A054123 -id:A054123 - cp's OEIS frontend

A054125 Sum of the arrays in A054123 and A054124.

2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 2, 4, 4, 4, 2, 2, 5, 6, 6, 5, 2, 2, 6, 9, 8, 9, 6, 2, 2, 7, 13, 12, 12, 13, 7, 2, 2, 8, 18, 19, 16, 19, 18, 8, 2, 2, 9, 24, 30, 24, 24, 30, 24, 9, 2, 2, 10, 31, 46, 39, 32, 39, 46, 31, 10, 2, 2, 11, 39, 68, 65, 48, 48, 65

Offset: 0

Clark Kimberling

Row sums are twice Fibonacci numbers, A006355(n+2).

			Rows:
  2;
  2,2;
  2,2,2;
  2,3,3,2;
  ...

A052509(n,k) = sum(m=0, k, binomial(n-k, m));
T(n,k) = if(k==0 || k==n, 2, A052509(n-1,k) + A052509(n-1,n-k)) \\ Jianing Song, May 30 2022

From Jianing Song, May 30 2022: (Start)
T(n,k) = 2 if k = 0 or k = n, A052509(n-1,k) + A052509(n-1,n-k) otherwise.
G.f.: Sum_{n>=0, 0<=k<=n} T(n,k) * x^n * y^k = (1-x^2*y) * (1/((1-x*y)*(1-x-x^2*y)) + 1/((1-x)*(1-x*y-x^2*y))). (End)

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

A052509 Knights-move Pascal triangle: T(n,k), n >= 0, 0 <= k <= n; T(n,0) = T(n,n) = 1, T(n,k) = T(n-1,k) + T(n-2,k-1) for k = 1,2,...,n-1, n >= 2.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 4, 2, 1, 1, 5, 7, 4, 2, 1, 1, 6, 11, 8, 4, 2, 1, 1, 7, 16, 15, 8, 4, 2, 1, 1, 8, 22, 26, 16, 8, 4, 2, 1, 1, 9, 29, 42, 31, 16, 8, 4, 2, 1, 1, 10, 37, 64, 57, 32, 16, 8, 4, 2, 1, 1, 11, 46, 93, 99, 63, 32, 16, 8, 4, 2, 1

Offset: 0

N. J. A. Sloane, Mar 17 2000

Also square array T(n,k) (n >= 0, k >= 0) read by antidiagonals: T(n,k) = Sum_{i=0..k} binomial(n,i).
As a number triangle read by rows, this is T(n,k) = Sum_{i=n-2*k..n-k} binomial(n-k,i), with T(n,k) = T(n-1,k) + T(n-2,k-1). Row sums are A000071(n+2). Diagonal sums are A023435(n+1). It is the reverse of the Whitney triangle A004070. - Paul Barry, Sep 04 2005
Also, twice number of orthants intersected by a generic k-dimensional subspace of R^n [Naiman and Scheinerman, 2017]. - N. J. A. Sloane, Mar 03 2018

			Triangle begins:
[0] 1;
[1] 1, 1;
[2] 1, 2,  1;
[3] 1, 3,  2,  1;
[4] 1, 4,  4,  2,  1;
[5] 1, 5,  7,  4,  2,  1;
[6] 1, 6, 11,  8,  4,  2, 1;
[7] 1, 7, 16, 15,  8,  4, 2, 1;
[8] 1, 8, 22, 26, 16,  8, 4, 2, 1;
[9] 1, 9, 29, 42, 31, 16, 8, 4, 2, 1;
As a square array, this begins:
  1  1  1  1  1  1 ...
  1  2  2  2  2  2 ...
  1  3  4  4  4  4 ...
  1  4  7  8  8  8 ...
  1  5 11 15 16 ...
  1  6 16 26 31 32 ...

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Clark Kimberling, Path-counting and Fibonacci numbers, Fib. Quart. 40 (4) (2002) 328-338, Example 1C.
Daniel Q. Naiman and Edward R. Scheinerman, Arbitrage and Geometry, arXiv:1709.07446 [q-fin.MF], 2017 [Contains the square array multiplied by 2].
Richard L. Ollerton and Anthony G. Shannon, Some properties of generalized Pascal squares and triangles, Fib. Q., 36 (1998), 98-109. See Tables 5 and 14.
D. J. Price, Some unusual series occurring in n-dimensional geometry, Math. Gaz., 30 (1946), 149-150.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A054123, A054124, A007318, A008949.
Row sums A000071; Diagonal sums A023435; Mirror A004070.
Columns give A000027, A000124, A000125, A000127, A006261, ...
Cf. A052509, A054123, A054124, A007318, A008949, A052553.
Partial sums across rows of (extended) Pascal's triangle A052553.

A052509:=Flat(List([0..100],n->List([0..n],k->Sum([0..n],m->Binomial(n-k,k-m))))); # Muniru A Asiru, Sat Feb 17 2018

a052509 n k = a052509_tabl !! n !! k
a052509_row n = a052509_tabl !! n
a052509_tabl = [1] : [1,1] : f [1] [1,1] where
   f row' row = rs : f row rs where
     rs = zipWith (+) ([0] ++ row' ++ [1]) (row ++ [0])
-- Reinhard Zumkeller, Nov 22 2012

[[(&+[Binomial(n-k, k-j): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, May 13 2019

a := proc(n::nonnegint, k::nonnegint) option remember: if k=0 then RETURN(1) fi: if k=n then RETURN(1) fi: a(n-1,k)+a(n-2,k-1) end: for n from 0 to 11 do for k from 0 to n do printf(`%d,`,a(n,k)) od: od: # James Sellers, Mar 17 2000
with(combinat): for s from 0 to 11 do for n from s to 0 by -1 do if n=0 or s-n=0 then printf(`%d,`,1) else printf(`%d,`,sum(binomial(n, i), i=0..s-n)) fi; od: od: # James Sellers, Mar 17 2000

Table[Sum[Binomial[n-k, k-m], {m, 0, n}], {n, 0, 10}, {k, 0, n}]
T[n_, k_] := Hypergeometric2F1[-k, -n + k, -k, -1];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Peter Luschny, Nov 28 2021 *)

T(n,k)=sum(m=0,n,binomial(n-k,k-m));
for(n=0,10,for(k=0,n,print1(T(n,k),", "););print();); /* show triangle */

[[sum(binomial(n-k, k-j) for j in (0..n)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, May 13 2019

T(n, k) = Sum_{m=0..n} binomial(n-k, k-m). - Wouter Meeussen, Oct 03 2002
From Werner Schulte, Feb 15 2018: (Start)
Referring to the square array T(i,j):
G.f. of row n: Sum_{k>=0} T(n,k) * x^k = (1+x)^n / (1-x).
G.f. of T(i,j): Sum_{k>=0, n>=0} T(n,k) * x^k * y^n = 1 / ((1-x)*(1-y-x*y)).
Let a_i(n) be multiplicative with a_i(p^e) = T(i, e), p prime and e >= 0, then Sum_{n>0} a_i(n)/n^s = (zeta(s))^(i+1) / (zeta(2*s))^i for i >= 0.
(End)
T(n, k) = hypergeom([-k, -n + k], [-k], -1). - Peter Luschny, Nov 28 2021
From Jianing Song, May 30 2022: (Start)
Referring to the triangle, G.f.: Sum_{n>=0, 0<=k<=n} T(n,k) * x^n * y^k = 1 / ((1-x*y)*(1-x-x^2*y)).
T(n,k) = 2^(n-k) for ceiling(n/2) <= k <= n. (End)

More terms from James Sellers, Mar 17 2000
Entry formed by merging two earlier entries. - N. J. A. Sloane, Jun 17 2007
Edited by Johannes W. Meijer, Jul 24 2011

A052553 Square array of binomial coefficients T(n,k) = binomial(n,k), n >= 0, k >= 0, read by upward antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 0, 0, 1, 3, 1, 0, 0, 1, 4, 3, 0, 0, 0, 1, 5, 6, 1, 0, 0, 0, 1, 6, 10, 4, 0, 0, 0, 0, 1, 7, 15, 10, 1, 0, 0, 0, 0, 1, 8, 21, 20, 5, 0, 0, 0, 0, 0, 1, 9, 28, 35, 15, 1, 0, 0, 0, 0, 0, 1, 10, 36, 56, 35, 6, 0, 0, 0, 0, 0, 0, 1, 11, 45, 84, 70, 21, 1, 0, 0, 0, 0, 0, 0, 1, 12, 55

Offset: 0

N. J. A. Sloane, Mar 17 2000

Another version of Pascal's triangle A007318.
As a triangle read by rows, it is (1,0,0,0,0,0,0,0,0,...) DELTA (0,1,-1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938 and it is the Riordan array (1/(1-x), x^2/(1-x)). The row sums of this triangle are F(n+1) = A000045(n+1). - Philippe Deléham, Dec 11 2011
As a triangle, binomial(n-k, k) is also the number of ways to add k pierced circles to a path graph P_n so that no two circles share a vertex (see Lemma 3.1 at page 5 in Owad and Tsvietkova). - Stefano Spezia, May 18 2022
For all n >= 0, k >= 0, the k-th homology group of the n-torus H_k(T^n) is the free abelian group of rank T(n,k) = binomial(n,k). See the Math Stack Exchange link below. - Jianing Song, Mar 13 2023

			Array begins:
  1, 0,  0,  0, 0, 0, ...
  1, 1,  0,  0, 0, 0, ...
  1, 2,  1,  0, 0, 0, ...
  1, 3,  3,  1, 0, 0, ...
  1, 4,  6,  4, 1, 0, ...
  1, 5, 10, 10, 5, 1, ...
As a triangle, this begins:
  1;
  1, 0;
  1, 1,  0;
  1, 2,  0, 0;
  1, 3,  1, 0, 0;
  1, 4,  3, 0, 0, 0;
  1, 5,  6, 1, 0, 0, 0;
  1, 6, 10, 4, 0, 0, 0, 0;
  ...

Vincenzo Librandi, Table of n, a(n) for n = 0..5459
Math Stack Exchange, Homology of the n-torus using the Künneth Formula
Nicholas Owad and Anastasiia Tsvietkova, Random meander model for links, arXiv:2205.03451 [math.GT], 2022.
Index entries for triangles and arrays related to Pascal's triangle

The official entry for Pascal's triangle is A007318. See also A026729 (the same array read by downward antidiagonals).
As a triangle without zeros: A011973.
Cf. A052509, A054123, A054124, A008949.

/* As triangle */ [[Binomial(n-k,k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Feb 08 2017

with(combinat): for s from 0 to 20 do for n from s to 0 by -1 do printf(`%d,`, binomial(n, s-n)) od:od: # James Sellers, Mar 17 2000

Flatten[ Table[ Binomial[n-k , k], {n, 0, 13}, {k, 0, n}]]  (* Jean-François Alcover, Dec 05 2012 *)

T(n,k) = binomial(n,k) \\ Charles R Greathouse IV, Feb 07 2017

As a triangle: T(n,k) = A026729(n,n-k).

G.f. of the triangular version: 1/(1-x-x^2*y). - R. J. Mathar, Aug 11 2015

A054124 Left Fibonacci row-sum array, n >= 0, 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 2, 4, 4, 1, 1, 1, 2, 4, 7, 5, 1, 1, 1, 2, 4, 8, 11, 6, 1, 1, 1, 2, 4, 8, 15, 16, 7, 1, 1, 1, 2, 4, 8, 16, 26, 22, 8, 1, 1, 1, 2, 4, 8, 16, 31, 42, 29, 9, 1, 1, 1, 2, 4, 8, 16, 32, 57, 64, 37, 10, 1, 1, 1, 2

Offset: 0

Clark Kimberling

Reflection of array in A054123 about vertical central line.
Starting with g(0) = {0}, generate g(n) for n > 0 inductively using these rules:
  (1)  if x is in g(n-1), then x+1 is in g(n); and
  (2)  if x is in g(n-1) and x < 2, then x/2 is in g(n).
Then g(1) = {1/1}, g(2) = {1/2,2/1}, g(3) = {1/4,3/2,3/1}, etc. The denominators in g(n) are 2^0, 2^1, ..., 2^(n-1), and T(n,k) is the number of occurrences of 2^k, for k = 0..n-1. - Clark Kimberling, Nov 09 2015
Variant of A004070 with an additional column of 1's on the left. - Jianing Song, May 30 2022

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Row sums: A000045. Central numbers: 1, 1, 2, 4, 8, ... (A000079).

First n numbers of n-th column for n >= 1 form the array in A008949.

a054124 n k = a054124_tabl !! n !! k
a054124_row n = a054124_tabl !! n
a054124_tabl = map reverse a054123_tabl
-- Reinhard Zumkeller, May 26 2015

t[, 0|1] = t[n, n_] = 1; t[n_, k_] := t[n, k] = t[n-1, k-1] + t[n-2, k-1]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 25 2013 *)

A052509(n,k) = sum(m=0, k, binomial(n-k, m));
T(n,k) = if(k==0, 1, A052509(n-1,n-k)) \\ Jianing Song, May 30 2022

T(n, 0) = T(n, n) = 1 for n >= 0; T(n, 1) = 1 for n >= 1; T(n, k) = T(n-1, k-1) + T(n-2, k-1) for k=2, 3, ..., n-1, n >= 3. [Corrected by Jianing Song, May 30 2022]

G.f.: Sum_{n>=0, 0<=k<=n} T(n,k) * x^n * y^k = (1-x^2*y) / ((1-x)*(1-x*y-x^2*y)). - Jianing Song, May 30 2022

A129713 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and starting with exactly k 1's (0<=k<=n). A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 5, 3, 2, 1, 1, 1, 8, 5, 3, 2, 1, 1, 1, 13, 8, 5, 3, 2, 1, 1, 1, 21, 13, 8, 5, 3, 2, 1, 1, 1, 34, 21, 13, 8, 5, 3, 2, 1, 1, 1, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1, 1, 89, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1, 1, 144, 89, 55, 34, 21, 13, 8, 5, 3, 2, 1, 1, 1, 233, 144

Offset: 0

Emeric Deutsch, May 12 2007

Row sums are the Fibonacci numbers (A000045). Sum(k*T(n,k), 0<=k<=n) = F(n+3)-2 = A001911(n).

			T(6,2) = 3 because we have 110110, 110111, 110101.
Triangle starts:
1;
1,1;
1,1,1;
2,1,1,1;
3,2,1,1,1;
5,3,2,1,1,1;
8,5,3,2,1,1,1;

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened

Cf. A000045, A001911.
Cf. A054123.
Cf. A007298. - Altug Alkan, May 03 2016

a129713 n k = a129713_tabl !! n !! k
a129713_row n = a129713_tabl !! n
a129713_tabl = [1] : [1, 1] : f [1] [1, 1] where
   f us vs = ws : f vs ws where
             ws = zipWith (+) (init us ++ [0, 0, 0]) (vs ++ [1])
-- Reinhard Zumkeller, May 26 2015

with(combinat): T:=proc(n,k) if k<=n-2 then fibonacci(n-k) elif k=n-1 or k=n then 1 else 0 fi end: for n from 0 to 15 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

nn=15;a=1/(1-y x);b=1/(1-x);Map[Select[#,#>0&]&,CoefficientList[Series[a (1+x)/(1-x^2b),{x,0,nn}],{x,y}]]//Grid (* Geoffrey Critzer, Dec 04 2013 *)

T(n,k) = F(n-k) if k<=n-2, T(n,n-1) = T(n,n) = 1, where F(j) are the Fibonacci numbers (F(0)=0, F(1)=1). G.f.: G(t,z) = (1-z^2)/[(1-z-z^2)(1-tz)].

a(n) = A007298(n+4) - A007298(n+3). - Altug Alkan, May 03 2016

A264200 Numerator of sum of numbers in set g(n) generated as in Comments.

Original entry on oeis.org

0, 1, 5, 19, 69, 235, 789, 2603, 8533, 27819, 90453, 293547, 951637, 3082923, 9983317, 32320171, 104617301, 338602667, 1095849301, 3546458795, 11477013845, 37141260971, 120193373525, 388957383339, 1258699445589, 4073250794155, 13181344109909, 42655780874923

Offset: 0

Clark Kimberling, Nov 09 2015

Starting with g(0) = {0}, generate g(n) for n > 0 inductively using these rules:
(1)  if x is in g(n-1), then x + 1 is in g(n); and
(2)  if x is in g(n-1) and x < 2, then x/2 is in g(n).
The sum of numbers in g(n) is a(n)/2^(n-1).

			g(0) = {0}, sum = 0.
g(1) = {1}, sum = 1.
g(2) = {1/2,2/1}, sum = 5/4.
g(3) = {1/4,3/2,3/1}, sum = 19/8.

Cf. A054123, A054124, A264201.

z = 30; x = 1/2; g[0] = {0}; g[1] = {1};
g[n_] := g[n] = Union[1 + g[n - 1], (1/2) Select[g[n - 1], # < 2 &]]
Table[g[n], {n, 0, z}]; Table[Total[g[n]], {n, 0, z}]
Numerator[Table[Total[g[n]], {n, 0, z}] ]

Conjecture: a(n) = 3*a(n-1) + 4*a(n-2) - 8*a(n-3) - 8*a(n-4).

cp's OEIS Frontend

A054125 Sum of the arrays in A054123 and A054124.

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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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A052509 Knights-move Pascal triangle: T(n,k), n >= 0, 0 <= k <= n; T(n,0) = T(n,n) = 1, T(n,k) = T(n-1,k) + T(n-2,k-1) for k = 1,2,...,n-1, n >= 2.

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A052553 Square array of binomial coefficients T(n,k) = binomial(n,k), n >= 0, k >= 0, read by upward antidiagonals.

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A054124 Left Fibonacci row-sum array, n >= 0, 0<=k<=n.

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A129713 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and starting with exactly k 1's (0<=k<=n). A Fibonacci binary word is a binary word having no 00 subword.

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