cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A221645 Square array read by antidiagonals: T(n,k^2) = A040000(n) (= 1,2,2,2,...) if n=0 (mod k), T(n,k) = 0 else, n>=0, k>=1.

Original entry on oeis.org

1, 2, 0, 2, 0, 0, 2, 0, 0, 1, 2, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Omar E. Pol, Feb 02 2013

Keywords

Comments

Other definition: Square array read by antidiagonal: T(n,k), n>=0, k>=1, in which column k lists the numbers A040000 (1, 2, 2, 2, 2...) interleaved with k^(1/2)-1 zeros, if k is a square otherwise column k lists only zeros.
The sum of elements of the n-th antidiagonal equals the number of divisors of n. In other words, the antidiagonal sums give A000005.

Examples

			First 16 elements of first 16 rows of the square array are
1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,1,...
2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,...
2,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,...
2,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,...
2,0,0,2,0,0,0,0,0,0,0,0,0,0,0,2,...
2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,...
2,0,0,2,0,0,0,0,2,0,0,0,0,0,0,0,...
2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,...
2,0,0,2,0,0,0,0,0,0,0,0,0,0,0,2,...
2,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,...
2,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,...
2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,...
2,0,0,2,0,0,0,0,2,0,0,0,0,0,0,2,...
2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,...
2,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,...
2,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,...
...
For n = 3 the sum of the third antidiagonal is 2+0+0 = d(3) = A000005(3) = 2.
For n = 16 the sum of the 16th antidiagonal is 2+0+0+2+0+0+0+0+0+0+0+0+0+0+0+1 = d(16) = A000005(16) = 5.
		

Crossrefs

Programs

  • PARI
    A221645(n,m)={my(t); issquare(m,&t)||return; n||return(1); n%t & return;2} \\ M. F. Hasler, Feb 02 2013

Formula

sum_{k=1...n} a(n-k,k) = A000005(n) for all n>0. - M. F. Hasler, Feb 02 2013

A292400 p-INVERT of (1,2,2,2,2,2,2,...) (A040000), where p(S) = (1 - S)^2.

Original entry on oeis.org

2, 7, 20, 57, 158, 431, 1160, 3089, 8154, 21367, 55644, 144137, 371638, 954335, 2441872, 6228129, 15839794, 40181095, 101690404, 256812121, 647303502, 1628647055, 4091042328, 10260849073, 25699419914, 64283165143, 160599382124, 400772669481, 999059833190
Offset: 0

Views

Author

Clark Kimberling, Sep 30 2017

Keywords

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

Crossrefs

Programs

  • Mathematica
    z = 60; s = x (x + 1)/(1 - x); p = (1 - s)^2;
    Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A040000 *)
    Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292400 *)

Formula

G.f.: -(((1 + x) (-2 + 3 x + x^2))/(-1 + 2 x + x^2)^2).
a(n) = 4*a(n-1) - 2*a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.

A339275 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the terms of A040000: 1, 2, 2, 2, ... interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

1, 2, 2, 1, 2, 0, 2, 2, 2, 0, 1, 2, 2, 0, 2, 0, 0, 2, 2, 2, 2, 0, 0, 1, 2, 2, 0, 0, 2, 0, 2, 0, 2, 2, 0, 0, 2, 0, 0, 2, 2, 2, 2, 0, 1, 2, 0, 0, 0, 0, 2, 2, 0, 0, 0, 2, 0, 2, 2, 0, 2, 2, 0, 0, 0, 2, 0, 0, 0, 2, 2, 2, 2, 0, 0, 1, 2, 0, 0, 2, 0, 0, 2, 2, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 2, 0, 0, 2, 0, 2
Offset: 1

Views

Author

Omar E. Pol, Dec 01 2020

Keywords

Comments

T(n,k) is also the number of horizontal line segments in the n-th level of the k-th largest double-staircase of the diagram defined in A335616 (see example).
The partial sums of column k give the k-th column of A338721.

Examples

			Triangle begins (rows 1..28):
1;
2;
2,  1;
2,  0;
2,  2;
2,  0,  1;
2,  2,  0;
2,  0,  0;
2,  2,  2;
2,  0,  0,  1;
2,  2,  0,  0;
2,  0,  2,  0;
2,  2,  0,  0;
2,  0,  0,  2;
2,  2,  2,  0,  1;
2,  0,  0,  0,  0;
2,  2,  0,  0,  0;
2,  0,  2,  2,  0;
2,  2,  0,  0,  0;
2,  0,  0,  0,  2;
2,  2,  2,  0,  0,  1;
2,  0,  0,  2,  0,  0;
2,  2,  0,  0,  0,  0;
2,  0,  2,  0,  0,  0;
2,  2,  0,  0,  2,  0;
2,  0,  0,  2,  0,  0;
2,  2,  2,  0,  0,  2;
2,  0,  0,  0,  0,  0,  1;
...
For an illustration of the rows of triangle consider the infinite "double-staircases" diagram defined in A335616.
The first 15 levels of the structure looks like this:
.
Level                         "Double-staircases" diagram
n                                          _
1                                        _|1|_
2                                      _|1 _ 1|_
3                                    _|1  |1|  1|_
4                                  _|1   _| |_   1|_
5                                _|1    |1 _ 1|    1|_
6                              _|1     _| |1| |_     1|_
7                            _|1      |1  | |  1|      1|_
8                          _|1       _|  _| |_  |_       1|_
9                        _|1        |1  |1 _ 1|  1|        1|_
10                     _|1         _|   | |1| |   |_         1|_
11                   _|1          |1   _| | | |_   1|          1|_
12                 _|1           _|   |1  | |  1|   |_           1|_
13               _|1            |1    |  _| |_  |    1|            1|_
14             _|1             _|    _| |1 _ 1| |_    |_             1|_
15            |1              |1    |1  | |1| |  1|    1|              1|
.
For n = 15, in the 15th level of the diagram we have that the first largest double-staircase has two horizontal steps, the second double-staircase has two steps, the third double-staircase has two steps, there are no steps in the fourth double-stairce and the fifth double-staircase has only one step, so the 15th row of triangle is [2, 2, 2, 0, 1].
		

Crossrefs

Column 1 is A040000.
Row sums give A335616.
Row n has length A003056(n).
Column k starts in row A000217(k).
The number of positive terms in row n is A001227(n).

A147623 The 3rd Witt transform of A040000.

Original entry on oeis.org

0, 2, 6, 12, 22, 34, 48, 66, 86, 108, 134, 162, 192, 226, 262, 300, 342, 386, 432, 482, 534, 588, 646, 706, 768, 834, 902, 972, 1046, 1122, 1200, 1282, 1366, 1452, 1542, 1634, 1728, 1826, 1926, 2028, 2134, 2242, 2352, 2466, 2582, 2700, 2822, 2946, 3072
Offset: 0

Views

Author

R. J. Mathar, Nov 08 2008

Keywords

Comments

The 2nd Witt transform of A040000 is represented by A042964.

Crossrefs

Programs

  • Magma
    [n le 2 select 1+(-1)^n else 4*(1+(n-2)^2) - Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Oct 24 2022
    
  • Mathematica
    CoefficientList[Series[2x(1+x)(1 +x^2)/((1-x)^3 (1+x+x^2)), {x,0,40}], x] (* Vincenzo Librandi, Dec 14 2012 *)
    LinearRecurrence[{2,-1,1,-2,1},{0,2,6,12,22},50] (* Harvey P. Dale, Jul 04 2021 *)
  • SageMath
    [2*(2*(1+3*n^2) -(2*chebyshev_U(n, -1/2) +chebyshev_U(n-1, -1/2)))/9 for n in range(41)] # G. C. Greubel, Oct 24 2022

Formula

G.f.: 2*x*(1+x)*(1+x^2)/((1-x)^3*(1+x+x^2)).
a(n) = 2*A071619(n).
From G. C. Greubel, Oct 24 2022: (Start)
a(n) = 4*(2 - 2*n + n^2) - a(n-1) - a(n-2).
a(n) = 2*(2*(1 + 3*n^2) - (2*A049347(n) + A049347(n-1)))/9. (End)

A192575 Triangle T(n,0) = A040000(n), T(n,k)=0 (odd-numbered columns); T(n,k) = (-1)^(k/2)*A110813(n-k/2-1,k/2-1) (even-numbered columns, k>0).

Original entry on oeis.org

1, 2, 0, 2, 0, -1, 2, 0, -3, 0, 2, 0, -5, 0, 1, 2, 0, -7, 0, 4, 0, 2, 0, -9, 0, 9, 0, -1, 2, 0, -11, 0, 16, 0, -5, 0, 2, 0, -13, 0, 25, 0, -14, 0, 1, 2, 0, -15, 0, 36, 0, -30, 0, 6, 0, 2, 0, -17, 0, 49, 0, -55, 0, 20, 0, -1
Offset: 0

Views

Author

Paul Curtz, Jul 04 2011

Keywords

Comments

A zero-padded variant of A110813, which provides more information.

Examples

			1;
2 0;
2 0  -1;
2 0  -3 0;
2 0  -5 0  1;
2 0  -7 0  4 0;
2 0  -9 0  9 0  -1;
2 0 -11 0 16 0  -5 0;
2 0 -13 0 25 0 -14 0 1;
2 0 -15 0 36 0 -30 0 6 0;
		

Crossrefs

Cf. A191662.

Formula

T(n,k) = T(n-1,k)-T(n-2,k-2), n>1.
T(n,2k+1)=0.
T(n,2k) = (-1)^k*binomial(n-k-1,k-1)*(2n-3k)/k , k>0. - R. J. Mathar, Aug 26 2011
T(n,0) = A040000(n).
sum_{k=0..n} T(n,k) = A057079(n).
sum_{k=0..n} |T(n,k)| = A000045(n+2). (See A129710).

A001333 Pell-Lucas numbers: numerators of continued fraction convergents to sqrt(2).

Original entry on oeis.org

1, 1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363, 8119, 19601, 47321, 114243, 275807, 665857, 1607521, 3880899, 9369319, 22619537, 54608393, 131836323, 318281039, 768398401, 1855077841, 4478554083, 10812186007, 26102926097, 63018038201, 152139002499, 367296043199
Offset: 0

Views

Author

Keywords

Comments

Number of n-step non-selfintersecting paths starting at (0,0) with steps of types (1,0), (-1,0) or (0,1) [Stanley].
Number of n steps one-sided prudent walks with east, west and north steps. - Shanzhen Gao, Apr 26 2011
Number of ternary strings of length n-1 with subwords (0,2) and (2,0) not allowed. - Olivier Gérard, Aug 28 2012
Number of symmetric 2n X 2 or (2n-1) X 2 crossword puzzle grids: all white squares are edge connected; at least 1 white square on every edge of grid; 180-degree rotational symmetry. - Erich Friedman
a(n+1) is the number of ways to put molecules on a 2 X n ladder lattice so that the molecules do not touch each other.
In other words, a(n+1) is the number of independent vertex sets and vertex covers in the n-ladder graph P_2 X P_n. - Eric W. Weisstein, Apr 04 2017
Number of (n-1) X 2 binary arrays with a path of adjacent 1's from top row to bottom row, see A359576. - R. H. Hardin, Mar 16 2002
a(2*n+1) with b(2*n+1) := A000129(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = -1.
a(2*n) with b(2*n) := A000129(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 2*b^2 = +1 (see Emerson reference).
Bisection: a(2*n) = T(n,3) = A001541(n), n >= 0 and a(2*n+1) = S(2*n,2*sqrt(2)) = A002315(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
Binomial transform of A077957. - Paul Barry, Feb 25 2003
For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 2. - Herbert Kociemba, Jun 02 2004
For n > 1, a(n) corresponds to the longer side of a near right-angled isosceles triangle, one of the equal sides being A000129(n). - Lekraj Beedassy, Aug 06 2004
Exponents of terms in the series F(x,1), where F is determined by the equation F(x,y) = xy + F(x^2*y,x). - Jonathan Sondow, Dec 18 2004
Number of n-words from the alphabet A={0,1,2} which two neighbors differ by at most 1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 30 2006
Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003 [Amended by Paul E. Black (paul.black(AT)nist.gov), Dec 18 2006]
Odd-indexed prime numerators are prime RMS numbers (A140480) and also NSW primes (A088165). - Ctibor O. Zizka, Aug 13 2008
The intermediate convergents to 2^(1/2) begin with 4/3, 10/7, 24/17, 58/41; essentially, numerators=A052542 and denominators here. - Clark Kimberling, Aug 26 2008
Equals right border of triangle A143966. Starting (1, 3, 7, ...) equals INVERT transform of (1, 2, 2, 2, ...) and row sums of triangle A143966. - Gary W. Adamson, Sep 06 2008
Inverse binomial transform of A006012; Hankel transform is := [1, 2, 0, 0, 0, 0, 0, 0, 0, ...]. - Philippe Deléham, Dec 04 2008
From Charlie Marion, Jan 07 2009: (Start)
In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
let a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2) and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
let b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2) and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
for example, if k=1 and n=3, then b(1,n)=a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
(End)
This sequence occurs in the lower bound of the order of the set of equivalent resistances of n equal resistors combined in series and in parallel (A048211). - Sameen Ahmed Khan, Jun 28 2010
Let M = a triangle with the Fibonacci series in each column, but the leftmost column is shifted upwards one row. A001333 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 2 types of other natural numbers. - Milan Janjic, Aug 13 2010
Equals the INVERTi transform of A055099. - Gary W. Adamson, Aug 14 2010
From L. Edson Jeffery, Apr 04 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(8,2) = (0 0 1 0)
(0 1 0 1)
(1 0 2 0)
(0 2 0 1).
Then a(n) = (1/4)*Trace(U^n). (See also A084130, A006012.)
(End)
For n >= 1, row sums of triangle
m/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....2
.2..|..1.....2.....4
.3..|..1.....4.....4.....8
.4..|..1.....4....12.....8....16
.5..|..1.....6....12....32....16....32
.6..|..1.....6....24....32....80....32....64
.7..|..1.....8....24....80....80...192....64...128
which is the triangle for numbers 2^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) is also the number of ways to place k non-attacking wazirs on a 2 X n board, summed over all k >= 0 (a wazir is a leaper [0,1]). - Vaclav Kotesovec, May 08 2012
The sequences a(n) and b(n) := A000129(n) are entries of powers of the special case of the Brahmagupta Matrix - for details see Suryanarayan's paper. Further, as Suryanarayan remark, if we set A = 2*(a(n) + b(n))*b(n), B = a(n)*(a(n) + 2*b(n)), C = a(n)^2 + 2*a(n)*b(n) + 2*b(n)^2 we obtain integral solutions of the Pythagorean relation A^2 + B^2 = C^2, where A and B are consecutive integers. - Roman Witula, Jul 28 2012
Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, .... - R. J. Mathar, Aug 10 2012
This sequence and A000129 give the diagonal numbers described by Theon of Smyrna. - Sture Sjöstedt, Oct 20 2012
a(n) is the top left entry of the n-th power of any of the following six 3 X 3 binary matrices: [1, 1, 1; 1, 1, 1; 1, 0, 0] or [1, 1, 1; 1, 1, 0; 1, 1, 0] or [1, 1, 1; 1, 0, 1; 1, 1, 0] or [1, 1, 1; 1, 1, 0; 1, 0, 1] or [1, 1, 1; 1, 0, 1; 1, 0, 1] or [1, 1, 1; 1, 0, 0; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
If p is prime, a(p) == 1 (mod p) (compare with similar comment for A000032). - Creighton Dement, Oct 11 2005, modified by Davide Colazingari, Jun 26 2016
a(n) = A000129(n) + A000129(n-1), where A000129(n) is the n-th Pell Number; e.g., a(6) = 99 = A000129(6) + A000129(5) = 70 + 29. Hence the sequence of fractions has the form 1 + A000129(n-1)/A000129(n), and the ratio A000129(n-1)/A000129(n)converges to sqrt(2) - 1. - Gregory L. Simay, Nov 30 2018
For n > 0, a(n+1) is the length of tau^n(1) where tau is the morphism: 1 -> 101, 0 -> 1. See Song and Wu. - Michel Marcus, Jul 21 2020
For n > 0, a(n) is the number of nonisomorphic quasitrivial semigroups with n elements, see Devillet, Marichal, Teheux. A292932 is the number of labeled quasitrivial semigroups. - Peter Jipsen, Mar 28 2021
a(n) is the permanent of the n X n tridiagonal matrix defined in A332602. - Stefano Spezia, Apr 12 2022
From Greg Dresden, May 08 2023: (Start)
For n >= 2, 4*a(n) is the number of ways to tile this T-shaped figure of length n-1 with two colors of squares and one color of domino; shown here is the figure of length 5 (corresponding to n=6), and it has 4*a(6) = 396 different tilings.
_
|| _
|||_|||
|_|
(End)
12*a(n) = number of walks of length n in the cyclic Kautz digraph CK(3,4). - Miquel A. Fiol, Feb 15 2024

Examples

			Convergents are 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/A000129.
The 15 3 X 2 crossword grids, with white squares represented by an o:
  ooo ooo ooo ooo ooo ooo ooo oo. o.o .oo o.. .o. ..o oo. .oo
  ooo oo. o.o .oo o.. .o. ..o ooo ooo ooo ooo ooo ooo .oo oo.
G.f. = 1 + x + 3*x^2 + 7*x^3 + 17*x^4 + 41*x^5 + 99*x^6 + 239*x^7 + 577*x^8 + ...
		

References

  • M. R. Bacon and C. K. Cook, Some properties of Oresme numbers and convolutions ..., Fib. Q., 62:3 (2024), 233-240.
  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 204.
  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.
  • J. Devillet, J.-L. Marichal, and B. Teheux, Classifications of quasitrivial semigroups, Semigroup Forum, 100 (2020), 743-764.
  • Maribel Díaz Noguera [Maribel Del Carmen Díaz Noguera], Rigoberto Flores, Jose L. Ramirez, and Martha Romero Rojas, Catalan identities for generalized Fibonacci polynomials, Fib. Q., 62:2 (2024), 100-111.
  • Kenneth Edwards and Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.
  • R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.5 The Fibonacci and Related Sequences, p. 288.
  • A. F. Horadam, R. P. Loh, and A. G. Shannon, Divisibility properties of some Fibonacci-type sequences, pp. 55-64 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979.
  • Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
  • Kin Y. Li, Math Problem Book I, 2001, p. 24, Problem 159.
  • I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 102, Problem 10.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics, Volume 1 (1986), p. 203, Example 4.1.2.
  • A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.
  • R. C. Tilley et al., The cell growth problem for filaments, Proc. Louisiana Conf. Combinatorics, ed. R. C. Mullin et al., Baton Rouge, 1970, 310-339.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 34.

Crossrefs

For denominators see A000129.
See A040000 for the continued fraction expansion of sqrt(2).
See also A078057 which is the same sequence without the initial 1.
Cf. also A002203, A152113.
Row sums of unsigned Chebyshev T-triangle A053120. a(n)= A054458(n, 0) (first column of convolution triangle).
Row sums of A140750, A160756, A135837.
Equals A034182(n-1) + 2 and A084128(n)/2^n. First differences of A052937. Partial sums of A052542. Pairwise sums of A048624. Bisection of A002965.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
Second row of the array in A135597.
Cf. A055099.
Cf. A028859, A001906 / A088305, A033303, A000225, A095263, A003945, A006356, A002478, A214260, A001911 and A000217 for other restricted ternary words.
Cf. Triangle A106513 (alternating row sums).
Equals A293004 + 1.
Cf. A033539, A332602, A086395 (subseq. of primes).

Programs

  • Haskell
    a001333 n = a001333_list !! n
    a001333_list = 1 : 1 : zipWith (+)
                           a001333_list (map (* 2) $ tail a001333_list)
    -- Reinhard Zumkeller, Jul 08 2012
    
  • Magma
    [n le 2 select 1 else 2*Self(n-1)+Self(n-2): n in [1..35]]; // Vincenzo Librandi, Nov 10 2018
    
  • Maple
    A001333 := proc(n) option remember; if n=0 then 1 elif n=1 then 1 else 2*procname(n-1)+procname(n-2) fi end;
    Digits := 50; A001333 := n-> round((1/2)*(1+sqrt(2))^n);
    with(numtheory): cf := cfrac (sqrt(2),1000): [seq(nthnumer(cf,i), i=0..50)];
    a:= n-> (M-> M[2, 1]+M[2, 2])(<<2|1>, <1|0>>^n):
    seq(a(n), n=0..33);  # Alois P. Heinz, Aug 01 2008
    A001333List := proc(m) local A, P, n; A := [1,1]; P := [1,1];
    for n from 1 to m - 2 do P := ListTools:-PartialSums([op(A), P[-2]]);
    A := [op(A), P[-1]] od; A end: A001333List(32); # Peter Luschny, Mar 26 2022
  • Mathematica
    Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[2], n]]], {n, 1, 40}], 1, 1] (* Stefan Steinerberger, Apr 08 2006 *)
    Table[((1 - Sqrt[2])^n + (1 + Sqrt[2])^n)/2, {n, 0, 29}] // Simplify (* Robert G. Wilson v, May 02 2006 *)
    a[0] = 1; a[1] = 1; a[n_] := a[n] = 2a[n - 1] + a[n - 2]; Table[a@n, {n, 0, 29}] (* Robert G. Wilson v, May 02 2006 *)
    Table[ MatrixPower[{{1, 2}, {1, 1}}, n][[1, 1]], {n, 0, 30}] (* Robert G. Wilson v, May 02 2006 *)
    a=c=0;t={b=1}; Do[c=a+b+c; AppendTo[t,c]; a=b;b=c,{n,40}]; t (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
    LinearRecurrence[{2, 1}, {1, 1}, 40] (* Vladimir Joseph Stephan Orlovsky, Mar 23 2009 *)
    Join[{1}, Numerator[Convergents[Sqrt[2], 30]]] (* Harvey P. Dale, Aug 22 2011 *)
    Table[(-I)^n ChebyshevT[n, I], {n, 10}] (* Eric W. Weisstein, Apr 04 2017 *)
    CoefficientList[Series[(-1 + x)/(-1 + 2 x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
    Table[Sqrt[(ChebyshevT[n, 3] + (-1)^n)/2], {n, 0, 20}] (* Eric W. Weisstein, Apr 17 2018 *)
  • PARI
    {a(n) = if( n<0, (-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [1, 1]}; /* Michael Somos, Sep 02 2012 */
    
  • PARI
    {a(n) = polchebyshev(n, 1, I) / I^n}; /* Michael Somos, Sep 02 2012 */
    
  • PARI
    a(n) = real((1 + quadgen(8))^n); \\ Michel Marcus, Mar 16 2021
    
  • PARI
    { for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[1, 1]; if (a > 10^(10^3 - 6), break); write("b001333.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 12 2009
    
  • Python
    from functools import cache
    @cache
    def a(n): return 1 if n < 2 else 2*a(n-1) + a(n-2)
    print([a(n) for n in range(32)]) # Michael S. Branicky, Nov 13 2022
  • Sage
    from sage.combinat.sloane_functions import recur_gen2
    it = recur_gen2(1,1,2,1)
    [next(it) for i in range(30)] ## Zerinvary Lajos, Jun 24 2008
    
  • Sage
    [lucas_number2(n,2,-1)/2 for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009
    

Formula

a(n) = A055642(A125058(n)). - Reinhard Zumkeller, Feb 02 2007
a(n) = 2a(n-1) + a(n-2);
a(n) = ((1-sqrt(2))^n + (1+sqrt(2))^n)/2.
a(n)+a(n+1) = 2 A000129(n+1). 2*a(n) = A002203(n).
G.f.: (1 - x) / (1 - 2*x - x^2) = 1 / (1 - x / (1 - 2*x / (1 + x))). - Simon Plouffe in his 1992 dissertation.
A000129(2n) = 2*A000129(n)*a(n). - John McNamara, Oct 30 2002
a(n) = (-i)^n * T(n, i), with T(n, x) Chebyshev's polynomials of the first kind A053120 and i^2 = -1.
a(n) = a(n-1) + A052542(n-1), n>1. a(n)/A052542(n) converges to sqrt(1/2). - Mario Catalani (mario.catalani(AT)unito.it), Apr 29 2003
E.g.f.: exp(x)cosh(x*sqrt(2)). - Paul Barry, May 08 2003
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k)2^k. - Paul Barry, May 13 2003
For n > 0, a(n)^2 - (1 + (-1)^(n))/2 = Sum_{k=0..n-1} ((2k+1)*A001653(n-1-k)); e.g., 17^2 - 1 = 288 = 1*169 + 3*29 + 5*5 + 7*1; 7^2 = 49 = 1*29 + 3*5 + 5*1. - Charlie Marion, Jul 18 2003
a(n+2) = A078343(n+1) + A048654(n). - Creighton Dement, Jan 19 2005
a(n) = A000129(n) + A000129(n-1) = A001109(n)/A000129(n) = sqrt(A001110(n)/A000129(n)^2) = ceiling(sqrt(A001108(n))). - Henry Bottomley, Apr 18 2000
Also the first differences of A000129 (the Pell numbers) because A052937(n) = A000129(n+1) + 1. - Graeme McRae, Aug 03 2006
a(n) = Sum_{k=0..n} A122542(n,k). - Philippe Deléham, Oct 08 2006
For another recurrence see A000129.
a(n) = Sum_{k=0..n} A098158(n,k)*2^(n-k). - Philippe Deléham, Dec 26 2007
a(n) = upper left and lower right terms of [1,1; 2,1]^n. - Gary W. Adamson, Mar 12 2008
If p[1]=1, and p[i]=2, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det A. - Milan Janjic, Apr 29 2010
For n>=2, a(n)=F_n(2)+F_(n+1)(2), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(-n) = (-1)^n * a(n). - Michael Somos, Sep 02 2012
Dirichlet g.f.: (PolyLog(s,1-sqrt(2)) + PolyLog(s,1+sqrt(2)))/2. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A000129(n) - A000129(n-1), where A000129(n) is the n-th Pell Number. Hence the continued fraction is of the form 1-(A000129(n-1)/A000129(n)). - Gregory L. Simay, Nov 09 2018
a(n) = (A000129(n+3) + A000129(n-3))/10, n>=3. - Paul Curtz, Jun 16 2021
a(n) = (A000129(n+6) - A000129(n-6))/140, n>=6. - Paul Curtz, Jun 20 2021
a(n) = round((1/2)*sqrt(Product_{k=1..n} 4*(1 + sin(k*Pi/n)^2))), for n>=1. - Greg Dresden, Dec 28 2021
a(n)^2 + a(n+1)^2 = A075870(n+1) = 2*(b(n)^2 + b(n+1)^2) for all n in Z where b(n) := A000129(n). - Michael Somos, Apr 02 2022
a(n) = 2*A048739(n-2)+1. - R. J. Mathar, Feb 01 2024
Sum_{n>=1} 1/a(n) = 1.5766479516393275911191017828913332473... - R. J. Mathar, Feb 05 2024
From Peter Bala, Jul 06 2025: (Start)
G.f.: Sum_{n >= 1} (-1)^(n+1) * x^(n-1) * Product_{k = 1..n} (1 - k*x)/(1 - 3*x + k*x^2).
The following series telescope:
Sum_{n >= 1} (-1)^(n+1)/(a(2*n) + 1/a(2*n)) = 1/4, since 1/(a(2*n) + 1/a(2*n)) = 1/A077445(n) + 1/A077445(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) - 1/a(2*n+1)) = 1/8, since. 1/(a(2*n+1) - 1/a(2*n+1)) = 1/(4*Pell(2*n)) + 1/(4*Pell(2*n+2)), where Pell(n) = A000129(n).
Sum_{n >= 1} (-1)^(n+1)/(a(2*n+1) + 9/a(2*n+1)) = 1/10, since 1/(a(2*n+1) + 9/a(2*n+1)) = b(n) + b(n+1), where b(n) = A001109(n)/(2*Pell(2*n-1)*Pell(2*n+1)).
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 1 - sqrt(2)/2 = A268682, since (-1)^(n+1)/(a(n)*a(n+1)) = Pell(n)/a(n) - Pell(n+1)/a(n+1). (End)

Extensions

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A002193 Decimal expansion of square root of 2.

Original entry on oeis.org

1, 4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5, 0, 4, 8, 8, 0, 1, 6, 8, 8, 7, 2, 4, 2, 0, 9, 6, 9, 8, 0, 7, 8, 5, 6, 9, 6, 7, 1, 8, 7, 5, 3, 7, 6, 9, 4, 8, 0, 7, 3, 1, 7, 6, 6, 7, 9, 7, 3, 7, 9, 9, 0, 7, 3, 2, 4, 7, 8, 4, 6, 2, 1, 0, 7, 0, 3, 8, 8, 5, 0, 3, 8, 7, 5, 3, 4, 3, 2, 7, 6, 4, 1, 5, 7
Offset: 1

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Author

Keywords

Comments

Sometimes called Pythagoras's constant.
Its continued fraction expansion is [1; 2, 2, 2, ...] (see A040000). - Arkadiusz Wesolowski, Mar 10 2012
The discovery of irrational numbers is attributed to Hippasus of Metapontum, who may have proved that sqrt(2) is not a rational number; thus sqrt(2) is often regarded as the earliest known irrational number. - Clark Kimberling, Oct 12 2017
From Clark Kimberling, Oct 12 2017: (Start)
In the first million digits,
0 occurs 99814 times;
1 occurs 99925 times;
2 occurs 100436 times;
3 occurs 100190 times;
4 occurs 100024 times;
5 occurs 100155 times;
6 occurs 99886 times;
7 occurs 100008 times;
8 occurs 100441 times;
9 occurs 100121 times. (End)
Diameter of a sphere whose surface area equals 2*Pi. More generally, the square root of x is also the diameter of a sphere whose surface area equals x*Pi. - Omar E. Pol, Nov 10 2018
Sqrt(2) = 1 + area of region bounded by y = sin x, y = cos x, and x = 0. - Clark Kimberling, Jul 03 2020
Also aspect ratio of the ISO 216 standard for paper sizes. - Stefano Spezia, Feb 24 2021
The standard deviation of a roll of a 5-sided die. - Mohammed Yaseen, Feb 23 2023
From Michal Paulovic, Mar 22 2023: (Start)
The length of a unit square diagonal.
The infinite tetration (power tower) sqrt(2)^(sqrt(2)^(sqrt(2)^(...))) equals 2 from the identity (x^(1/x))^((x^(1/x))^((x^(1/x))^(...))) = x where 1/e <= x <= e. (End)

Examples

			1.41421356237309504880168872420969807856967187537694807317667...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 24, 182.
  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, Section 1.1.
  • David Flannery, The Square Root of 2, Copernicus Books Springer-Praxis Pub. 2006.
  • Martin Gardner, Gardner's Workout, Chapter 2 "The Square Root of 2=1.414213562373095..." pp. 9-19 A. K. Peters MA 2002.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.4 Irrational Numbers and §4.4 Powers and Roots, pp. 84, 145.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 64-67.
  • B. Rittaud, Le fabuleux destin de sqrt(2), Le Pommier, Paris 2006.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 60, page 605.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, pp. 34-35.

Crossrefs

Cf. A004539 (binary version).

Programs

  • Haskell
    -- After Michael B. Porter's PARI program.
    a002193 n = a002193_list !! (n-1)
    a002193_list = w 2 0 where
    w x r = dig : w (100 * (x - (20 * r + dig) * dig)) (10 * r + dig)
    where dig = head (dropWhile (\d -> (20 * r + d) * d < x) [0..]) - 1
    -- Reinhard Zumkeller, Nov 22 2013
  • Maple
    Digits:=100; evalf(sqrt(2)); # Wesley Ivan Hurt, Dec 04 2013
  • Mathematica
    RealDigits[N[2^(1/2), 128]] (* Vladimir Joseph Stephan Orlovsky, Dec 25 2008 *)
  • Maxima
    fpprec: 100$ ev(bfloat(sqrt(2))); /* Martin Ettl, Oct 17 2012 */
    
  • PARI
    default(realprecision, 20080); x=sqrt(2); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b002193.txt", n, " ", d)); \\ Harry J. Smith, Apr 21 2009
    
  • PARI
    r=0; x=2; /* Digit-by-digit method */
    for(digits=1,100,{d=0;while((20*r+d)*d <= x,d++);
    d--; /* while loop overshoots correct digit */
    print(d);x=100*(x-(20*r+d)*d);r=10*r+d}) \\ Michael B. Porter, Oct 20 2009
    
  • PARI
    \\ Works in v2.15.0; n = 100 decimal places
    my(n=100); digits(floor(10^n*quadgen(8))) \\ Michal Paulovic, Mar 22 2023
    

Formula

Sqrt(2) = 14 * Sum_{n >= 0} (A001790(n)/2^A005187(floor(n/2)) * 10^(-2n-1)) where A001790(n) are numerators in expansion of 1/sqrt(1-x) and the denominators in expansion of 1/sqrt(1-x) are 2^A005187(n). 14 = 2*7, see A010503 (expansion of 1/sqrt(2)). - Gerald McGarvey, Jan 01 2005
Limit_{n -> +oo} (1/n)*(Sum_{k = 1..n} frac(sqrt(1+zeta(k+1)))) = 1/(1+sqrt(2)). - Yalcin Aktar, Jul 14 2005
sqrt(2) = 2 + n*A167199(n-1)/A167199(n) as n -> infinity (conjecture). - Mats Granvik, Oct 30 2009
sqrt(2) = limit as n goes to infinity of A179807(n+1)/A179807(n) - 1. - Mats Granvik, Feb 15 2011
sqrt(2) = Product_{l=0..k-1} 2*cos((2*l+1)*Pi/(4*k)) = (Product_{l=0..k-1} R(2*l+1,rho(4*k)) - 1), identical for k >= 1, with the row polynomials R(n, x) from A127672 and rho(4*k) := 2*cos(Pi/(4*k)) is the length ratio (smallest diagonal)/side in a regular (4*k)-gon. From the product formula given in a Oct 21 2013 formula contribution to A056594, with n -> 2*k, using cos(Pi-alpha) = - cos(alpha) to obtain 2 for the square of the present product. - Wolfdieter Lang, Oct 22 2013
If x = sqrt(2), 1/log(x - 1) + 1/log(x + 1) = 0. - Kritsada Moomuang, Jul 10 2020
From Amiram Eldar, Jul 25 2020: (Start)
Equals Product_{k>=0} (1 + (-1)^k/(2*k + 1)).
Equals Sum_{k>=0} binomial(2*k,k)/8^k. (End)
Equals i^(1/2) + i^(-1/2). - Gary W. Adamson, Jul 11 2022
Equals (sqrt(2) + (sqrt(2) + (sqrt(2) + ...)^(1/3))^(1/3))^(1/3). - Michal Paulovic, Mar 22 2023
Equals 1 + Sum_{k>=1} (-1)^(k-1)/(2^(2*k)*(2*k - 1))*binomial(2*k,k) [Newton]. - Stefano Spezia, Oct 15 2024
From Antonio Graciá Llorente, Dec 19 2024: (Start)
Equals Sum_{k>=0} 2*k*binomial(2*k,k)/8^k.
Equals Product_{k>=2} k/sqrt(k^2 + 1).
Equals Product_{k>=0} (6*k + 3)/((6*k + 3) - (-1)^k).
Equals Product_{k>=1} (2*k + 1)/((2*k + 1) + (-1)^k).
Equals Product_{k>=0} ((4*k + 3)*(4*k + 1 + (-1)^k))/((4*k + 1)*(4*k + 3 + (-1)^k)). (End)
Equals hypergeom([1/2, 1/2], [1/2], 1/2). - Stefano Spezia, Jan 05 2025

A003945 Expansion of g.f. (1+x)/(1-2*x).

Original entry on oeis.org

1, 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, 6291456, 12582912, 25165824, 50331648, 100663296, 201326592, 402653184, 805306368, 1610612736, 3221225472, 6442450944, 12884901888
Offset: 0

Views

Author

Keywords

Comments

Coordination sequence for infinite tree with valency 3.
Number of Hamiltonian cycles in K_3 X P_n.
Number of ternary words of length n avoiding aa, bb, cc.
For n > 0, row sums of A029635. - Paul Barry, Jan 30 2005
Binomial transform is {1, 4, 13, 40, 121, 364, ...}, see A003462. - Philippe Deléham, Jul 23 2005
Convolved with the Jacobsthal sequence A001045 = A001786: (1, 4, 12, 32, 80, ...). - Gary W. Adamson, May 23 2009
Equals (n+1)-th row sums of triangle A161175. - Gary W. Adamson, Jun 05 2009
a(n) written in base 2: a(0) = 1, a(n) for n >= 1: 11, 110, 11000, 110000, ..., i.e.: 2 times 1, (n-1) times 0 (see A003953(n)). - Jaroslav Krizek, Aug 17 2009
INVERTi transform of A003688. - Gary W. Adamson, Aug 05 2010
An elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 42, 138, 162 and 168, lead to this sequence. For the corner squares these vectors lead to the companion sequence A083329. - Johannes W. Meijer, Aug 15 2010
A216022(a(n)) != 2 and A216059(a(n)) != 3. - Reinhard Zumkeller, Sep 01 2012
Number of length-n strings of 3 letters with no two adjacent letters identical. The general case (strings of r letters) is the sequence with g.f. (1+x)/(1-(r-1)*x). - Joerg Arndt, Oct 11 2012
Sums of pairs of rows of Pascal's triangle A007318, T(2n,k)+T(2n+1,k); Sum_{n>=1} A000290(n)/a(n) = 4. - John Molokach, Sep 26 2013

Crossrefs

Essentially same as A007283 (3*2^n) and A042950.
Generating functions of the form (1+x)/(1-k*x) for k=1 to 12: A040000, A003945, A003946, A003947, A003948, A003949, A003950, A003951, A003952.
Generating functions of the form (1+x)/(1-k*x) for k=13 to 30: A170732, A170733, A170734, A170735, A170736, A170737, A170738, A170739, A170740, A170741, A170742, A170743, A170744, A170745, A170746, A170747, A170748.
Generating functions of the form (1+x)/(1-k*x) for k=31 to 50: A170749, A170750, A170751, A170752, A170753, A170754, A170755, A170756, A170757, A170758, A170759, A170760, A170761, A170762, A170763, A170764, A170765, A170766, A170767, A170768, A170769.
Cf. A003688.

Programs

  • Maple
    k := 3; if n = 0 then 1 else k*(k-1)^(n-1); fi;
  • Mathematica
    Join[{1}, 3*2^Range[0, 60]] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
    Table[2^n+Floor[2^(n-1)], {n,0,30}] (* Martin Grymel, Oct 17 2012 *)
    CoefficientList[Series[(1+x)/(1-2x),{x,0,40}],x] (* or *) LinearRecurrence[ {2},{1,3},40] (* Harvey P. Dale, May 04 2017 *)
  • PARI
    a(n)=if(n,3<Charles R Greathouse IV, Jan 12 2012

Formula

a(0) = 1; for n > 0, a(n) = 3*2^(n-1).
a(n) = 2*a(n-1), n > 1; a(0)=1, a(1)=3.
More generally, the g.f. (1+x)/(1-k*x) produces the sequence [1, 1 + k, (1 + k)*k, (1 + k)*k^2, ..., (1+k)*k^(n-1), ...], with a(0) = 1, a(n) = (1+k)*k^(n-1) for n >= 1. Also a(n+1) = k*a(n) for n >= 1. - Zak Seidov and N. J. A. Sloane, Dec 05 2009
The g.f. (1+x)/(1-k*x) produces the sequence with closed form (in PARI notation) a(n)=(n>=0)*k^n+(n>=1)*k^(n-1). - Jaume Oliver Lafont, Dec 05 2009
Binomial transform of A000034. a(n) = (3*2^n - 0^n)/2. - Paul Barry, Apr 29 2003
a(n) = Sum_{k=0..n} (n+k)*binomial(n, k)/n. - Paul Barry, Jan 30 2005
a(n) = Sum_{k=0..n} A029653(n, k)*x^k for x = 1. - Philippe Deléham, Jul 10 2005
Binomial transform of A000034. Hankel transform is {1,-3,0,0,0,...}. - Paul Barry, Aug 29 2006
a(0) = 1, a(n) = 2 + Sum_{k=0..n-1} a(k) for n >= 1. - Joerg Arndt, Aug 15 2012
a(n) = 2^n + floor(2^(n-1)). - Martin Grymel, Oct 17 2012
E.g.f.: (3*exp(2*x) - 1)/2. - Stefano Spezia, Jan 31 2023

Extensions

Edited by N. J. A. Sloane, Dec 04 2009

A011655 Period 3: repeat [0, 1, 1].

Original entry on oeis.org

0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1
Offset: 0

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Comments

A binary m-sequence: expansion of reciprocal of x^2+x+1 (mod 2).
A Chebyshev transform of the Jacobsthal numbers A001045: if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))*A(x/(1+x^2)). - Paul Barry, Feb 16 2004
This is the r = 1 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.
This is the Fibonacci sequence (A000045) modulo 2. - Stephen Jordan (sjordan(AT)mit.edu), Sep 10 2007
For n > 0: a(n) = A084937(n-1) mod 2. - Reinhard Zumkeller, Dec 16 2007
This is also the Lucas numbers (A000032) mod 2. In general, this is the parity of any Lucas sequence associated with any pair (P,Q) when P and Q are odd; i.e., a(n) = U_n(P,Q) mod 2 = V_n(P,Q) mod 2. See Ribenboim. - Rick L. Shepherd, Feb 07 2009
Starting with offset 1: (1, 1, 0, 1, 1, 0, ...) = INVERTi transform of the tribonacci sequence A001590 starting (1, 2, 3, 6, 11, 20, 37, ...). - Gary W. Adamson, May 04 2009
From Reinhard Zumkeller, Nov 30 2009: (Start)
Characteristic function of numbers coprime to 3.
a(n) = 1 - A079978(n); a(A001651(n)) = 1; a(A008585(n)) = 0;
A000212(n) = Sum_{k=0..n} a(k)*(n-k). (End)
Sum_{k>0} a(k)/k/2^k = log(7)/3. - Jaume Oliver Lafont, Jun 01 2010
The sequence is the principal Dirichlet character of the reduced residue system mod 3 (the other is A102283). Associated Dirichlet L-functions are L(2,chi) = Sum_{n>=1} a(n)/n^2 = 4*Pi^2/27 = A214549, and L(3,chi) = Sum_{n>=1} a(n)/n^3 = 1.157536... = -(psi''(1/3) + psi''(2/3))/54 where psi'' is the tetragamma function. [Jolley eq 309 and arXiv:1008.2547, L(m = 3, r = 1, s)]. - R. J. Mathar, Jul 15 2010
a(n+1), n >= 0, is the sequence of the row sums of the Riordan triangle A158454. - Wolfdieter Lang, Dec 18 2010
Removing the first two elements and keeping the offset at 0, this is a periodic sequence (1, 0, 1, 1, 0, 1, ...). Its INVERTi transform is (1, -1, 2, -2, 2, -2, ...) with period (2,-2). - Gary W. Adamson, Jan 21 2011
Column k = 1 of triangle in A198295. - Philippe Deléham, Jan 31 2012
The set of natural numbers, A000027: (1, 2, 3, ...); is the INVERT transform of the signed periodic sequence (1, 1, 0, -1, -1, 0, 1, 1, 0, ...). - Gary W. Adamson, Apr 28 2013
Any integer sequence s(n) = |s(n-1) - s(n-2)| (equivalently, max(s(n-1), s(n-2)) - min(s(n-1), s(n-2))) for n > i + 1 with s(i) = j and s(i+1) = k, where j and k are not both 0, is or eventually becomes a multiple of this sequence, namely, the sequence repeat gcd(j, k), gcd(j, k), 0 (at some offset). In particular, if j and k are coprime, then s(n) is or eventually becomes this sequence (see, e.g., A110044). - Rick L. Shepherd, Jan 21 2014
For n >= 1, a(n) is also the characteristic function for rational g-adic integers (+n/3)A001651).%20See%20the%20definition%20in%20the%20Mahler%20reference,%20p.%207%20and%20also%20p.%2010.%20-%20_Wolfdieter%20Lang">g and also (-n/3)_g for all integers g >= 2 without a factor 3 (A001651). See the definition in the Mahler reference, p. 7 and also p. 10. - _Wolfdieter Lang, Jul 11 2014
Characteristic function for A007908(n+1) being divisible by 3. a(n) = bit flipped A007908(n+1) (mod 3) = bit flipped A079978(n). - Wolfdieter Lang, Jun 12 2017
Also Jacobi or Kronecker symbol (n/9) (or (n/9^e) for all e >= 1). - Jianing Song, Jul 09 2018
The binomial trans. is 0, 1, 3, 6, 11, 21, 42, 85, 171, 342,.. (see A024495). - R. J. Mathar, Feb 25 2023

Examples

			G.f. = x + x^2 + x^4 + x^5 + x^7 + x^8 + x^10 + x^11 + x^13 + x^14 + x^16 + x^17 + ...
		

References

  • S. W. Golomb, Shift-Register Sequences, Holden-Day, San Francisco, 1967.
  • H. D. Lueke, Korrelationssignale, Springer 1992, pp. 43-48.
  • F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes, Elsevier/North Holland, 1978, p. 408.
  • K. Mahler, p-adic numbers and their functions, 2nd ed., Cambridge University press, 1981.
  • Paulo Ribenboim, The Little Book of Big Primes. Springer-Verlag, NY, 1991, p. 46. [Rick L. Shepherd, Feb 07 2009]

Crossrefs

Partial sums of A057078 give A011655(n+1).
Cf. A035191 (Mobius transform), A001590, A002487, A049347.
Cf. A000027, A000045, A004523 (partial sums), A057078 (first differences).
Cf. A007908, A079978 (bit flipped).
Cf. A011656 - A011751 for other binary m-sequences.
Cf. A002264.

Programs

Formula

G.f.: (x + x^2) / (1 - x^3) = Sum_{k>0} (x^k - x^(3*k)).
G.f.: x / (1 - x / (1 + x / (1 + x / (1 - 2*x / (1 + x))))). - Michael Somos, Apr 02 2012
a(n) = a(n+3) = a(-n), a(3*n) = 0, a(3*n + 1) = a(3*n + 2) = 1 for all n in Z.
a(n) = (1/2)*( (-1)^(floor((2n + 4)/3)) + 1 ). - Mario Catalani (mario.catalani(AT)unito.it), Oct 22 2003
a(n) = Fibonacci(n) mod 2. - Paul Barry, Nov 12 2003
a(n) = (2/3)*(1 - cos(2*Pi*n/3)). - Ralf Stephan, Jan 06 2004
a(n) = 1 - a(n-1)*a(n-2), a(n) = n for n < 2. - Reinhard Zumkeller, Feb 28 2004
a(n) = 2*(1 - T(n, -1/2))/3 with Chebyshev's polynomials T(n, x) of the first kind; see A053120. - Wolfdieter Lang, Oct 18 2004
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)*A001045(n-2k)/(n-k). - Paul Barry, Oct 31 2004
a(n) = A002487(n) mod 2. - Paul Barry, Jan 14 2005
From Bruce Corrigan (scentman(AT)myfamily.com), Aug 08 2005: (Start)
a(n) = n^2 mod 3.
a(n) = (1/3)*(2 - (r^n + r^(2*n))) where r = (-1 + sqrt(-3))/2. (End)
From Michael Somos, Sep 23 2005: (Start)
Euler transform of length 3 sequence [ 1, -1, 1].
Moebius transform is length 3 sequence [ 1, 0, -1].
Multiplicative with a(3^e) = 0^e, a(p^e) = 1 otherwise. (End)
From Hieronymus Fischer, Jun 27 2007: (Start)
a(n) = (4/3)*(|sin(Pi*(n-2)/3)| + |sin(Pi*(n-1)/3)|)*|sin(Pi*n/3)|.
a(n) = ((n+1) mod 3 + 1) mod 2 = (1 - (-1)^(n - 3*floor((n+1)/3)))/2. (End)
a(n) = 2 - a(n-1) - a(n-2) for n > 1. - Reinhard Zumkeller, Apr 13 2008
a(2*n+1) = a(n+1) XOR a(n), a(2*n) = a(n), a(1) = 1, a(0) = 0. - Reinhard Zumkeller, Dec 27 2008
Sum_{n>=1} a(n)/n^s = (1-1/3^s)*Riemann_zeta(s), s > 1. - R. J. Mathar, Jul 31 2010
a(n) = floor((4*n-5)/3) mod 2. - Gary Detlefs, May 15 2011
a(n) = (a(n-1) - a(n-2))^2 with a(0) = 0, a(1) = 1. - Francesco Daddi, Aug 02 2011
Convolution of A040000 with A049347. - R. J. Mathar, Jul 21 2012
G.f.: Sum_{k>0} x^A001651(k). - L. Edson Jeffery, Dec 05 2012
G.f.: x/(G(0) - x^2) where G(k) = 1 - x/(x + 1/(1 - x/G(k+1))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 15 2013
For the general case: The characteristic function of numbers that are not multiples of m is a(n) = floor((n-1)/m) - floor(n/m) + 1, with m,n > 0. - Boris Putievskiy, May 08 2013
a(n) = sign(n mod 3). - Wesley Ivan Hurt, Jun 22 2013
a(n) = A000035(A000032(n)) = A000035(A000045(n)). - Omar E. Pol, Oct 28 2013
a(n) = (-n mod 3)^((n-1) mod 3). - Wesley Ivan Hurt, Apr 16 2015
a(n) = (2/3) * (1 - sin((Pi/6) * (4*n + 3))) for n >= 0. - Werner Schulte, Jul 20 2017
a(n) = a(n-1) XOR a(n-2) with a(0) = 0, a(1) = 1. - Chunqing Liu, Dec 18 2022
a(n) = floor((n+2)/3) - floor(n/3) = A002264(n+2) - A002264(n). - Aaron J Grech, Jul 30 2024
E.g.f.: 2*(exp(x) - exp(-x/2)*cos(sqrt(3)*x/2))/3. - Stefano Spezia, Mar 30 2025
Dirichlet g.f.: zeta(s)*(1-1/3^s). - R. J. Mathar, Aug 10 2025

Extensions

Better name from Omar E. Pol, Oct 28 2013

A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1
Offset: 0

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T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

Examples

			Triangle T(n, k) begins:
  n\k     0     1     2     3     4     5    6   7   8  9
  0:      1
  1:      1     1
  2:      2     3     1
  3:      5     9     5     1
  4:     14    28    20     7     1
  5:     42    90    75    35     9     1
  6:    132   297   275   154    54    11    1
  7:    429  1001  1001   637   273    77   13   1
  8:   1430  3432  3640  2548  1260   440  104  15   1
  9:   4862 11934 13260  9996  5508  2244  663 135  17  1
  ... Reformatted by _Wolfdieter Lang_, Dec 21 2015
From _Paul Barry_, Feb 17 2011: (Start)
Production matrix begins
  1, 1,
  1, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =
  2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

Crossrefs

Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).

Programs

  • Magma
    /* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015
    
  • Maple
    T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006
    T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end:
    seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023
  • Mathematica
    Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *)
    Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *)
    Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)
  • PARI
    a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)
    trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))
    trianglerows(10) \\ Felix Fröhlich, Jun 24 2016
  • Sage
    # Algorithm of L. Seidel (1877)
    # Prints the first n rows of the triangle
    def A039599_triangle(n) :
        D = [0]*(n+2); D[1] = 1
        b = True ; h = 1
        for i in range(2*n-1) :
            if b :
                for k in range(h,0,-1) : D[k] += D[k-1]
                h += 1
            else :
                for k in range(1,h, 1) : D[k] += D[k+1]
            if b : print([D[z] for z in (1..h-1)])
            b = not b
    A039599_triangle(10)  # Peter Luschny, May 01 2012
    

Formula

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Extensions

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009
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