1)/2-k, 1). - Peter Luschny, May 10 2016

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

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

A024718 a(n) = (1/2)(1 + Sum_{k=0..n} binomial(2k, k)).

Original entry on oeis.org

1, 2, 5, 15, 50, 176, 638, 2354, 8789, 33099, 125477, 478193, 1830271, 7030571, 27088871, 104647631, 405187826, 1571990936, 6109558586, 23782190486, 92705454896, 361834392116, 1413883873976, 5530599237776, 21654401079326, 84859704298202, 332818970772254

Offset: 0

Clark Kimberling

nonn

Total number of leaves in all rooted ordered trees with at most n edges. - Michael Somos, Feb 14 2006
Also: Number of UH-free Schroeder paths of semilength n with horizontal steps only at level less than two [see Yan]. - R. J. Mathar, May 24 2008
Hankel transform is A010892. - Paul Barry, Apr 28 2009
Binomial transform of A005773. - Philippe Deléham, Dec 13 2009
Number of vertices all of whose children are leaves in all ordered trees with n+1 edges. Example: a(3) = 15; for an explanation see David Callan's comment in A001519. - Emeric Deutsch, Feb 12 2015

			G.f. = 1 + 2*x + 5*x^2 + 15*x^3 + 50*x^4 + 176*x^5 + 638*x^6 + ...

Michael De Vlieger, Table of n, a(n) for n = 0..1664
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Sherry H. F. Yan, Schröder Paths and Pattern Avoiding Partitions, arXiv:0805.2465 [math.CO], 2008-2009.

Partial sums of A088218.
Bisection of A086905.
Second column of triangle A102541.
Cf. A000108, A006134, A024494, A079309.

A024718 := n -> (binomial(2*n, n)*hypergeom([1, -n], [1/2 - n], 1/4) + 1)/2:
seq(simplify(A024718(n)), n = 0..26); # Peter Luschny, Dec 15 2024

Table[Sum[Binomial[2k-1,k-1],{k,0,n}],{n,0,100}] (* Emanuele Munarini, May 18 2018 *)

a(n) = (1 + sum(k=0, n, binomial(2*k, k)))/2; \\ Michel Marcus, May 18 2018

a(n) = A079309(n) + 1.
G.f.: 1/((1 - x)*(2 - C)), where C = g.f. for the Catalan numbers A000108. - N. J. A. Sloane, Aug 30 2002
Given g.f. A(x), then x * A(x - x^2) is the g.f. of A024494. - Michael Somos, Feb 14 2006
G.f.: (1 + 1 / sqrt(1 - 4*x)) / (2 - 2*x). - Michael Somos, Feb 14 2006
D-finite with recurrence: n*a(n) - (5*n-2)*a(n-1) + 2*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 02 2012
Remark: The above recurrence is true (it can be easily proved by differentiating the generating function). Notice that it is the same recurrence satisfied by the partial sums of the central binomial coefficients (A006134). - Emanuele Munarini, May 18 2018
0 = a(n)*(16*a(n+1) - 22*a(n+2) + 6*a(n+3)) + a(n+1)*(-18*a(n+1) + 27*a(n+2) - 7*a(n+3)) + a(n+2)*(-3*a(n+2) + a(n+3)) for all n in Z if a(n) = 1/2 for n < 0. - Michael Somos, Apr 23 2014
a(n) = ((1 - I/sqrt(3))/2 - binomial(2*n+1, n)*hypergeom([n+3/2, 1], [n+2], 4)). - Peter Luschny, May 18 2018
a(n) = [x^n] 1/((1-x+x^2) * (1-x)^n). - Seiichi Manyama, Apr 06 2024

Name edited by Petros Hadjicostas, Aug 04 2020

A030662 Number of combinations of n things from 1 to n at a time, with repeats allowed.

Original entry on oeis.org

1, 5, 19, 69, 251, 923, 3431, 12869, 48619, 184755, 705431, 2704155, 10400599, 40116599, 155117519, 601080389, 2333606219, 9075135299, 35345263799, 137846528819, 538257874439, 2104098963719, 8233430727599, 32247603683099, 126410606437751, 495918532948103

Offset: 1

Donald Mintz (djmintz(AT)home.com)

Add terms of an increasingly bigger diamond-shaped part of Pascal's triangle:
.......................... 1
............ 1 .......... 1 1
.. 1 ...... 1 1 ........ 1 2 1
. 1 1 =5 . 1 2 1 =19 .. 1 3 3 1 =69
.. 2 ...... 3 3 ........ 4 6 4
............ 6 ......... 10 10
.......................... 20
-  Ralf Stephan, May 17 2004
The prime p divides a((p-1)/2) for p in A002144 (Pythagorean primes). - Alexander Adamchuk, Jul 04 2006
Also, number of square submatrices of a square matrix. - Jono Henshaw (jjono(AT)hotmail.com), Apr 22 2008
Partial sums of A051924. - J. M. Bergot, Jun 22 2013
Number of partitions with Ferrers diagrams that fit in an n X n box (excluding the empty partition of 0). - Michael Somos, Jun 02 2014
Also number of non-descending sequences with length and last number are less or equal to n, and also the number of integer partitions (of any positive integer) with length and largest part are less or equal to n. - Zlatko Damijanic, Dec 06 2024

			G.f. = x + 5*x^2 + 19*x^3 + 69*x^4 + 251*x^5 + 923*x^6 + 3431*x^7 + ...

T. D. Noe, Table of n, a(n) for n = 1..500
Narcisse G. Bell Bogmis, Guy R. Biyogmam, Hesam Safa, and Calvin Tcheka, Upper bounds on the dimension of the Schur Lie-multiplier of Lie-nilpotent Leibniz n-algebras, arXiv:2403.14884 [math.RA], 2024. See p. 7.
Joseph D. Horton and Andrew Kurn, Counting sequences with complete increasing subsequences, Congress Numerantium, 33 (1981), 75-80. MR 681905
Milan Janjic and Boris Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
Milan Janjic and Boris Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
Raimundas Vidunas, Counting derangements and Nash equilibria, Ann. Comb. 21, No. 1, 131-152 (2017).
Jianqiang Zhao, Uniform Approach to Double Shuffle and Duality Relations of Various q-Analogs of Multiple Zeta Values via Rota-Baxter Algebras, arXiv preprint arXiv:1412.8044 [math.NT], 2014.

Cf. A000984, A001700, A002144, A051924, A091908, A144660, A322938.
Column k=2 of A047909.
Central column of triangle A014473.
Right-hand column 2 of triangle A102541.

[(n+1)*Catalan(n)-1: n in [1..40]]; // G. C. Greubel, Apr 07 2024

seq(sum((binomial(n,m))^2,m=1..n),n=1..23); # Zerinvary Lajos, Jun 19 2008
f:=n->add( add( binomial(i+j,i), i=0..n),j=0..n); [seq(f(n),n=0..12)]; # N. J. A. Sloane, Jan 31 2009

Table[Sum[Sum[(2n-i-j)!/(n-i)!/(n-j)!,{i,1,n}],{j,1,n}],{n,1,20}] (* Alexander Adamchuk, Jul 04 2006 *)
a[n_] := 2*(2*n-1)!/(n*(n-1)!^2)-1; Table[a[n], {n, 1, 26}] (* Jean-François Alcover, Oct 11 2012, from first formula *)

a(n)=binomial(2*n,n)-1 \\ Charles R Greathouse IV, Jun 26 2013

from math import comb
def a(n): return comb(2*n, n) - 1
print([a(n) for n in range(1, 27)]) # Michael S. Branicky, Jul 11 2023

def a(n) : return binomial(2*n,n) - 1
[a(n) for n in (1..26)] # Peter Luschny, Apr 21 2012

a(n) = A000984(n) - 1.
a(n) = 2*A001700(n-1) - 1.
a(n) = 2*(2*n-1)!/(n!*(n-1)!)-1.
a(n) = Sum_{k=1..n} binomial(n, k)^2. - Benoit Cloitre, Aug 20 2002
a(n) = Sum_{j=0..n} Sum_{i=j..n+j} binomial(i, j). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jul 23 2003
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-1} binomial(i+j, i). - N. J. A. Sloane, Jan 31 2009
Also for n>1: a(n)=(2*n)!/(n!)^2-1. - Hugo Pfoertner, Feb 10 2004
a(n) = Sum_{j=1..n} Sum_{i=1..n} (2n-i-j)!/((n-i)!*(n-j)!). - Alexander Adamchuk, Jul 04 2006
a(n) = A115112(n) + 1. - Jono Henshaw (jjono(AT)hotmail.com), Apr 22 2008
G.f.: Q(0)*(1-4*x)/x - 1/x/(1-x), where Q(k)= 1 + 4*(2*k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
D-finite with recurrence: n*a(n) +2*(-3*n+2)*a(n-1) +(9*n-14)*a(n-2) +2*(-2*n+5)*a(n-3)=0. - R. J. Mathar, Jun 25 2013
0 = a(n)*(+16*a(n+1) - 70*a(n+2) + 68*a(n+3) - 14*a(n+4)) + a(n+1)*(-2*a(n+1) + 61*a(n+2) - 96*a(n+3) + 23*a(n+4)) + a(n+2)*(-6*a(n+2) + 31*a(n+3)  - 10*a(n+4)) + a(n+3)*(-2*a(n+3) + a(n+4)) for all n in Z. - Michael Somos, Jun 02 2014
From Ilya Gutkovskiy, Jan 25 2017: (Start)
O.g.f.: (1 - x - sqrt(1 - 4*x))/((1 - x)*sqrt(1 - 4*x)).
E.g.f.: exp(x)*(exp(x)*BesselI(0,2*x) - 1). (End)
a(n) = 3*n*Sum_{k=1..n} (-1)^(k+1)/(2*n+k)*binomial(2*n+k,n-k). - Vladimir Kruchinin, Jul 29 2025
a(n) = n * binomial(2*n, n) * Sum_{k = 1..n} 1/(k*binomial(n+k, k)). - Peter Bala, Aug 05 2025

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

Johannes W. Meijer, Aug 26 2013

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

			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

Andrew Howroyd, Table of n, a(n) for n = 0..989

Cf. A034851, A102541, A228570, A192928.

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

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 *)

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

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).

A238009 Number T(n,k) of equivalence classes of ways of placing k 2 X 2 tiles in an n X 3 rectangle under all symmetry operations of the rectangle; irregular triangle T(n,k), n>=2, 0<=k<=floor(n/2), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 2, 4, 1, 3, 8, 3, 1, 3, 12, 8, 1, 4, 18, 22, 6, 1, 4, 24, 40, 22, 1, 5, 32, 73, 66, 10, 1, 5, 40, 112, 146, 48, 1, 6, 50, 172, 292, 174, 20, 1, 6, 60, 240, 516, 448, 116, 1, 7, 72, 335, 860, 1020, 464, 36, 1, 7, 84, 440, 1340, 2016, 1360, 256

Offset: 2

Christopher Hunt Gribble, Feb 16 2014

			The first 19 rows of T(n,k) are:
   n\k 0  1   2    3    4     5     6     7     8    9  10
   2   1  1
   3   1  1
   4   1  2   2
   5   1  2   4
   6   1  3   8    3
   7   1  3  12    8
   8   1  4  18   22    6
   9   1  4  24   40   22
  10   1  5  32   73   66    10
  11   1  5  40  112  146    48
  12   1  6  50  172  292   174    20
  13   1  6  60  240  516   448   116
  14   1  7  72  335  860  1020   464    36
  15   1  7  84  440 1340  2016  1360   256
  16   1  8  98  578 2010  3716  3400  1168    72
  17   1  8 112  728 2890  6336  7432  3840   584
  18   1  9 128  917 4046 10326 14864 10600  2920  136
  19   1  9 144 1120 5502 16016 27536 25344 10600 1280
  20   1 10 162 1368 7336 24066 48188 54992 31800 7080 272

Christopher Hunt Gribble and Andrew Howroyd, Rows n=2..60 of T(n,k) flattened (Rows n=2..20 from Christopher Hunt Gribble)
Christopher Hunt Gribble, C++ program

Cf. A034851, A226048, A102541, A226290, A228570, A225812, A238189, A238190, A228572, A228022, A231145, A231473, A231568, A232440, A228165, A238550, A238551, A238552, A228166, A238555, A238556, A228167, A238557, A238558, A238559, A228168, A238581, A238582, A238583, A228169, A238586, A238592.

T(n,k)={(2^k*binomial(n-1*k,k) + ((k%2==0)+(n%2==0||k%2==0)+(k==0)) * 2^((k+1)\2)*binomial((n-1*k-(k%2)-(n%2))/2,k\2))/4}
for(n=2,20,for(k=0,floor(n/2), print1(T(n,k), ", "));print) \\ Andrew Howroyd, May 29 2017

Corrected C++ program and xrefs added by Christopher Hunt Gribble, Apr 25 2015

A231145 Number T(n,k) of equivalence classes of ways of placing k 2 X 2 tiles in an n X 5 rectangle under all symmetry operations of the rectangle; irregular triangle T(n,k), n>=2, 0<=k<=n-n%2, read by rows.

Original entry on oeis.org

1, 2, 2, 1, 2, 4, 1, 4, 13, 10, 4, 1, 4, 23, 35, 23, 1, 6, 40, 101, 125, 54, 10, 1, 6, 58, 206, 403, 336, 106, 1, 8, 83, 392, 1056, 1438, 956, 240, 25, 1, 8, 109, 641, 2281, 4424, 4718, 2409, 473, 1, 10, 142, 1011, 4429, 11370, 17252, 14478, 6094, 1020, 70

Offset: 2

Christopher Hunt Gribble, Feb 23 2014

			The first 8 rows of T(n,k) are:
.\ k  0     1     2     3     4     5     6     7     8
n
2     1     2     2
3     1     2     4
4     1     4    13    10     4
5     1     4    23    35    23
6     1     6    40   101   125    54    10
7     1     6    58   206   403   336   106
8     1     8    83   392  1056  1438   956   240    25
9     1     8   109   641  2281  4424  4718  2409   473

Andrew Howroyd, Table of n, a(n) for n = 2..967 (terms n=2..71 from Christopher Hunt Gribble)
Christopher Hunt Gribble, C++ program

Cf. A034851, A226048, A102541, A226290, A238009, A228570, A225812, A238189, A238190, A228572, A228022, A231473, A231568, A232440, A228165, A238550, A238551, A238552, A228166, A238555, A238556, A228167, A238557, A238558, A238559, A228168, A238581, A238582, A238583, A228169, A238586, A238592.

Terms corrected and xrefs updated by Christopher Hunt Gribble, Apr 26 2015

A231473 Number T(n,k) of equivalence classes of ways of placing k 3 X 3 tiles in an n X 5 rectangle under all symmetry operations of the rectangle; irregular triangle T(n,k), n>=3, 0<=k<=floor(n/3), read by rows.

Original entry on oeis.org

1, 2, 1, 2, 1, 4, 1, 4, 4, 1, 6, 9, 1, 6, 18, 1, 8, 28, 10, 1, 8, 42, 28, 1, 10, 57, 76, 1, 10, 76, 140, 25, 1, 12, 96, 254, 107, 1, 12, 120, 392, 321, 1, 14, 145, 600, 731, 70, 1, 14, 174, 840, 1462, 366, 1, 16, 204, 1170, 2610, 1308, 1, 16, 238, 1540, 4350, 3416, 196

Offset: 3

Christopher Hunt Gribble, Feb 23 2014

			The first 11 rows of T(n,k) are:
.\ k    0      1      2      3      4
n
3       1      2
4       1      2
5       1      4
6       1      4      4
7       1      6      9
8       1      6     18
9       1      8     28     10
10      1      8     42     28
11      1     10     57     76
12      1     10     76    140     25
13      1     12     96    254    107

Andrew Howroyd, Table of n, a(n) for n = 3..974
Christopher Hunt Gribble, C++ program

Cf. A034851, A226048, A102541, A226290, A238009, A228570, A225812, A238189, A238190, A228572, A228022, A231145, A231568, A232440, A228165, A238550-A238552, A228166, A238555, A238556, A228167, A238557-A238559, A228168, A238581-A238583, A228169, A238586, A238592.

T[n_, k_] := ((3^k + 1)*Binomial[n - 2k, k] + Boole[EvenQ[k] || OddQ[n]]*(3^(Quotient[(k + 1), 2]) + 3^Quotient[k, 2]) Binomial[(n - 2k - Mod[n, 2])/2, Quotient[k, 2]])/4; Table[T[n, k], {n, 3, 20}, {k, 0, Floor[n/3]}] // Flatten (* Jean-François Alcover, Oct 06 2017, after Andrew Howroyd *)

T(n,k)={((3^k+1)*binomial(n-2*k,k) + (k%2==0||n%2==1) * (3^((k+1)\2)+3^(k\2)) * binomial((n-2*k-(n%2))/2,k\2))/4}
for(n=3,20,for(k=0,floor(n/3), print1(T(n,k), ", "));print) \\ Andrew Howroyd, May 29 2017

Terms corrected and xrefs updated by Christopher Hunt Gribble, Apr 26 2015

Terms a(40) and beyond from Andrew Howroyd, May 29 2017

A231568 Number T(n,k) of equivalence classes of ways of placing k 4 X 4 tiles in an n X 5 rectangle under all symmetry operations of the rectangle; irregular triangle T(n,k), n>=4, 0<=k<=floor(n/4), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 1, 3, 4, 1, 4, 8, 1, 4, 12, 1, 5, 18, 3, 1, 5, 24, 8, 1, 6, 32, 22, 1, 6, 40, 40, 1, 7, 50, 73, 6, 1, 7, 60, 112, 22, 1, 8, 72, 172, 66, 1, 8, 84, 240, 146, 1, 9, 98, 335, 292, 10, 1, 9, 112, 440, 516, 48, 1, 10, 128, 578, 860, 174

Offset: 4

Christopher Hunt Gribble, Feb 23 2014

			The first 14 rows of T(n,k) are:
.\  k    0      1      2      3     4
n
4        1      1
5        1      1
6        1      2
7        1      2
8        1      3      2
9        1      3      4
10       1      4      8
11       1      4     12
12       1      5     18      3
13       1      5     24      8
14       1      6     32     22
15       1      6     40     40
16       1      7     50     73     6
17       1      7     60    112    22

Andrew Howroyd, Table of n, a(n) for n = 4..989
Christopher Hunt Gribble, C++ program

Cf. A034851, A226048, A102541, A226290, A238009, A228570, A225812, A238189, A238190, A228572, A228022, A231145, A231473, A232440, A228165, A238550, A238551, A238552, A228166, A238555, A238556, A228167, A238557, A238558, A238559, A228168, A238581, A238582, A238583, A228169, A238586, A238592.

T[n_, k_] := (2^k Binomial[n - 3k, k] + (Boole[EvenQ[k]] + Boole[EvenQ[n] || EvenQ[k]] + Boole[k == 0]) 2^Quotient[k+1, 2] Binomial[(n - 3k - Mod[k, 2] - Mod[n, 2])/2, Quotient[k, 2]])/4; Table[T[n, k], {n, 4, 20}, {k, 0, Floor[n/4]}] // Flatten (* Jean-François Alcover, Oct 06 2017, after Andrew Howroyd *)

T(n,k)={(2^k*binomial(n-3*k,k) + ((k%2==0)+(n%2==0||k%2==0)+(k==0)) * 2^((k+1)\2)*binomial((n-3*k-(k%2)-(n%2))/2,k\2))/4}
for(n=2,20,for(k=0,floor(n/4), print1(T(n,k), ", "));print) \\ Andrew Howroyd, May 29 2017

Terms extended and xrefs updated by Christopher Hunt Gribble, Apr 26 2015

Terms a(32) and beyond from Andrew Howroyd, May 29 2017

cp's OEIS Frontend

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).

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A011973 Irregular triangle read by rows: T(n,k) = binomial(n-k, k), n >= 0, 0 <= k <= floor(n/2); or, coefficients of (one version of) Fibonacci polynomials.

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

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A024718 a(n) = (1/2)*(1 + Sum_{k=0..n} binomial(2*k, k)).

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A030662 Number of combinations of n things from 1 to n at a time, with repeats allowed.

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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).

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A024718 a(n) = (1/2)(1 + Sum_{k=0..n} binomial(2k, k)).