A025581 -id:A025581 - cp's OEIS frontend

A002262 Triangle read by rows: T(n,k) = k, 0 <= k <= n, in which row n lists the first n+1 nonnegative integers.

0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13

Offset: 0

Angele Hamel (amh(AT)maths.soton.ac.uk)

The point with coordinates (x = A025581(n), y = A002262(n)) sweeps out the first quadrant by upwards antidiagonals. N. J. A. Sloane, Jul 17 2018
Old name: Integers 0 to n followed by integers 0 to n+1 etc.
a(n) = n - the largest triangular number <= n. - Amarnath Murthy, Dec 25 2001
The PARI functions t1, t2 can be used to read a square array T(n,k) (n >= 0, k >= 0) by antidiagonals downwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002
Values x of unique solution pair (x,y) to equation T(x+y) + x = n, where T(k)=A000217(k). - Lekraj Beedassy, Aug 21 2004
a(A000217(n)) = 0; a(A000096(n)) = n. - Reinhard Zumkeller, May 20 2009
Concatenation of the set representation of ordinal numbers, where the n-th ordinal number is represented by the set of all ordinals preceding n, 0 being represented by the empty set. - Daniel Forgues, Apr 27 2011
An integer sequence is nonnegative if and only if it is a subsequence of this sequence. - Charles R Greathouse IV, Sep 21 2011
a(A195678(n)) = A000040(n) and a(m) <> A000040(n) for m < A195678(n), an example of the preceding comment. - Reinhard Zumkeller, Sep 23 2011
A sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements of the sequence A. A002262 is reluctant sequence of 0,1,2,3,... The nonnegative integers, A001477. - Boris Putievskiy, Dec 12 2012

			From _Daniel Forgues_, Apr 27 2011: (Start)
Examples of set-theoretic representation of ordinal numbers:
  0: {}
  1: {0} = {{}}
  2: {0, 1} = {0, {0}} = {{}, {{}}}
  3: {0, 1, 2} = {{}, {0}, {0, 1}} = ... = {{}, {{}}, {{}, {{}}}} (End)
From _Omar E. Pol_, Jul 15 2012: (Start)
  0;
  0, 1;
  0, 1, 2;
  0, 1, 2, 3;
  0, 1, 2, 3, 4;
  0, 1, 2, 3, 4, 5;
  0, 1, 2, 3, 4, 5, 6;
  0, 1, 2, 3, 4, 5, 6, 7;
  0, 1, 2, 3, 4, 5, 6, 7, 8;
  0, 1, 2, 3, 4, 5, 6, 7, 8, 9;
  0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10;
(End)

Charles R Greathouse IV, Rows n = 0..100, flattened
Boris Putievskiy, Transformations [Of] Integer Sequences And Pairing Functions, arXiv preprint arXiv:1212.2732 [math.CO], 2012.
Michael Somos, Sequences used for indexing triangular or square arrays

Cf. A002024, A002260, A004736, A025581, A025675, A025682.
Cf. A025691, A048645, A053186, A053645, A056558, A127324.
As a sequence, essentially same as A048151.
Cf. A060510 (parity).

a002262 n k = a002262_tabl !! n !! k
a002262_row n = a002262_tabl !! n
a002262_tabl = map (enumFromTo 0) [0..]
a002262_list = concat a002262_tabl
-- Reinhard Zumkeller, Aug 05 2015, Jul 13 2012, Mar 07 2011

seq(seq(i,i=0..n),n=0..14); # Peter Luschny, Sep 22 2011
A002262 := n -> n - binomial(floor((1/2)+sqrt(2*(1+n))),2);

m[n_]:= Floor[(-1 + Sqrt[8n - 7])/2]
b[n_]:= n - m[n] (m[n] + 1)/2
Table[m[n], {n, 1, 105}]     (* A003056 *)
Table[b[n], {n, 1, 105}]     (* A002260 *)
Table[b[n] - 1, {n, 1, 120}] (* A002262 *)
(* Clark Kimberling, Jun 14 2011 *)
Flatten[Table[k, {n, 0, 14}, {k, 0, n}]] (* Alonso del Arte, Sep 21 2011 *)
Flatten[Table[Range[0,n], {n,0,15}]] (* Harvey P. Dale, Jan 31 2015 *)

PARI
```
a(n)=n-binomial(round(sqrt(2+2*n)),2)
```

t1(n)=n-binomial(floor(1/2+sqrt(2+2*n)),2) /* A002262, this sequence */

t2(n)=binomial(floor(3/2+sqrt(2+2*n)),2)-(n+1) /* A025581, cf. comment by Somos for reading arrays by antidiagonals */

concat(vector(15,n,vector(n,i,i-1)))  \\ M. F. Hasler, Sep 21 2011

apply( {A002262(n)=n-binomial(sqrtint(8*n+8)\/2,2)}, [0..99]) \\ M. F. Hasler, Oct 20 2022

for i in range(16):
    for j in range(i):
        print(j, end=", ") # Mohammad Saleh Dinparvar, May 13 2020

from math import comb, isqrt
def a(n): return n - comb((1+isqrt(8+8*n))//2, 2)
print([a(n) for n in range(105)]) # Michael S. Branicky, May 07 2023

a(n) = A002260(n) - 1.
a(n) = n - (trinv(n)*(trinv(n)-1))/2; trinv := n -> floor((1+sqrt(1+8*n))/2) (cf. A002024); # gives integral inverses of triangular numbers
a(n) = n - A000217(A003056(n)) = n - A057944(n). - Lekraj Beedassy, Aug 21 2004
a(n) = A140129(A023758(n+2)). - Reinhard Zumkeller, May 14 2008
a(n) = f(n,1) with f(n,m) = if nReinhard Zumkeller, May 20 2009
a(n) = (1/2)*(t - t^2 + 2*n), where t = floor(sqrt(2*n+1) + 1/2) = round(sqrt(2*n+1)). - Ridouane Oudra, Dec 01 2019
a(n) = ceiling((-1 + sqrt(9 + 8*n))/2) * (1 - ((1/2)*ceiling((1 + sqrt(9 + 8*n))/2))) + n. - Ryan Jean, Sep 03 2022
G.f.: x*y/((1 - x)*(1 - x*y)^2). - Stefano Spezia, Feb 21 2024

New name from Omar E. Pol, Jul 15 2012

Typo in definition fixed by Reinhard Zumkeller, Aug 05 2015

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

A174344 List of x-coordinates of point moving in clockwise square spiral.

Original entry on oeis.org

0, 1, 1, 0, -1, -1, -1, 0, 1, 2, 2, 2, 2, 1, 0, -1, -2, -2, -2, -2, -2, -1, 0, 1, 2, 3, 3, 3, 3, 3, 3, 2, 1, 0, -1, -2, -3, -3, -3, -3, -3, -3, -3, -2, -1, 0, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 3, 2, 1, 0, -1, -2, -3, -4, -4, -4, -4, -4, -4, -4, -4, -4, -3, -2

Offset: 1

Nikolas Garofil (nikolas(AT)garofil.be), Mar 16 2010

sign

Also, list of x-coordinates of point moving in counterclockwise square spiral.
This spiral, in either direction, is sometimes called the "Ulam spiral", but "square spiral" is a better name. (Ulam looked at the positions of the primes, but of course the spiral itself must be much older.) - N. J. A. Sloane, Jul 17 2018
Graham, Knuth and Patashnik give an exercise and answer on mapping n to square spiral x,y coordinates, and back x,y to n.  They start 0 at the origin and first segment North so their y(n) is a(n+1).  In their table of sides, it can be convenient to take n-4*k^2 so the ranges split at -m, 0, m. - Kevin Ryde, Sep 16 2019

			Here is the beginning of the clockwise square spiral. Sequence gives x-coordinate of the n-th point.
.
  20--21--22--23--24--25
   |                   |
  19   6---7---8---9  26
   |   |           |   |
  18   5   0---1  10  27
   |   |       |   |   |
  17   4---3---2  11  28
   |               |   |
  16--15--14--13--12  29
                       |
  35--34--33--32--32--30
.
Given the offset equal to 1, a(n) gives the x-coordinate of the point labeled n-1 in the above drawing. - _M. F. Hasler_, Nov 03 2019

Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, Addison-Wesley, 1989, chapter 3, Integer Functions, exercise 40 page 99 and answer page 498.

Peter Kagey, Table of n, a(n) for n = 1..10000
Seppo Mustonen, Ulam spiral in color [Interactive web page]
Seppo Mustonen, Ulam spiral in color [Local copy of a snapshot of the page]
Hugo Pfoertner, Visualization of spiral using Plot 2, May 29 2018
N. J. A. Sloane, Ulam spiral in color.
Aaron Snook, Augmented Integer Linear Recurrences, Thesis, 2012.
Index entries for sequences related to coordinates of 2D curves

Cf. A180714. A268038 (or A274923) gives sequence of y-coordinates.
The (x,y) coordinates for a point sweeping a quadrant by antidiagonals are (A025581, A002262). - N. J. A. Sloane, Jul 17 2018
See A296030 for the pairs (A174344(n), A274923(n)). - M. F. Hasler, Oct 20 2019
The diagonal rays are: A002939 (2*n*(2*n-1): 0, 2, 12, 30, ...), A016742 = (4n^2: 0, 4, 16, 36, ...), A002943 (2n(2n+1): 0, 6, 20, 42, ...), A033996 = (4n(n+1): 0, 8, 24, 48, ...). - M. F. Hasler, Oct 31 2019

function SquareSpiral(len)
    x, y, i, j, N, n, c = 0, 0, 0, 0, 0, 0, 0
    for k in 0:len-1
        print("$x, ") # or print("$y, ") for A268038.
        if n == 0
            c += 1; c > 3 && (c =  0)
            c == 0 && (i = 0; j =  1)
            c == 1 && (i = 1; j =  0)
            c == 2 && (i = 0; j = -1)
            c == 3 && (i = -1; j = 0)
            c in [1, 3] && (N += 1)
            n = N
        end
        n -= 1
        x, y = x + i, y + j
end end
SquareSpiral(75) # Peter Luschny, May 05 2019

fx:=proc(n) option remember; local k; if n=1 then 0 else
k:=floor(sqrt(4*(n-2)+1)) mod 4;
fx(n-1) + sin(k*Pi/2); fi; end;
[seq(fx(n),n=1..120)]; # Based on Seppo Mustonen's formula. - N. J. A. Sloane, Jul 11 2016

a[n_]:=a[n]=If[n==0,0,a[n-1]+Sin[Mod[Floor[Sqrt[4*(n-1)+1]],4]*Pi/2]]; Table[a[n],{n,0,50}] (* Seppo Mustonen, Aug 21 2010 *)

L=0; d=1;
for(r=1,9,d=-d;k=floor(r/2)*d;for(j=1,L++,print1(k,", "));forstep(j=k-d,-floor((r+1)/2)*d+d,-d,print1(j,", "))) \\ Hugo Pfoertner, Jul 28 2018

a(n) = n--; my(m=sqrtint(n),k=ceil(m/2)); n -= 4*k^2; if(n<0, if(n<-m, k, -k-n), if(nKevin Ryde, Sep 16 2019

apply( A174344(n)={my(m=sqrtint(n-=1), k=m\/2); if(n < 4*k^2-m, k, 0 > n -= 4*k^2, -k-n, n < m, -k, n-3*k)}, [1..99]) \\ M. F. Hasler, Oct 20 2019

# Based on Kevin Ryde's PARI script
import math
def A174344(n):
    n -= 1
    m = math.isqrt(n)
    k = math.ceil(m/2)
    n -= 4*k*k
    if n < 0: return k if n < -m else -k-n
    return -k if n < m else n-3*k # David Radcliffe, Aug 04 2025

a(1) = 0, a(n) = a(n-1) + sin(floor(sqrt(4n-7))*Pi/2). For a corresponding formula for the y-coordinate, replace sin with cos. - Seppo Mustonen, Aug 21 2010 with correction by Peter Kagey, Jan 24 2016

a(n) = A010751(A037458(n-1)) for n>1. - William McCarty, Jul 29 2021

Link corrected by Seppo Mustonen, Sep 05 2010

Definition clarified by N. J. A. Sloane, Dec 20 2012

A274923 List of y-coordinates of point moving in counterclockwise square spiral.

Original entry on oeis.org

0, 0, 1, 1, 1, 0, -1, -1, -1, -1, 0, 1, 2, 2, 2, 2, 2, 1, 0, -1, -2, -2, -2, -2, -2, -2, -1, 0, 1, 2, 3, 3, 3, 3, 3, 3, 3, 2, 1, 0, -1, -2, -3, -3, -3, -3, -3, -3, -3, -3, -2, -1, 0, 1, 2, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 2, 1, 0, -1, -2, -3, -4, -4, -4, -4, -4, -4, -4, -4, -4, -4, -3, -2, -1, 0

Offset: 1

N. J. A. Sloane, Jul 11 2016

This spiral, in either direction, is sometimes called the "Ulam spiral, but "square spiral" is a better name. (Ulam looked at the positions of the primes, but of course the spiral itself must be much older.) - N. J. A. Sloane, Jul 17 2018

Graham, Knuth and Patashnik give an exercise and answer on mapping n to square spiral x,y coordinates, and back x,y to n. They start 0 at the origin and first segment North so a(n) is their -x(n-1). In their table of sides, it can be convenient to take n-4*k^2 so the ranges split at -m, 0, m. - Kevin Ryde, Sep 17 2019

Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, Addison-Wesley, 1989, chapter 3, Integer Functions, exercise 40 page 99 and answer page 498.

Alois P. Heinz, Table of n, a(n) for n = 1..10000
Visualization of spiral using Plot 2. - _Hugo Pfoertner_, May 29 2018
Index entries for sequences related to coordinates of 2D curves

Cf. A268038 (negated), A317186 (indices of 0's).
Cf. A174344 (x-coordinates).
The (x,y) coordinates for a point sweeping a quadrant by antidiagonals are (A025581, A002262). - N. J. A. Sloane, Jul 17 2018
A296030 gives pairs (x = A174344(n), y = a(n)). - M. F. Hasler, Oct 20 2019
The diagonal rays of the square spiral (coordinates (+-n,+-n)) are: A002939 (2n(2n-1): 0, 2, 12, 30, ...), A016742 = (4n^2: 0, 4, 16, 36, ...), A002943 (2n(2n+1): 0, 6, 20, 42, ...), A033996 = (4n(n+1): 0, 8, 24, 48, ...). - M. F. Hasler, Oct 31 2019

fy:=proc(n) option remember; local k; if n=1 then 0 else
k:=floor(sqrt(4*(n-2)+1)) mod 4;
fy(n-1) - cos(k*Pi/2); fi; end;
[seq(fy(n),n=1..120)]; # Based on Seppo Mustonen's formula in A174344.

a[n_] := a[n] = If[n == 0, 0, a[n-1] - Cos[Mod[Floor[Sqrt[4*(n-1)+1]], 4]* Pi/2]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jun 11 2018, after Seppo Mustonen *)

L=1;d=1;
for(r=1,9,d=-d;k=floor(r/2)*d;for(j=1,L++,print1(k,", "));forstep(j=k-d,-floor((r+1)/2)*d+d,-d,print1(j,", "))) \\ Hugo Pfoertner, Jul 28 2018

a(n) = n--; my(m=sqrtint(n), k=ceil(m/2)); n -= 4*k^2; if(n<0, if(n<-m, 3*k+n, k), if(nKevin Ryde, Sep 17 2019

apply( A274923(n)={my(m=sqrtint(n-=1), k=m\/2); if(m <= n -= 4*k^2, -k, n >= 0, k-n, n >= -m, k, 3*k+n)}, [1..99]) \\ M. F. Hasler, Oct 20 2019

# Based on Kevin Ryde's PARI script
import math
def A274923(n):
    n -= 1
    m = math.isqrt(n)
    k = math.ceil(m/2)
    n -= 4*k*k
    if n < 0: return 3*k+n if n < -m else k
    return k-n if n < m else -k # David Radcliffe, Aug 04 2025

A085478 Triangle read by rows: T(n, k) = binomial(n + k, 2*k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 15, 7, 1, 1, 15, 35, 28, 9, 1, 1, 21, 70, 84, 45, 11, 1, 1, 28, 126, 210, 165, 66, 13, 1, 1, 36, 210, 462, 495, 286, 91, 15, 1, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1

Offset: 0

Philippe Deléham, Aug 14 2003

Coefficient array for Morgan-Voyce polynomial b(n,x). A053122 (unsigned) is the coefficient array for B(n,x). Reversal of A054142. - Paul Barry, Jan 19 2004
This triangle is formed from even-numbered rows of triangle A011973 read in reverse order. - Philippe Deléham, Feb 16 2004
T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k+1 peaks. T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k peaks at height >= 2. T(n,k) is the number of directed column-convex polyominoes of area n+1, having k+1 columns. - Emeric Deutsch, May 31 2004
Riordan array (1/(1-x), x/(1-x)^2). - Paul Barry, May 09 2005
The triangular matrix a(n,k) = (-1)^(n+k)*T(n,k) is the matrix inverse of A039599. - Philippe Deléham, May 26 2005
The n-th row gives absolute values of coefficients of reciprocal of g.f. of bottom-line of n-wave sequence. - Floor van Lamoen (fvlamoen(AT)planet.nl), Sep 24 2006
Unsigned version of A129818. - Philippe Deléham, Oct 25 2007
T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k >=1 (height(alpha) = |Im(alpha)|) and of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Oct 02 2008
A085478 is jointly generated with A078812 as a triangular array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n>1, u(n,x) = u(n-1,x)+x*v(n-1)x and v(n,x) = u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012
Per Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016
Subtriangle of the triangle given by (0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 26 2012
T(n,k) is also the number of compositions (ordered partitions) of 2*n+1 into 2*k+1 parts which are all odd. Proof: The o.g.f. of column k, x^k/(1-x)^(2*k+1) for k >= 0, is the o.g.f. of the odd-indexed members of the sequence with o.g.f. (x/(1-x^2))^(2*k+1) (bisection, odd part). Thus T(n,k) is obtained from the sum of the multinomial numbers A048996 for the partitions of 2*n+1 into 2*k+1 parts, all of which are odd. E.g., T(3,1) = 3 + 3 from the numbers for the partitions [1,1,5] and [1,3,3], namely 3!/(2!*1!) and 3!/(1!*2!), respectively. The number triangle with the number of these partitions as entries is A152157. - Wolfdieter Lang, Jul 09 2012
The matrix elements of the inverse are T^(-1)(n,k) = (-1)^(n+k)*A039599(n,k). - R. J. Mathar, Mar 12 2013
T(n,k) = A258993(n+1,k) for k = 0..n-1. - Reinhard Zumkeller, Jun 22 2015
The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the algebraic function F(x)*G(x)^n about 0, where F(x) = (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) and G(x) = ((1 + sqrt(1 + 4*x))/2)^2. For example, for n = 4, (1 + sqrt(1 + 4*x))/(2*sqrt(1 + 4*x)) * ((1 + sqrt(1 + 4*x))/2)^8 = (x^4  + 10*x^3 + 15*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 23 2018
Row n also gives the coefficients of the characteristc polynomial of the tridiagonal n X n matrix M_n given in A332602: Phi(n, x) := Det(M_n - x*1_n) = Sum_{k=0..n} T(n, k)*(-x)^k, for n >= 0, with Phi(0, x) := 1. - Wolfdieter Lang, Mar 25 2020
It appears that the largest root of the n-th degree polynomial is equal to the sum of the distinct diagonals of a (2*n+1)-gon including the edge, 1. The largest root of x^3 - 6*x^2 + 5*x - 1 is 5.048917... = the sum of (1 + 1.80193... + 2.24697...). Alternatively, the largest root of the n-th degree polynomial is equal to the square of sigma(2*n+1). Check: 5.048917... is the square of sigma(7), 2.24697.... Given N = 2*n+1, sigma(N) (N odd) can be defined as 1/(2*sin(Pi/(2*N))). Relating to the 9-gon, the largest root of x^4 - 10*x^3 + 15*x^2 - 7*x + 1 is 8.290859..., = the sum of (1 + 1.879385... + 2.532088... + 2.879385...), and is the square of sigma(9), 2.879385... Refer to A231187 for a further clarification of sigma(7). - Gary W. Adamson, Jun 28 2022
For n >=1, the n-th row is given by the coefficients of the minimal polynomial of -4*sin(Pi/(4*n + 2))^2. - Eric W. Weisstein, Jul 12 2023
Denoting this lower triangular array by L, then L * diag(binomial(2*k,k)^2) * transpose(L) is the LDU factorization of A143007, the square array of crystal ball sequences for the A_n X A_n lattices. - Peter Bala, Feb 06 2024
T(n, k) is the number of occurrences of the periodic substring (01)^k in the periodic string (01)^n (see Proposition 4.7 at page 7 in Fang). - Stefano Spezia, Jun 09 2024

			Triangle begins as:
  1;
  1    1;
  1    3    1;
  1    6    5    1;
  1   10   15    7    1;
  1   15   35   28    9    1;
  1   21   70   84   45   11    1;
  1   28  126  210  165   66   13    1;
  1   36  210  462  495  286   91   15    1;
  1   45  330  924 1287 1001  455  120   17    1;
  1   55  495 1716 3003 3003 1820  680  153   19    1;
...
From _Philippe Deléham_, Mar 26 2012: (Start)
(0, 1, 0, 1, 0, 0, 0, ...) DELTA (1, 0, 1, -1, 0, 0, 0, ...) begins:
  1
  0, 1
  0, 1,  1
  0, 1,  3,   1
  0, 1,  6,   5,   1
  0, 1, 10,  15,   7,   1
  0, 1, 15,  35,  28,   9,  1
  0, 1, 21,  70,  84,  45, 11,  1
  0, 1, 28, 126, 210, 165, 66, 13, 1. (End)

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
J.-P. Allouche and M. Mendès France, Stern-Brocot polynomials and power series, arXiv preprint arXiv:1202.0211 [math.NT], 2012. - _N. J. A. Sloane_, May 10 2012
Peter Bala, A 4-parameter family of embedded Riordan arrays
E. Barcucci, A. Del Lungo, S. Fezzi, and R. Pinzani, Nondecreasing Dyck paths and q-Fibonacci numbers, Discrete Math., 170, 1997, 211-217.
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, J. Integer Sequ., Vol. 8 (2005), Article 05.4.5.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Paul Barry, Notes on the Hankel transform of linear combinations of consecutive pairs of Catalan numbers, arXiv:2011.10827 [math.CO], 2020.
Paul Barry, The second production matrix of a Riordan array, arXiv:2011.13985 [math.CO], 2020.
Paul Barry and Aoife Hennessy, Notes on a Family of Riordan Arrays and Associated Integer Hankel Transforms , JIS 12 (2009) 09.5.3.
Eduardo H. M. Brietzke, Recurrence Relations for Sums of Binomial Coefficients and Some Generalizations, Journal of Integer Sequences, Vol. 27 (2024), Article 24.3.4.
E. Czabarka et al, Enumerations of peaks and valleys on non-decreasing Dyck paths, Disc. Math. 341 (2018) 2789-2807, Theorem 3.
Emeric Deutsch and H. Prodinger, A bijection between directed column-convex polyominoes and ordered trees of height at most three, Theoretical Comp. Science, 307, 2003, 319-325.
James East and Nicholas Ham, Lattice paths and submonoids of Z^2, arXiv:1811.05735 [math.CO], 2018.
Wenjie Fang, Maximal number of subword occurrences in a word, arXiv:2406.02971 [math.CO], 2024.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving full transformations, Semigroup Forum 72 (2006), 51-62. - _Abdullahi Umar_, Oct 02 2008
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
Yidong Sun, Numerical Triangles and Several Classical Sequences, Fib. Quart. 43, no. 4, (2005) 359-370.
Eric Weisstein's World of Mathematics, Morgan-Voyce Polynomials

Row sums: A001519. Signed versions: A123970, A129818.
Cf. A000108, A001006, A007318, A098158, A183160, A078812, A091042.
Cf. A258993, A025581, A051162, A054142, A332602.
Cf. A231187, A143007.

Flat(List([0..12], n-> List([0..n], k-> Binomial(n+k, 2*k) ))); # G. C. Greubel, Aug 01 2019

a085478 n k = a085478_tabl !! n !! k
a085478_row n = a085478_tabl !! n
a085478_tabl = zipWith (zipWith a007318) a051162_tabl a025581_tabl
-- Reinhard Zumkeller, Jun 22 2015

[Binomial(n+k, 2*k): k in [0..n], n in [0..12]]; // G. C. Greubel, Aug 01 2019

T := (n,k) -> binomial(n+k,2*k): seq(seq(T(n,k), k=0..n), n=0..11);

(* First program *)
u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x];
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A085478 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A078812 *) (*Clark Kimberling, Feb 25 2012 *)
(* Second program *)
Table[Binomial[n + k, 2 k], {n, 0, 12}, {k, 0, n}] // Flatten (* G. C. Greubel, Aug 01 2019 *)
CoefficientList[Table[Fibonacci[2 n + 1, Sqrt[x]], {n, 0, 10}], x] // Flatten (* Eric W. Weisstein, Jul 03 2023 *)
Join[{{1}}, CoefficientList[Table[MinimalPolynomial[-4 Sin[Pi/(4 n + 2)]^2, x], {n, 20}], x]] (* Eric W. Weisstein, Jul 12 2023 *)

PARI
```
T(n,k) = binomial(n+k,n-k)
```

[[binomial(n+k,2*k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Aug 01 2019

T(n, k) = (n+k)!/((n-k)!*(2*k)!).
G.f.: (1-z)/((1-z)^2-tz). - Emeric Deutsch, May 31 2004
Row sums are A001519 (Fibonacci(2n+1)). Diagonal sums are A011782. Binomial transform of A026729 (product of lower triangular matrices). - Paul Barry, Jun 21 2004
T(n, 0) = 1, T(n, k) = 0 if n=0} T(n-1-j, k-1)*(j+1). T(0, 0) = 1, T(0, k) = 0 if k>0; T(n, k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j). For the column k, g.f.: Sum_{n>=0} T(n, k)*x^n = (x^k) / (1-x)^(2*k+1). - Philippe Deléham, Feb 15 2004
Sum_{k=0..n} T(n,k)*x^(2*k) = A000012(n), A001519(n+1), A001653(n), A078922(n+1), A007805(n), A097835(n), A097315(n), A097838(n), A078988(n), A097841(n), A097727(n), A097843(n), A097730(n), A098244(n), A097733(n), A098247(n), A097736(n), A098250(n), A097739(n), A098253(n), A097742(n), A098256(n), A097767(n), A098259(n), A097770(n), A098262(n), A097773(n), A098292(n), A097776(n) for x=0,1,2,...,27,28 respectively. - Philippe Deléham, Dec 31 2007
T(2*n,n) = A005809(n). - Philippe Deléham, Sep 17 2009
A183160(n) = Sum_{k=0..n} T(n,k)*T(n,n-k). - Paul D. Hanna, Dec 27 2010
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Feb 06 2012
O.g.f. for column k: x^k/(1-x)^(2*k+1), k >= 0. [See the o.g.f. of the triangle above, and a comment on compositions. - Wolfdieter Lang, Jul 09 2012]
E.g.f.: (2/sqrt(x + 4))*sinh((1/2)*t*sqrt(x + 4))*cosh((1/2)*t*sqrt(x)) = t + (1 + x)*t^3/3! + (1 + 3*x + x^2)*t^5/5! + (1 + 6*x + 5*x^2 + x^3)*t^7/7! + .... Cf. A091042. - Peter Bala, Jul 29 2013
T(n, k) = A065941(n+3*k, 4*k) = A108299(n+3*k, 4*k) = A194005(n+3*k, 4*k). - Johannes W. Meijer, Sep 05 2013
Sum_{k=0..n} (-1)^k*T(n,k)*A000108(k) = A000007(n) for n >= 0. - Werner Schulte, Jul 12 2017
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A001006(n) for n >= 0. - Werner Schulte, Jul 12 2017
From  Peter Bala, Jun 26 2025: (Start)
The n-th row polynomial b(n, x) = (-1)^n * U(2*n, (i/2)*sqrt(x)), where U(n,x) is the n-th Chebyshev polynomial of the second kind.
b(n, x) = (-1)^n * Dir(n, -1 - x/2), where Dir(n, x) is the n-th row polynomial of the triangle A244419.
b(n, -1 - x) is the n-th row polynomial of A098493. (End)

A099884 XOR difference triangle of the powers of 2, read by rows; Square array A(row,col): A(0,col) = 2^col, A(row,col) = A048724(A(row-1, col)) for row > 0, read by descending antidiagonals.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 8, 12, 10, 15, 16, 24, 20, 30, 17, 32, 48, 40, 60, 34, 51, 64, 96, 80, 120, 68, 102, 85, 128, 192, 160, 240, 136, 204, 170, 255, 256, 384, 320, 480, 272, 408, 340, 510, 257, 512, 768, 640, 960, 544, 816, 680, 1020, 514, 771, 1024, 1536, 1280, 1920

Offset: 0

Paul D. Hanna, Oct 28 2004

Define an "XOR difference triangle" for a sequence A by the following process. Start with A in the leftmost column. Generate the next column by performing the XOR operation between adjacent terms of the prior column. Repeat this process to generate the XOR difference triangle for A. Further, we define the "XOR BINOMIAL transform" of A as the main diagonal in the XOR difference triangle for A. The XOR BINOMIAL transform is its self-inverse. Let a sequence B be the XOR BINOMIAL transform of A, then we may express B by: B(n) = SumXOR_{k=0..n} A047999(n,k)*A(k), which is equivalent to: B(n) = (C(n,0)mod 2)*A(0) XOR (C(n,1)mod 2)*A(1) XOR (C(n,2)mod 2)*A(2) XOR ... XOR (X(n,n)mod 2)*A(n), where the coefficients are C(n,k)(mod 2) = A047999(n,k).

This sequence is a rearrangement of the numbers which are 2^k times distinct Fermat numbers (numbers of the form 2^(2^m) + 1). This matches the sizes of polygons constructible with compass and straightedge (A003401) up to 2^32+1, which is the first nonprime Fermat number. - Franklin T. Adams-Watters, Jun 16 2006

			The main diagonal equals A001317 (Pascal's triangle mod 2 in decimal):
{1,3,5,15,17,51,85,255,257,771,1285,3855,...}, and defines the XOR BINOMIAL transform of the powers of 2.
Rows begin:
  1;
  2, 3;
  4, 6, 5;
  8, 12, 10, 15;
  16, 24, 20, 30, 17;
  32, 48, 40, 60, 34, 51;
  64, 96, 80, 120, 68, 102, 85;
  128, 192, 160, 240, 136, 204, 170, 255;
  256, 384, 320, 480, 272, 408, 340, 510, 257;
  512, 768, 640, 960, 544, 816, 680, 1020, 514, 771;
  1024, 1536, 1280, 1920, 1088, 1632, 1360, 2040, 1028, 1542, 1285;
  2048, 3072, 2560, 3840, 2176, 3264, 2720, 4080, 2056, 3084, 2570, 3855;
  ...
From _Antti Karttunen_, Sep 19 2016: (Start)
Viewed as a square array, the top left corner looks like this:
     1,    2,     4,     8,    16,     32,     64,    128
     3,    6,    12,    24,    48,     96,    192,    384
     5,   10,    20,    40,    80,    160,    320,    640
    15,   30,    60,   120,   240,    480,    960,   1920
    17,   34,    68,   136,   272,    544,   1088,   2176
    51,  102,   204,   408,   816,   1632,   3264,   6528
    85,  170,   340,   680,  1360,   2720,   5440,  10880
   255,  510,  1020,  2040,  4080,   8160,  16320,  32640
   257,  514,  1028,  2056,  4112,   8224,  16448,  32896
   771, 1542,  3084,  6168, 12336,  24672,  49344,  98688
  1285, 2570,  5140, 10280, 20560,  41120,  82240, 164480
  3855, 7710, 15420, 30840, 61680, 123360, 246720, 493440
  4369, 8738, 17476, 34952, 69904, 139808, 279616, 559232
  ...
(End)
The square array shown above can be viewed as a subtable of a multiplication table with particular relevance to the carryless multiplication defined by A048720, as the first column gives the A048720 powers of 3 (and the first row gives powers of 2, which are the same as in standard arithmetic). - _Peter Munn_, Jan 13 2020

Paul D. Hanna, First 45 Rows of Triangle, in flattened form.
SeqFan-mailing list, Discussion of about related array A255483
Index entries for sequences related to polynomials in ring GF(2)[X]

Essentially GF(2)[X] analog of table A036561. - Antti Karttunen, Jan 18 2020
Cf. A047999, A158875 (row sums).
Cf. A000215, A003401, A048675, A048720, A048724, A066117, A248663, A255483, A276586.
Cf. A000079 (first column of triangular table, the topmost row of square array).
Cf. A001317 (the rightmost diagonal of triangular table, the leftmost column of square array).
Cf. A099885, A117998 (central diagonals).
Cf. A276618 (transpose), A091202, A193231.

a[n_]:= Sum[Mod[Binomial[n, i], 2]*2^i, {i, 0, n}]; T[n_, k_]:=2^(n - k)a[k]; Table[T[n, k], {n, 0, 20}, {k, 0, n}] // Flatten (* Indranil Ghosh, Apr 11 2017 *)

{T(n,k)=local(B);B=0;for(i=0,k,B=bitxor(B,binomial(k,i)%2*2^(n-i)));B}
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

from sympy import binomial
def a(n):
    return sum((binomial(n, i)%2)*2**i for i in range(n + 1))
def T(n, k): return 2**(n - k)*a(k)
for n in range(21): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017

(define (A099884 n) (A099884bi (A002262 n) (A025581 n)))
;; Then use either this recurrence:
(define (A099884bi row col) (if (zero? row) (A000079 col) (A048724 (A099884bi (- row 1) col))))
;; or this one:
(define (A099884bi row col) (if (zero? col) (A001317 row) (* 2 (A099884bi row (- col 1)))))
;; Antti Karttunen, Sep 19 2016

T(n, k) = 2^(n-k)*A001317(k). T(n, n) = A001317(n) = SumXOR_{k=0..n} A047999(n, k)*2^k, where SumXOR is the analog of summation under the binary XOR operation.
From Antti Karttunen, Sep 19 2016: (Start)
When viewed as a square array A(row,col), with row >= 0, col >= 0, the following recurrences and formulas are valid:
A(0,col) = A000079(col), for row > 0, A(row,col) = A048724(A(row-1, col)).
A(row,0) = A001317(row), for col > 0, A(row,col) = 2*A(row,col-1).
A(row,col) = A248663(A066117(row+1,col+1)) = A048675(A255483(row,col+1)).
(End)
With the definitions from Antti Karttunen above, A(row+1, col) = A048720(3, A(row, col)). - Peter Munn, Jan 13 2020
A(n,k) = A193231(A(k,n)) = A091202(A036561(n,k)). - Antti Karttunen, Jan 18 2020

Square array interpretation added as a second, alternative description by Antti Karttunen, Sep 19 2016

A051775 Table T(n,m) = Nim-product of n and m, read by antidiagonals, for n >= 0, m >= 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 3, 3, 0, 0, 4, 1, 1, 4, 0, 0, 5, 8, 2, 8, 5, 0, 0, 6, 10, 12, 12, 10, 6, 0, 0, 7, 11, 15, 6, 15, 11, 7, 0, 0, 8, 9, 13, 2, 2, 13, 9, 8, 0, 0, 9, 12, 14, 14, 7, 14, 14, 12, 9, 0, 0, 10, 14, 4, 10, 8, 8, 10, 4, 14, 10, 0, 0, 11, 15, 7, 11

Offset: 0

N. J. A. Sloane, Dec 19 1999

Note on an algorithm, R. J. Mathar, May 29 2011: (Start)
Let N* denote the Nim-product and N+ the Nim-sum (A003987) of two numbers, and let * and + denote the usual multiplication and addition.
To compute n N* m, write n and m separately as Nim-sums with the aid of the binary representation of n = n0 + n1*2 + n2*4 + n3*8 + n4*16.. and m = m0 + m1*2 + m2*4 + m3*8 + m4*16... . Because Nim-summation is the same as the binary XOR-function, the + may then be replaced by N+ in both sums:
n = Nim-sum_i 2^a(i) and m = Nim-sum_j 2^b(j) with two integer sequences a(i) and b(j).
Because N+ and N* are the operations in a field, N+ and N* are distributive, which is used to write the product over the sums as a double-Nim-sum over Nim-products:
n N* m = Nim-sum_{i,j} 2^a(i) N* 2^b(j) .
What remains is to compute the Nim-products of powers of 2.
Splitting a(i) and b(j) separately into (ordinary) products of Fermat numbers A001146 (i.e., writing a(i) and b(j) in binary), and noting that the ordinary product of distinct Fermat numbers equals the Nim-product of distinct Fermat numbers,
2^a(i) N* 2^b(j) = 2^(2^A0) N* 2^(2^A1) N* ... N* 2^(2^B0) N* 2^(2^B1) N* ... for two binary integer sequences A and B.
This finite product is regrouped by pairing the cases for the same bit in the A-sequence and in the B-sequence. If the bit is set in both sequences, use that the Nim-square of a Fermat number is 3/2 times (ordinary multiple of) that Fermat number; if the bit is set only in one of the two sequences, use (again) that the Nim-product of distinct Fermat numbers is the ordinary product.
Due to the potential presence of the Nim-squares, this leaves in general a Nim-product which is treated by recursion.
This algorithm is implemented in the Maple program in the b-file. nimprodP2() calculates the Nim-product of two powers of 2. (End)

			The table begins:
  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 ...
  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 ...
  0  2  3  1  8 10 11  9 12 14 15 13  4  6  7  5 ...
  0  3  1  2 12 15 13 14  4  7  5  6  8 11  9 10 ...
  0  4  8 12  6  2 14 10 11 15  3  7 13  9  5  1 ...
  0  5 10 15  2  7  8 13  3  6  9 12  1  4 11 14 ...
  (...)

J. H. Conway, On Numbers and Games, Academic Press, p. 52.

R. J. Mathar, Table of n, a(n) for n = 0..1890
Tilman Piesk, 256x256 table and dual matrix
Rémy Sigrist, Colored representation of T(x,y) for x = 0..1023 and y = 0..1023 (where the hue is function of T(x,y))
Wikipedia, Nimber
Index entries for sequences related to Nim-multiplication

Cf. A051776, A003987.

We continue from A003987: to compute a Nim-multiplication table using (a) an addition table AT := array(0..NA, 0..NA) and (b) a nimsum procedure for larger values; MT := array(0..N,0..N); for a from 0 to N do MT[a,0] := 0; MT[0,a] := 0; MT[a,1] := a; MT[1,a] := a; od: for a from 2 to N do for b from a to N do t1 := {}; for i from 0 to a-1 do for j from 0 to b-1 do u1 := MT[i,b]; u2 := MT[a,j];
if u1<=NA and u2<=NA then u12 := AT[u1,u2]; else u12 := nimsum(u1,u2); fi; u3 := MT[i,j]; if u12<=NA and u3<=NA then u4 := AT[u12,u3]; else u4 := nimsum(u12,u3); fi; t1 := { op(t1), u4}; #t1 := { op(t1), AT[ AT[ MT[i,b], MT[a,j] ], MT[i,j] ] }; od; od;
t2 := sort(convert(t1,list)); j := nops(t2); for i from 1 to nops(t2) do if t2[i] <> i-1 then j := i-1; break; fi; od; MT[a,b] := j; MT[b,a] := j; od; od;

NP_table=Map(); NP(x,y)={ if(x<2 || y<2, x*y, mapisdefined(NP_table, if(y>x, [x,y]=[y,x], [x,y])), mapget(NP_table,[x,y]), x==3, y-1, x==2, 3, my(F=4); until(!F *= F, if(x<2*F, F=if(x>F, bitxor(NP(F,y), NP(x-F,y)), yi, bitxor(NP(t,i), NP(t,y-i)), NP(F\2*3, NP(t/F,i/F))); break(3))); if(y==t, F=NP(F\2*3, NP(t/F,t/F)); break(2))); if(x<2*t, F=bitxor(NP(t,y), NP(x-t,y)); break(2)))); mapput(NP_table,[x,y], F); F)} \\ M. F. Hasler, Jan 18 2021
A051775(n,m="")={if(m!="", NP(n,m), NP((1+m=sqrtint(8*n+1)\/2)*m/2-n-1, n-m*(m-1)/2))} \\ Then A051775(n) = a(n) [flattened sequence, cf. A025581 & A002262], A051775(n,m) = T(n,m): for example, {matrix(6,15,m,n, A051775(m,n))} - M. F. Hasler, Jan 22 2021

A004247 Multiplication table read by antidiagonals: T(i,j) = ij (i>=0, j>=0). Alternatively, multiplication triangle read by rows: P(i,j) = j(i-j) (i>=0, 0<=j<=i).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 4, 3, 0, 0, 4, 6, 6, 4, 0, 0, 5, 8, 9, 8, 5, 0, 0, 6, 10, 12, 12, 10, 6, 0, 0, 7, 12, 15, 16, 15, 12, 7, 0, 0, 8, 14, 18, 20, 20, 18, 14, 8, 0, 0, 9, 16, 21, 24, 25, 24, 21, 16, 9, 0, 0, 10, 18, 24, 28, 30, 30, 28, 24, 18, 10, 0, 0, 11, 20, 27, 32, 35, 36, 35, 32, 27, 20, 11, 0, 0, 12, 22, 30, 36, 40, 42, 42, 40, 36, 30

Offset: 0

David W. Wilson

Table of x*y, where (x,y) = (0,0),(0,1),(1,0),(0,2),(1,1),(2,0),...
Or, triangle read by rows, in which row n gives the numbers 0, n*1, (n-1)*2, (n-2)*3, ..., 2*(n-1), 1*n, 0.
Letting T(n,k) be the (k+1)st entry in the (n+1)st row (same numbering used for Pascal's triangle), T(n,k) is the dimension of the space of all k-dimensional subspaces of a (fixed) n-dimensional real vector space. - Paul Boddington, Oct 21 2003
From Dennis P. Walsh, Nov 10 2009: (Start)
Triangle P(n,k), 0<=k<=n, equals n^2 x the variance of a binary data set with k zeros and (n-k) ones. [For the case when n=0, let the variance of the empty set be defined as 0.]
P(n,k) is also the number of ways to form an opposite-sex dance couple from k women and (n-k) men. (End)
P(n,k) is the number of negative products of two numbers from a set of n real numbers, k of which are negative. - Logan Pipes, Jul 08 2021

			As the triangle P, sequence begins:
  0;
  0,0;
  0,1,0;
  0,2,2,0;
  0,3,4,3,0;
  0,4,6,6,4,0,;
  0,5,8,9,8,5,0;
  ...
From _Dennis P. Walsh_, Nov 10 2009: (Start)
P(5,2)=T(2,3)=6 since the variance of the data set <0,0,1,1,1> equals 6/25.
P(5,2)=6 since, with 2 women, say Alice and Betty, and with 3 men, say Charles, Dennis, and Ed, the dance couple is one of the following: {Alice, Charles}, {Alice, Dennis}, {Alice, Ed}, {Betty, Charles}, {Betty, Dennis} and {Betty, Ed}. (End)

T. D. Noe, Rows n = 0..50 of triangle, flattened
Dennis Walsh, Variance bounds on binary data sets

See A003991 for another version with many more comments.

Cf. A002262, A025581, A003056, A004197, A003984, A048720, A325820, A000292 (row sums of triangle), A002620.

seq(seq(k*(n-k),k=0..n),n=0..13); # Dennis P. Walsh, Nov 10 2009

Table[(x - y) y, {x, 0, 13}, {y, 0, x}] // Flatten (* Robert G. Wilson v, Oct 06 2007 *)

T(i,j)=i*j \\ Charles R Greathouse IV, Jun 23 2017

a(n) = A002262(n) * A025581(n). - Antti Karttunen
From Ridouane Oudra, Dec 14 2019: (Start)
a(n) = A004197(n)*A003984(n).
a(n) = (3/4 + n)*t^2 - (1/4)*t^4 - (1/2)*t - n^2 - n, where t = floor(sqrt(2*n+1)+1/2). (End)
P(n,k) = (P(n-1,k-1) + P(n-1,k) + n) / 2. - Robert FERREOL, Jan 16 2020
P(n,floor(n/2)) = A002620(n). - Logan Pipes, Jul 08 2021
From Stefano Spezia, Aug 19 2024: (Start)
G.f. as array: x*y/((1 - x)^2*(1 - y)^2).
E.g.f. as array: exp(x+y)*x*y. (End)

Edited by N. J. A. Sloane, Sep 30 2007

A056558 Third tetrahedral coordinate, i.e., tetrahedron with T(t,n,k)=k; succession of growing finite triangles with increasing values towards bottom right.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 0, 1, 2, 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 6, 0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5

Offset: 0

Henry Bottomley, Jun 26 2000

nonn

Alternatively, write n = C(i,3)+C(j,2)+C(k,1) with i>j>k>=0; sequence gives k values. See A194847 for further information about this interpretation.
If {(X,Y,Z)} are triples of nonnegative integers with X>=Y>=Z ordered by X, Y and Z, then X=A056556(n), Y=A056557(n) and Z=A056558(n).
This is a 'Matryoshka doll' sequence with alpha=0 (cf. A000292 and A000178). - Peter Luschny, Jul 14 2009

			First triangle: [0]; second triangle: [0; 0 1]; third triangle: [0; 0 1; 0 1 2]; ...

D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.3, Eq. (20), p. 360.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Together with A056559 and A056560 might enable reading "by antidiagonals" of cube arrays as 3-dimensional analog of A002262 and A025581 with square arrays. Also cf. A000292, A056556, A056557.
See also A194847, A194848, A194849.
Cf. A002262, A127324, A000217.

import Data.List (inits)
a056558 n = a056558_list !! n
a056558_list = concatMap (concat . init . inits . enumFromTo 0) [0..]
-- Reinhard Zumkeller, Jun 01 2015

seq(seq(seq(i,i=0..k),k=0..n),n=0..6); # Peter Luschny, Sep 22 2011

Table[i, {k, 0, 7}, {j, 0, k}, {i, 0, j}] // Flatten  (* Robert G. Wilson v, Sep 27 2011 *)

T(t,n,k)=k \\ Charles R Greathouse IV, Feb 22 2017

from math import isqrt, comb
from sympy import integer_nthroot
def A056558(n): return (r:=n-comb((m:=integer_nthroot(6*(n+1),3)[0])+(n>=comb(m+2,3))+1,3))-comb((k:=isqrt(m:=r+1<<1))+(m>k*(k+1)),2) # Chai Wah Wu, Nov 04 2024

a(n) = n-A056556(n)*(A056556(n)+1)*(A056556(n)+2)/6-A056557(n)*(A056557(n)+1)/2 = n-A000292(A056556(n)-1)-A000217(A056557(n)) = A056557(n)-A056560(n).

a(n+1) = A056556(n)==a(n) ? 0 : A056557(n)==a(n) ? 0 : a(n)+1. - Graeme McRae, Jan 09 2007

A004070 Table of Whitney numbers W(n,k) read by antidiagonals, where W(n,k) is maximal number of pieces into which n-space is sliced by k hyperplanes, n >= 0, k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 2, 4, 4, 1, 1, 2, 4, 7, 5, 1, 1, 2, 4, 8, 11, 6, 1, 1, 2, 4, 8, 15, 16, 7, 1, 1, 2, 4, 8, 16, 26, 22, 8, 1, 1, 2, 4, 8, 16, 31, 42, 29, 9, 1, 1, 2, 4, 8, 16, 32, 57, 64, 37, 10, 1, 1, 2, 4, 8, 16, 32, 63, 99, 93, 46, 11, 1, 1, 2, 4, 8, 16, 32, 64, 120, 163

Offset: 0

N. J. A. Sloane

As a number triangle, this is given by T(n,k)=sum{j=0..n, C(n,j)(-1)^(n-j)sum{i=0..j, C(j+k,i-k)}}. - Paul Barry, Aug 23 2004
As a number triangle, this is the Riordan array (1/(1-x), x(1+x)) with T(n,k)=sum{i=0..n, binomial(k,i-k)}. Diagonal sums are then A023434(n+1). - Paul Barry, Feb 16 2005
Form partial sums across rows of square array of binomial coefficients A026729; see also A008949. - Philippe Deléham, Aug 28 2005
Square array A026729 -> Partial sums across rows
1 0 0 0 0 0 0 . . . . 1 1 1 1 1 1 1 . . . . . .
1 1 0 0 0 0 0 . . . . 1 2 2 2 2 2 2 . . . . . .
1 2 1 0 0 0 0 . . . . 1 3 4 4 4 4 4 . . . . . .
1 3 3 1 0 0 0 . . . . 1 4 7 8 8 8 8 . . . . . .
For other Whitney numbers see A007799.
W(n,k) is the number of length k binary sequences containing no more than n 1's. - Geoffrey Critzer, Mar 15 2010
From Emeric Deutsch, Jun 15 2010: (Start)
Viewed as a number triangle, T(n,k) is the number of internal nodes of the Fibonacci tree of order n+2 at level k. A Fibonacci tree of order n (n>=2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node.
(End)
Named after the American mathematician Hassler Whitney (1907-1989). - Amiram Eldar, Jun 13 2021

			Table W(n,k) begins:
  1 1 1 1  1  1  1 ...
  1 2 3 4  5  6  7 ...
  1 2 4 7 11 16 22 ...
  1 2 4 8 15 26 42 ...
W(2,4) = 11 because there are 11 length 4 binary sequences containing no more than 2 1's: {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 1, 0}, {0, 0, 1, 1}, {0, 1, 0, 0}, {0, 1, 0, 1}, {0, 1, 1, 0}, {1, 0, 0, 0}, {1, 0, 0, 1}, {1, 0, 1, 0}, {1, 1, 0, 0}. - _Geoffrey Critzer_, Mar 15 2010
Table T(n, k) begins:
  1
  1  1
  1  2  1
  1  2  3  1
  1  2  4  4  1
  1  2  4  7  5  1
  1  2  4  8 11  6  1
...

Donald E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.

Gustav Burosch, Hans-Dietrich O.F. Gronau, Jean-Marie Laborde and Ingo Warnke, On posets of m-ary words, Discrete Math., Vol. 152, No. 1-3 (1996), pp. 69-91. MR1388633 (97e:06002)
Matteo Cervetti and Luca Ferrari, Pattern avoidance in the matching pattern poset, arXiv:2009.01024 [math.CO], 2020.
Matteo Cervetti and Luca Ferrari, Enumeration of Some Classes of Pattern Avoiding Matchings, with a Glimpse into the Matching Pattern Poset, Annals of Combinatorics (2022).
Richard K. Guy, Letter to N. J. A. Sloane, Apr 1975.
Yasuichi Horibe, An entropy view of Fibonacci trees, Fibonacci Quarterly, Vol. 20, No. 2 (1982), pp. 168-178.
Robin Pemantle and Mark C. Wilson, Twenty Combinatorial Examples of Asymptotics Derived from Multivariate Generating Functions, SIAM Rev., Vol. 50, No. 2 (2008), pp. 199-272. See p. 233.

Cf. A007799. As a triangle, mirror A052509.
Rows converge to powers of two (A000079). Subdiagonals include A000225, A000295, A002662, A002663, A002664, A035038, A035039, A035040, A035041, A035042. Antidiagonal sums are A000071.
Rows are: A000012, A000027, A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862, A008863.
Cf. A178522, A178524.

Transpose[ Table[Table[Sum[Binomial[n, k], {k, 0, m}], {m, 0, 15}], {n, 0, 15}]] // Grid (* Geoffrey Critzer, Mar 15 2010 *)
T[ n_, k_] := Sum[ Binomial[n, j] (-1)^(n - j) Sum[ Binomial[j + k, i - k], {i, 0, j}], {j, 0, n}]; (* Michael Somos, May 31 2016 *)

/* array read by antidiagonals up coordinate index functions */
t1(n) = binomial(floor(3/2 + sqrt(2+2*n)), 2) - (n+1); /* A025581 */
t2(n) = n - binomial(floor(1/2 + sqrt(2+2*n)), 2); /* A002262 */
/* define the sequence array function for A004070 */
W(n, k) = sum(i=0, n, binomial(k, i));
/* visual check ( origin 0,0 ) */
printp(matrix(7, 7, n, k, W(n-1, k-1)));
/* print the sequence entries by antidiagonals going up ( origin 0,0 ) */
print1("S A004070 "); for(n=0, 32, print1(W(t1(n), t2(n))","));
print1("T A004070 "); for(n=33, 61, print1(W(t1(n), t2(n))","));
print1("U A004070 "); for(n=62, 86, print1(W(t1(n), t2(n))",")); /* Michael Somos, Apr 28 2000 */

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

W(n, k) = Sum_{i=0..n} binomial(k, i). - Bill Gosper
W(n, k) = if k=0 or n=0 then 1 else W(n, k-1)+W(n-1, k-1). - David Broadhurst, Jan 05 2000
The table W(n,k) = A000012 * A007318(transform), where A000012 = (1; 1,1; 1,1,1; ...). - Gary W. Adamson, Nov 15 2007
E.g.f. for row n: (1 + x + x^2/2! + ... + x^n/n!)* exp(x). - Geoffrey Critzer, Mar 15 2010
G.f.: 1 / (1 - x - x*y*(1 - x^2)) = Sum_{0 <= k <= n} x^n * y^k * T(n, k). - Michael Somos, May 31 2016
W(n, n) = 2^n. - Michael Somos, May 31 2016
From Jianing Song, May 30 2022: (Start)
T(n, 0) = T(n, n) = 1 for n >= 0; T(n, k) = T(n-1, k-1) + T(n-2, k-1) for k=1, 2, ..., n-1, n >= 2.
T(n, k) = Sum_{m=0..n-k} binomial(k, m).
T(n,k) = 2^k for 0 <= k <= floor(n/2). (End)

More terms from Larry Reeves (larryr(AT)acm.org), Mar 20 2000

cp's OEIS Frontend

A002262 Triangle read by rows: T(n,k) = k, 0 <= k <= n, in which row n lists the first n+1 nonnegative integers.

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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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A174344 List of x-coordinates of point moving in clockwise square spiral.

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A274923 List of y-coordinates of point moving in counterclockwise square spiral.

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A085478 Triangle read by rows: T(n, k) = binomial(n + k, 2*k).

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A004247 Multiplication table read by antidiagonals: T(i,j) = ij (i>=0, j>=0). Alternatively, multiplication triangle read by rows: P(i,j) = j(i-j) (i>=0, 0<=j<=i).