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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A008574 a(0) = 1, thereafter a(n) = 4n.

Original entry on oeis.org

1, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 144, 148, 152, 156, 160, 164, 168, 172, 176, 180, 184, 188, 192, 196, 200, 204, 208, 212, 216, 220, 224, 228, 232
Offset: 0

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Author

N. J. A. Sloane; entry revised Aug 24 2014

Keywords

Comments

Number of squares on the perimeter of an (n+1) X (n+1) board. - Jon Perry, Jul 27 2003
Coordination sequence for square lattice (or equivalently the planar net 4.4.4.4).
Apparently also the coordination sequence for the planar net 3.4.6.4. - Darrah Chavey, Nov 23 2014
From N. J. A. Sloane, Nov 26 2014: (Start)
I confirm that this is indeed the coordination sequence for the planar net 3.4.6.4. The points at graph distance n from a fixed point in this net essentially lie on a hexagon (see illustration in link).
If n = 3k, k >= 1, there are 2k + 1 nodes on each edge of the hexagon. This counts the corners of the hexagon twice, so the number of points in the shell is 6(2k + 1) - 6 = 4n. If n = 3k + 1, the numbers of points on the six edges of the hexagon are 2k + 2 (4 times) and 2k + 1 (twice), for a total of 12k + 10 - 6 = 4n. If n = 3k + 2 the numbers are 2k + 2 (4 times) and 2k + 3 twice, and again we get 4n points.
The illustration shows shells 0 through 12, as well as the hexagons formed by shells 9 (green, 36 points), 10 (black, 40 points), 11 (red, 44 points), and 12 (blue, 48 points).
It is clear from the net that this period-3 structure continues forever, and establishes the theorem.
In contrast, for the 4.4.4.4 planar net, the successive shells are diamonds instead of hexagons, and again the n-th shell (n > 0) contains 4n points.
Of course the two nets are very different, since 4.4.4.4 has the symmetry of the square, while 3.4.6.4 has only mirror symmetry (with respect to a point), and has the symmetry of a regular hexagon with respect to the center of any of the 12-gons. (End)
Also the coordination sequence for a 6.6.6.6 point in the 3-transitive tiling {4.6.6, 6.6.6, 6.6.6.6}, see A265045, A265046. - N. J. A. Sloane, Dec 27 2015
Also the coordination sequence for 2-dimensional cyclotomic lattice Z[zeta_4].
Susceptibility series H_1 for 2-dimensional Ising model (divided by 2).
Also the Engel expansion of exp^(1/4); cf. A006784 for the Engel expansion definition. - Benoit Cloitre, Mar 03 2002
This sequence differs from A008586, multiples of 4, only in its initial term. - Alonso del Arte, Apr 14 2011
Number of 2 X n binary matrices avoiding simultaneously the right angled numbered polyomino patterns (ranpp) (00,0), (00;1) and (10;1). An occurrence of a ranpp (xy;z) in a matrix A=(a(i,j)) is a triple (a(i1,j1), a(i1,j2), a(i2,j1)) where i1 < i2 and j1 < j2 and these elements are in same relative order as those in the triple (x,y,z). - Sergey Kitaev, Nov 11 2004
Central terms of the triangle in A118013. - Reinhard Zumkeller, Apr 10 2006
Also the coordination sequence for the htb net. - N. J. A. Sloane, Mar 31 2018
This is almost certainly also the coordination sequence for Dual(3.3.4.3.4) with respect to a tetravalent node. - Tom Karzes, Apr 01 2020
Minimal number of segments (equivalently, corners) in a rook circuit of a 2n X 2n board (maximal number is A085622). - Ruediger Jehn, Jan 02 2021

Examples

			From _Omar E. Pol_, Aug 20 2011 (Start):
Illustration of initial terms as perimeters of squares (cf. Perry's comment above):
.                                         o o o o o o
.                             o o o o o   o         o
.                   o o o o   o       o   o         o
.           o o o   o     o   o       o   o         o
.     o o   o   o   o     o   o       o   o         o
. o   o o   o o o   o o o o   o o o o o   o o o o o o
.
. 1    4      8        12         16           20
(End)
		

Crossrefs

Cf. A001844 (partial sums), A008586, A054275, A054410, A054389, A054764.
Convolution square of A040000.
Row sums of A130323 and A131032.
List of coordination sequences for uniform planar nets: A008458 (the planar net 3.3.3.3.3.3), A008486 (6^3), A008574 (4.4.4.4 and 3.4.6.4), A008576 (4.8.8), A008579(3.6.3.6), A008706 (3.3.3.4.4), A072154 (4.6.12), A219529(3.3.4.3.4), A250120 (3.3.3.3.6), A250122 (3.12.12).
List of coordination sequences for Laves tilings (or duals of uniform planar nets): [3,3,3,3,3.3] = A008486; [3.3.3.3.6] = A298014, A298015, A298016; [3.3.3.4.4] = A298022, A298024; [3.3.4.3.4] = A008574, A296368; [3.6.3.6] = A298026, A298028; [3.4.6.4] = A298029, A298031, A298033; [3.12.12] = A019557, A298035; [4.4.4.4] = A008574; [4.6.12] = A298036, A298038, A298040; [4.8.8] = A022144, A234275; [6.6.6] = A008458.
Coordination sequences for the 20 2-uniform tilings in the order in which they appear in the Galebach catalog, together with their names in the RCSR database (two sequences per tiling): #1 krt A265035, A265036; #2 cph A301287, A301289; #3 krm A301291, A301293; #4 krl A301298, A298024; #5 krq A301299, A301301; #6 krs A301674, A301676; #7 krr A301670, A301672; #8 krk A301291, A301293; #9 krn A301678, A301680; #10 krg A301682, A301684; #11 bew A008574, A296910; #12 krh A301686, A301688; #13 krf A301690, A301692; #14 krd A301694, A219529; #15 krc A301708, A301710; #16 usm A301712, A301714; #17 krj A219529, A301697; #18 kre A301716, A301718; #19 krb A301720, A301722; #20 kra A301724, A301726.
See also A265045, A265046.

Programs

  • Haskell
    a008574 0 = 1; a008574 n = 4 * n
    a008574_list = 1 : [4, 8 ..]  -- Reinhard Zumkeller, Apr 16 2015
  • Mathematica
    f[0] = 1; f[n_] := 4 n; Array[f, 59, 0] (* or *)
    CoefficientList[ Series[(1 + x)^2/(1 - x)^2, {x, 0, 58}], x] (* Robert G. Wilson v, Jan 02 2011 *)
    Join[{1},Range[4,232,4]] (* Harvey P. Dale, Aug 19 2011 *)
    a[ n_] := 4 n + Boole[n == 0]; (* Michael Somos, Jan 07 2019 *)
  • PARI
    {a(n) = 4*n + !n}; /* Michael Somos, Apr 16 2007 */
    

Formula

Binomial transform is A000337 (dropping the 0 there). - Paul Barry, Jul 21 2003
Euler transform of length 2 sequence [4, -2]. - Michael Somos, Apr 16 2007
G.f.: ((1 + x) / (1 - x))^2. E.g.f.: 1 + 4*x*exp(x). - Michael Somos, Apr 16 2007
a(-n) = -a(n) unless n = 0. - Michael Somos, Apr 16 2007
G.f.: exp(4*atanh(x)). - Jaume Oliver Lafont, Oct 20 2009
a(n) = a(n-1) + 4, n > 1. - Vincenzo Librandi, Dec 31 2010
a(n) = A005408(n-1) + A005408(n), n > 1. - Ivan N. Ianakiev, Jul 16 2012
a(n) = 4*n = A008586(n), n >= 1. - Tom Karzes, Apr 01 2020

A010701 Constant sequence: the all 3's sequence.

Original entry on oeis.org

3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
Offset: 0

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Author

Keywords

Comments

Decimal expansion of 1/3. - Raymond Wang, Mar 06 2010
Continued fraction expansion of (3+sqrt(13))/2. - Bruno Berselli, Mar 15 2011

Examples

			1/3 = 0.33333333333333333333333333333333333333333333... - _Bruno Berselli_, Mar 21 2014
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.3.1, p. 281.

Crossrefs

Programs

Formula

G.f.: 3/(1-x). - Bruno Berselli, Mar 15 2011
E.g.f.: 3*e^x. - Vincenzo Librandi, Jan 24 2012
a(n) = A040000(n) + A054977(n). - Reinhard Zumkeller, May 06 2012
a(n) = 3*A000012(n). - Michel Marcus, Dec 18 2015
a(n) = floor(1/(n - cot(1/n))). - Clark Kimberling, Mar 10 2020
Equals Sum_{k>=1} (1/4)^k (as a constant). - Michel Marcus, Jun 11 2020
Equals Sum_{k>=2} (k-1)/binomial(2*k,k) (as a constant). - Amiram Eldar, Jun 05 2021
Equals Sum_{k>=1} (-1)^(k+1)/2^k. - Michal Paulovic, Mar 02 2023

A029635 The (1,2)-Pascal triangle (or Lucas triangle) read by rows.

Original entry on oeis.org

2, 1, 2, 1, 3, 2, 1, 4, 5, 2, 1, 5, 9, 7, 2, 1, 6, 14, 16, 9, 2, 1, 7, 20, 30, 25, 11, 2, 1, 8, 27, 50, 55, 36, 13, 2, 1, 9, 35, 77, 105, 91, 49, 15, 2, 1, 10, 44, 112, 182, 196, 140, 64, 17, 2, 1, 11, 54, 156, 294, 378, 336, 204, 81, 19, 2, 1, 12, 65, 210, 450, 672, 714, 540, 285, 100
Offset: 0

Views

Author

Keywords

Comments

This is also called Vieta's array. - N. J. A. Sloane, Nov 22 2017
Dropping the first term and changing the boundary conditions to T(n,1)=n, T(n,n-1)=2 (n>=2), T(n,n)=1 yields the number of nonterminal symbols (which generate strings of length k) in a certain context-free grammar in Chomsky normal form that generates all permutations of n symbols. Summation over k (1<=k<=n) results in A003945. For the number of productions of this grammar: see A090327. Example: 1; 2, 1; 3, 2, 1; 4, 5, 2, 1; 5, 9, 7, 2, 1; 6, 14, 16, 9, 2, 1; In addition to the example of A090327 we have T(3,3)=#{S}=1, T(3,2)=#{D,E}=2 and T(3,1)=#{A,B,C}=3. - Peter R. J. Asveld, Jan 29 2004
Much as the original Pascal triangle gives the Fibonacci numbers as sums of its diagonals, this triangle gives the Lucas numbers (A000032) as sums of its diagonals; see Posamentier & Lehmann (2007). - Alonso del Arte, Apr 09 2012
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
It appears that for the infinite set of (1,N) Pascal's triangles, the binomial transform of the n-th row (n>0), followed by zeros, is equal to the n-th partial sum of (1, N, N, N, ...). Example: for the (1,2) Pascal's triangle, the binomial transform of the second row followed by zeros, i.e., of (1, 3, 2, 0, 0, 0, ...), is equal to the second partial sum of (1, 2, 2, 2, ...) = (1, 4, 9, 16, ...). - Gary W. Adamson, Aug 11 2015
Given any (1,N) Pascal triangle, let the binomial transform of the n-th row (n>1) followed by zeros be Q(x). It appears that the binomial transform of the (n-1)-th row prefaced by a zero is Q(n-1). Example: In the (1,2) Pascal triangle the binomial transform of row 3: (1, 4, 5, 2, 0, 0, 0, ...) is A000330 starting with 1: (1, 5, 14, 30, 55, 91, ...). The binomial transform of row 2 prefaced by a zero and followed by zeros, i.e., of (0, 1, 3, 2, 0, 0, 0, ...) is (0, 1, 5, 14, 30, 55, ...). - Gary W. Adamson, Sep 28 2015
It appears that in the array accompanying each (1,N) Pascal triangle (diagonals of the triangle), the binomial transform of (..., 1, N, 0, 0, 0, ...) preceded by (n-1) zeros generates the n-th row of the array (n>0). Then delete the zeros in the result. Example: in the (1,2) Pascal triangle, row 3 (1, 5, 14, 30, ...) is the binomial transform of (0, 0, 1, 2, 0, 0, 0, ...) with the resulting zeros deleted. - Gary W. Adamson, Oct 11 2015
Read as a square array (similar to the Example section Sq(m,j), but with Sq(0,0)=0 and Sq(m,j)=P(m+1,j) otherwise), P(n,k) are the multiplicities of the eigenvalues, lambda_n = n(n+k-1), of the Laplacians on the unit k-hypersphere, given by Teo (and Choi) as P(n,k) = (2n-k+1)(n+k-2)!/(n!(k-1)!). P(n,k) is also the numerator of a Dirichlet series for the Minakashisundarum-Pleijel zeta function for the sphere. Also P(n,k) is the dimension of the space of homogeneous, harmonic polynomials of degree k in n variables (Shubin, p. 169). For relations to Chebyshev polynomials and simple Lie algebras, see A034807. - Tom Copeland, Jan 10 2016
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, Jan 15 2016

Examples

			Triangle begins:
  [0] [2]
  [1] [1, 2]
  [2] [1, 3,  2]
  [3] [1, 4,  5,  2]
  [4] [1, 5,  9,  7,   2]
  [5] [1, 6, 14, 16,   9,  2]
  [6] [1, 7, 20, 30,  25, 11,  2]
  [7] [1, 8, 27, 50,  55, 36, 13,  2]
  [8] [1, 9, 35, 77, 105, 91, 49, 15, 2]
.
Read as a square, the array begins:
  n\k| 0  1   2    3    4    5
  --------------------------------------
  0 |  2  2   2    2    2    2   A040000
  1 |  1  3   5    7    9   11   A005408
  2 |  1  4   9   16   25   36   A000290
  3 |  1  5  14   30   55   91   A000330
  4 |  1  6  20   50  105  196   A002415
  5 |  1  7  27   77  182  378   A005585
  6 |  1  8  35  112  294  672   A040977
		

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers. New York: Prometheus Books (2007): 97 - 105.
  • M. Shubin and S. Andersson, Pseudodifferential Operators and Spectral Theory, Springer Series in Soviet Mathematics, 1987.

Crossrefs

Cf. A003945 (row sums), A007318, A034807, A061896, A029653 (row-reversed), A157000.
Sums along ascending antidiagonals give Lucas numbers, n>0.

Programs

  • Haskell
    a029635 n k = a029635_tabl !! n !! k
    a029635_row n = a029635_tabl !! n
    a029635_tabl = [2] : iterate
       (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,2]
    -- Reinhard Zumkeller, Mar 12 2012, Feb 23 2012
    
  • Maple
    T := proc(n, k) option remember;
    if n = k then 2 elif k = 0 then 1 else T(n-1, k-1) + T(n-1, k) fi end:
    for n from 0 to 8 do seq(T(n, k), k = 0..n) od;  # Peter Luschny, Dec 22 2024
  • Mathematica
    t[0, 0] = 2; t[n_, k_] := If[k < 0 || k > n, 0, Binomial[n, k] + Binomial[n-1, k-1]]; Flatten[Table[t[n, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, May 03 2011 *)
    (* The next program cogenerates A029635 and A029638. *)
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := u[n - 1, x] + v[n - 1, x]
    v[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1
    Table[Factor[u[n, x]], {n, 1, z}]
    Table[Factor[v[n, x]], {n, 1, z}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]   (* A029638  *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]   (* A029635 *)
    (* Clark Kimberling, Feb 20 2012 *)
    Table[Binomial[n,k]+Binomial[n-1,k-1],{n,0,20},{k,0,n}]//Flatten (* Harvey P. Dale, Feb 08 2024 *)
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, (n==0) + binomial(n, k) + binomial(n-1, k-1))}; /* Michael Somos, Jul 15 2003 */
    
  • Sage
    # uses[riordan_array from A256893]
    riordan_array((2-x)/(1-x), x/(1-x), 8) # Peter Luschny, Nov 09 2019

Formula

From Henry Bottomley, Apr 26 2002; (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k).
T(n, k) = C(n, k) + C(n-1, k-1).
T(n, k) = C(n, k)*(n + k)/n.
T(n, k) = A007318(n, k) + A007318(n-1, k-1).
T(n, k) = A061896(n + k, k) but with T(0, 0) = 1 and T(1, 1) = 2.
Row sum is floor(3^2(n-1)) i.e., A003945. (End)
G.f.: 1 + (1 + x*y) / (1 - x - x*y). - Michael Somos, Jul 15 2003
G.f. for n-th row: (x+2*y)*(x+y)^(n-1).
O.g.f. for row n: (1+x)/(1-x)^(n+1). The entries in row n are the nonzero entries in column n of A053120 divided by 2^(n-1). - Peter Bala, Aug 14 2008
T(2n, n) - T(2n, n+1)= Catalan(n)= A000108(n). - Philippe Deléham, Mar 19 2009
With T(0, 0) = 1 : Triangle T(n, k), read by rows, given by [1,0,0,0,0,0,...] DELTA [2,-1,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 10 2011
With T(0, 0) = 1, as in the Example section below, this is known as Vieta's array. The LU factorization of the square array is given by Yang and Leida, equation 20. - Peter Bala, Feb 11 2012
For n > 0: T(n, k) = A097207(n-1, k), 0 <= k < n. - Reinhard Zumkeller, Mar 12 2012
For n > 0: T(n, k) = A029600(n, k) - A007318(n, k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Riordan array ((2-x)/(1-x), x/(1-x)). - Philippe Deléham, Mar 15 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 4*x + 5*x^2/2! + 2*x^3/3!) = 1 + 5*x + 14*x^2/2! + 30*x^3/3! + 55*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
For n>=1: T(n, 0) + T(n, 1) + T(n, 2) = A000217(n+1). T(n, n-2) = (n-1)^2. - Bob Selcoe, Mar 29 2016:

Extensions

More terms from David W. Wilson
a(0) changed to 2 (was 1) by Daniel Forgues, Jul 06 2010

A194959 Fractalization of (1 + floor(n/2)).

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Sep 06 2011

Keywords

Comments

Suppose that p(1), p(2), p(3), ... is an integer sequence satisfying 1 <= p(n) <= n for n >= 1. Define g(1)=(1) and for n > 1, form g(n) from g(n-1) by inserting n so that its position in the resulting n-tuple is p(n). The sequence f obtained by concatenating g(1), g(2), g(3), ... is clearly a fractal sequence, here introduced as the fractalization of p. The interspersion associated with f is here introduced as the interspersion fractally induced by p, denoted by I(p); thus, the k-th term in the n-th row of I(p) is the position of the k-th n in f. Regarded as a sequence, I(p) is a permutation of the positive integers; its inverse permutation is denoted by Q(p).
...
Example: Let p=(1,2,2,3,3,4,4,5,5,6,6,7,7,...)=A008619. Then g(1)=(1), g(2)=(1,2), g(3)=(1,3,2), so that
f=(1,1,2,1,3,2,1,3,4,2,1,3,5,4,2,1,3,5,6,4,2,1,...)=A194959; and I(p)=A057027, Q(p)=A064578.
The interspersion I(P) has the following northwest corner, easily read from f:
1 2 4 7 11 16 22
3 6 10 15 21 28 36
5 8 12 17 23 30 38
9 14 20 27 35 44 54
...
Following is a chart of selected p, f, I(p), and Q(p):
p f I(p) Q(p)
Count odd numbers up to n, then even numbers down from n. - Franklin T. Adams-Watters, Jan 21 2012
This sequence defines the square array A(n,k), n > 0 and k > 0, read by antidiagonals and the triangle T(n,k) = A(n+1-k,k) for 1 <= k <= n read by rows (see Formula and Example). - Werner Schulte, May 27 2018

Examples

			The sequence p=A008619 begins with 1,2,2,3,3,4,4,5,5,..., so that g(1)=(1). To form g(2), write g(1) and append 2 so that in g(2) this 2 has position p(2)=2: g(2)=(1,2). Then form g(3) by inserting 3 at position p(3)=2: g(3)=(1,3,2), and so on. The fractal sequence A194959 is formed as the concatenation g(1)g(2)g(3)g(4)g(5)...=(1,1,2,1,3,2,1,3,4,2,1,3,5,4,2,...).
From _Werner Schulte_, May 27 2018: (Start)
This sequence seen as a square array read by antidiagonals:
  n\k: 1  2  3  4  5   6   7   8   9  10  11  12 ...
  ===================================================
   1   1  2  2  2  2   2   2   2   2   2   2   2 ... (see A040000)
   2   1  3  4  4  4   4   4   4   4   4   4   4 ... (see A113311)
   3   1  3  5  6  6   6   6   6   6   6   6   6 ...
   4   1  3  5  7  8   8   8   8   8   8   8   8 ...
   5   1  3  5  7  9  10  10  10  10  10  10  10 ...
   6   1  3  5  7  9  11  12  12  12  12  12  12 ...
   7   1  3  5  7  9  11  13  14  14  14  14  14 ...
   8   1  3  5  7  9  11  13  15  16  16  16  16 ...
   9   1  3  5  7  9  11  13  15  17  18  18  18 ...
  10   1  3  5  7  9  11  13  15  17  19  20  20 ...
  etc.
This sequence seen as a triangle read by rows:
  n\k:  1  2  3  4  5   6   7   8   9  10  11  12  ...
  ======================================================
   1    1
   2    1  2
   3    1  3  2
   4    1  3  4  2
   5    1  3  5  4  2
   6    1  3  5  6  4   2
   7    1  3  5  7  6   4   2
   8    1  3  5  7  8   6   4   2
   9    1  3  5  7  9   8   6   4   2
  10    1  3  5  7  9  10   8   6   4   2
  11    1  3  5  7  9  11  10   8   6   4   2
  12    1  3  5  7  9  11  12  10   8   6   4   2
  etc.
(End)
		

References

  • Clark Kimberling, "Fractal sequences and interspersions," Ars Combinatoria 45 (1997) 157-168.

Crossrefs

Cf. A000142, A000217, A005408, A005843, A008619, A057027, A064578, A209229, A210535, A219977; A000012 (col 1), A157532 (col 2), A040000 (row 1), A113311 (row 2); A194029 (introduces the natural fractal sequence and natural interspersion of a sequence - different from those introduced at A194959).
Cf. A003558 (g permutation order), A102417 (index), A330081 (on bits), A057058 (inverse).

Programs

  • Mathematica
    r = 2; p[n_] := 1 + Floor[n/r]
    Table[p[n], {n, 1, 90}]  (* A008619 *)
    g[1] = {1}; g[n_] := Insert[g[n - 1], n, p[n]]
    f[1] = g[1]; f[n_] := Join[f[n - 1], g[n]]
    f[20] (* A194959 *)
    row[n_] := Position[f[30], n];
    u = TableForm[Table[row[n], {n, 1, 5}]]
    v[n_, k_] := Part[row[n], k];
    w = Flatten[Table[v[k, n - k + 1], {n, 1, 13},
    {k, 1, n}]]  (* A057027 *)
    q[n_] := Position[w, n]; Flatten[
    Table[q[n], {n, 1, 80}]]  (* A064578 *)
    Flatten[FoldList[Insert[#1, #2, Floor[#2/2] + 1] &, {}, Range[10]]] (* Birkas Gyorgy, Jun 30 2012 *)
  • PARI
    T(n,k) = min(k<<1-1,(n-k+1)<<1); \\ Kevin Ryde, Oct 09 2020

Formula

From Werner Schulte, May 27 2018 and Jul 10 2018: (Start)
Seen as a triangle: It seems that the triangle T(n,k) for 1 <= k <= n (see Example) is the mirror image of A210535.
Seen as a square array A(n,k) and as a triangle T(n,k):
A(n,k) = 2*k-1 for 1 <= k <= n, and A(n,k) = 2*n for 1 <= n < k.
A(n+1,k+1) = A(n,k+1) + A(n,k) - A(n-1,k) for k > 0 and n > 1.
A(n,k) = A(k,n) - 1 for n > k >= 1.
P(n,x) = Sum_{k>0} A(n,k)*x^(k-1) = (1-x^n)*(1-x^2)/(1-x)^3 for n >= 1.
Q(y,k) = Sum_{n>0} A(n,k)*y^(n-1) = 1/(1-y) for k = 1 and Q(y,k) = Q(y,1) + P(k-1,y) for k > 1.
G.f.: Sum_{n>0, k>0} A(n,k)*x^(k-1)*y^(n-1) = (1+x)/((1-x)*(1-y)*(1-x*y)).
Sum_{k=1..n} A(n+1-k,k) = Sum_{k=1..n} T(n,k) = A000217(n) for n > 0.
Sum_{k=1..n} (-1)^(k-1) * A(n+1-k,k) = Sum_{k=1..n} (-1)^(k-1) * T(n,k) = A219977(n-1) for n > 0.
Product_{k=1..n} A(n+1-k,k) = Product_{k=1..n} T(n,k) = A000142(n) for n > 0.
A(n+m,n) = A005408(n-1) for n > 0 and some fixed m >= 0.
A(n,n+m) = A005843(n) for n > 0 and some fixed m > 0.
Let A_m be the upper left part of the square array A(n,k) with m rows and m columns. Then det(A_m) = 1 for some fixed m > 0.
The P(n,x) satisfy the recurrence equation P(n+1,x) = P(n,x) + x^n*P(1,x) for n > 0 and initial value P(1,x) = (1+x)/(1-x).
Let B(n,k) be multiplicative with B(n,p^e) = A(n,e+1) for e >= 0 and some fixed n > 0. That yields the Dirichlet g.f.: Sum_{k>0} B(n,k)/k^s = (zeta(s))^3/(zeta(2*s)*zeta(n*s)).
Sum_{k=1..n} A(k,n+1-k)*A209229(k) = 2*n-1. (conjectured)
(End)
From Kevin Ryde, Oct 09 2020: (Start)
T(n,k) = 2*k-1 if 2*k-1 <= n, or 2*(n+1-k) if 2*k-1 > n. [Lévy, chapter 1 section 1 equations (a),(b)]
Fixed points T(n,k)=k for k=1 and k = (2/3)*(n+1) when an integer. [Lévy, chapter 1 section 2 equation (3)]
(End)

Extensions

Name corrected by Franklin T. Adams-Watters, Jan 21 2012

A299254 Coordination sequence for 3D uniform tiling formed by stacking parallel layers of the 3^4.6 2D tiling (cf. A250120).

Original entry on oeis.org

1, 7, 21, 45, 79, 122, 175, 237, 309, 391, 482, 583, 693, 813, 943, 1082, 1231, 1389, 1557, 1735, 1922, 2119, 2325, 2541, 2767, 3002, 3247, 3501, 3765, 4039, 4322, 4615, 4917, 5229, 5551, 5882, 6223, 6573, 6933, 7303, 7682, 8071, 8469, 8877, 9295, 9722, 10159, 10605, 11061, 11527, 12002
Offset: 0

Views

Author

N. J. A. Sloane, Feb 06 2018

Keywords

References

  • B. Grünbaum, Uniform tilings of 3-space, Geombinatorics, 4 (1994), 49-56. See tiling #17.

Crossrefs

Partial sums: A299260.
The 28 uniform 3D tilings: cab: A299266, A299267; crs: A299268, A299269; fcu: A005901, A005902; fee: A299259, A299265; flu-e: A299272, A299273; fst: A299258, A299264; hal: A299274, A299275; hcp: A007899, A007202; hex: A005897, A005898; kag: A299256, A299262; lta: A008137, A299276; pcu: A005899, A001845; pcu-i: A299277, A299278; reo: A299279, A299280; reo-e: A299281, A299282; rho: A008137, A299276; sod: A005893, A005894; sve: A299255, A299261; svh: A299283, A299284; svj: A299254, A299260; svk: A010001, A063489; tca: A299285, A299286; tcd: A299287, A299288; tfs: A005899, A001845; tsi: A299289, A299290; ttw: A299257, A299263; ubt: A299291, A299292; bnn: A007899, A007202. See the Proserpio link in A299266 for overview.

Programs

  • Mathematica
    LinearRecurrence[{2, -1, 0, 0, 1, -2, 1}, {1, 7, 21, 45, 79, 122, 175, 237}, 50] (* Paolo Xausa, Jan 16 2025 *)
  • PARI
    Vec((1 + x)*(1 + x + x^2)*(1 + 3*x + 3*x^3 + x^4) / ((1 - x)^3*(1 + x + x^2 + x^3 + x^4)) + O(x^60)) \\ Colin Barker, Feb 07 2018

Formula

G.f.: (x^2+x+1)*(x^4+3*x^3+3*x+1)*(x+1) / ((x^4+x^3+x^2+x+1)*(1-x)^3). (This is the product of the g.f.'s for A250120 and A040000. - N. J. A. Sloane, Nov 10 2018)
a(n) = 2*a(n-1) - a(n-2) + a(n-5) - 2*a(n-6) + a(n-7) for n>7. - Colin Barker, Feb 07 2018
a(n) = 2*((sqrt(5) - 5)*(5 + 12*n^2) - (sqrt(5) - 1)*cos(2*n*Pi/5) + (sqrt(5) - 1)*cos(4*n*Pi/5))/(5*(sqrt(5) - 5)) for n > 0. - Stefano Spezia, Jun 06 2024

A097806 Riordan array (1+x, 1) read by rows.

Original entry on oeis.org

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

Views

Author

Paul Barry, Aug 25 2004

Keywords

Comments

Pair sum operator. Columns have g.f. (1+x)*x^k. Row sums are A040000. Diagonal sums are (1,1,1,....). Riordan inverse is (1/(1+x), 1). A097806 = B*A059260^(-1), where B is the binomial matrix.
Triangle T(n,k), 0<=k<=n, read by rows given by [1, -1, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, May 01 2007
Table T(n,k) read by antidiagonals. T(n,1) = 1, T(n,2) = 1, T(n,k) = 0, k > 2. - Boris Putievskiy, Jan 17 2013

Examples

			Rows begin {1}, {1,1}, {0,1,1}, {0,0,1,1}...
From _Boris Putievskiy_, Jan 17 2013: (Start)
The start of the sequence as table:
1..1..0..0..0..0..0...
1..1..0..0..0..0..0...
1..1..0..0..0..0..0...
1..1..0..0..0..0..0...
1..1..0..0..0..0..0...
1..1..0..0..0..0..0...
1..1..0..0..0..0..0...
. . .
The start of the sequence as triangle array read by rows:
  1;
  1, 1;
  0, 1, 1;
  0, 0, 1, 1;
  0, 0, 0, 1, 1;
  0, 0, 0, 0, 1, 1;
  0, 0, 0, 0, 0, 1, 1;
  0, 0, 0, 0, 0, 0, 1, 1; . . .
Row number r (r>4) contains (r-2) times '0' and 2 times '1'. (End)
		

Programs

  • Magma
    [k eq n or k eq n-1 select 1 else 0: k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 11 2019
    
  • Maple
    A097806 := proc(n,k)
        if k =n or k=n-1 then
            1;
        else
            0;
        end if;
    end proc: # R. J. Mathar, Jun 20 2015
  • Mathematica
    Table[Boole[n <= # <= n+1] & /@ Range[n+1], {n, 0, 15}] // Flatten (* or *)
    Table[Floor[(# +2)/(n+2)] & /@ Range[n+1], {n, 0, 15}] // Flatten (* Michael De Vlieger, Jul 21 2016 *)
  • PARI
    T(n, k) = if(k==n || k==n-1, 1, 0); \\ G. C. Greubel, Jul 11 2019
    
  • Sage
    def T(n, k):
        if (k==n or k==n-1): return 1
        else: return 0
    [[T(n, k) for k in (0..n)] for n in (0..15)] # G. C. Greubel, Jul 11 2019

Formula

T(n, k) = if(n=k or n-k=1, 1, 0).
a(n) = A103451(n+1). - Philippe Deléham, Oct 16 2007
From Boris Putievskiy, Jan 17 2013: (Start)
a(n) = floor((A002260(n)+2)/(A003056(n)+2)), n > 0.
a(n) = floor((i+2)/(t+2)), n > 0,
where i=n-t*(t+1)/2, t=floor((-1+sqrt(8*n-7))/2). (End)
G.f.: (1+x)/(1-x*y). - R. J. Mathar, Aug 11 2015

A054977 a(0)=2, a(n)=1 for n >= 1.

Original entry on oeis.org

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

Views

Author

Henry Gould, May 29 2000

Keywords

Comments

Arises in Gilbreath-Proth conjecture; see A036262.
a(n) is also the continued fraction for (3+sqrt(5))/2. - Enrique Pérez Herrero, May 16 2010
a(n) is also the denominator for odd Bernoulli Numbers. - Enrique Pérez Herrero, Jul 17 2010
a(n) = 3 - A040000(n); a(n) = A182579(n+1,1). - Reinhard Zumkeller, May 07 2012
From Paul Curtz, Feb 04 2014: (Start)
Difference table of a(n):
2, 1, 1, 1, 1, 1, 1, ...
-1, 0, 0, 0, 0, 0, 0, ...
1, 0, 0, 0, 0, 0, 0, ...
-1, 0, 0, 0, 0, 0, 0, ...
1, 0, 0, 0, 0, 0, 0, ...
-1, 0, 0, 0, 0, 0, 0, ... .
a(n) is an autosequence of second kind. Its inverse binomial transform is the signed sequence with the main diagonal (here A000038) double of the following diagonal (here A000007). Here the other diagonals are also A000007.
b(n) = A000032(n) - a(n) = 0, 0, 2, 3, 6, 10, 17, 28, ... = 0, followed by A001610(n) is the autosequence of second kind preceding A000032(n).
The corresponding autosequence of first kind, 0 followed by 1's, is A057427(n).
The Akiyama-Tanigawa transform applied to a(n) yields a(n).
(End)
Harmonic or factorial (base) expansion of e, cf. MathWorld link. - M. F. Hasler, Nov 25 2018
Decimal expansion of 19/90. - Elmo R. Oliveira, Aug 09 2024

Crossrefs

Programs

Formula

a(n) = A027642(2*n+1). - Enrique Pérez Herrero, Jul 17 2010
G.f.: (2-x)/(1-x). - Wolfdieter Lang, Oct 05 2014
Sum_{k>=1} a(n)/n! = exp(1). - G. C. Greubel, Nov 26 2018

A042964 Numbers that are congruent to 2 or 3 mod 4.

Original entry on oeis.org

2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 67, 70, 71, 74, 75, 78, 79, 82, 83, 86, 87, 90, 91, 94, 95, 98, 99, 102, 103, 106, 107, 110, 111, 114, 115, 118, 119, 122, 123, 126, 127
Offset: 1

Views

Author

Keywords

Comments

Also numbers m such that binomial(m+2, m) mod 2 = 0. - Hieronymus Fischer, Oct 20 2007
Also numbers m such that floor(1+(m/2)) mod 2 = 0. - Hieronymus Fischer, Oct 20 2007
Partial sums of the sequence 2, 1, 3, 1, 3, 1, 3, 1, 3, 1, ... which has period 2. - Hieronymus Fischer, Oct 20 2007
In groups of four add and divide by two the odd and even numbers. - George E. Antoniou, Dec 12 2001
From Jeremy Gardiner, Jan 22 2006: (Start)
Comments on the "mystery calculator". There are 6 cards.
Card 0: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, ... (A005408 sequence).
Card 1: 2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, ... (this sequence).
Card 2: 4, 5, 6, 7, 12, 13, 14, 15, 20, 21, 22, 23, 28, 29, 30, 31, 36, 37, 38, 39, ... (A047566).
Card 3: 8, 9, 10, 11, 12, 13, 14, 15, 24, 25, 26, 27, 28, 29, 30, 31, 40, 41, 42, ... (A115419).
Card 4: 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 48, 49, 50, ... (A115420).
Card 5: 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ... (A115421).
The trick: You secretly select a number between 1 and 63 from one of the cards. You indicate to me the cards on which that number appears; I tell you the number you selected!
The solution: I add together the first term from each of the indicated cards. The total equals the selected number. The numbers in each sequence all have a "1" in the same position in their binary expansion. Example: You indicate cards 1, 3 and 5. Your selected number is 2 + 8 + 32 = 42.
Numbers having a 1 in position 1 of their binary expansion. One of the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. (End)
Complement of A042948. - Reinhard Zumkeller, Oct 03 2008
Also the 2nd Witt transform of A040000 [Moree]. - R. J. Mathar, Nov 08 2008
In general, sequences of numbers congruent to {a,a+i} mod k will have a closed form of (k-2*i)*(2*n-1+(-1)^n)/4+i*n+a, from offset 0. - Gary Detlefs, Oct 29 2013
Union of A004767 and A016825; Fixed points of A098180. - Wesley Ivan Hurt, Jan 14 2014, Oct 13 2015

Crossrefs

Programs

  • Magma
    [2*n+((-1)^(n-1)-1)/2 : n in [1..100]]; // Wesley Ivan Hurt, Oct 13 2015
    
  • Magma
    [n: n in [1..150] | n mod 4 in [2, 3]]; // Vincenzo Librandi, Oct 13 2015
    
  • Maple
    A042964:=n->2*n+((-1)^(n-1)-1)/2; seq(A042964(n), n=1..100); # Wesley Ivan Hurt, Jan 07 2014
  • Mathematica
    Flatten[Table[4n + {2, 3}, {n, 0, 31}]] (* Alonso del Arte, Feb 07 2013 *)
    Select[Range[200],MemberQ[{2,3},Mod[#,4]]&] (* or *) LinearRecurrence[ {1,1,-1},{2,3,6},90] (* Harvey P. Dale, Nov 28 2018 *)
  • PARI
    a(n)=2*n+2-n%2
    
  • PARI
    Vec((2+x+x^2)/((1-x)*(1-x^2)) + O(x^100)) \\ Altug Alkan, Oct 13 2015

Formula

a(n) = A047406(n)/2.
From Michael Somos, Jan 12 2000: (Start)
G.f.: x*(2+x+x^2)/((1-x)*(1-x^2)).
a(n) = a(n-1) + 2 + (-1)^n. (End)
a(n) = 2n if n is odd, otherwise n = 2n - 1. - Amarnath Murthy, Oct 16 2003
a(n) = (3 + (-1)^(n-1))/2 + 2*(n-1) = 2n + 2 - (n mod 2). - Hieronymus Fischer, Oct 20 2007
A133872(a(n)) = 0. - Reinhard Zumkeller, Oct 03 2008
a(n) = 4*n - a(n-1) - 3 (with a(1) = 2). - Vincenzo Librandi, Nov 17 2010
a(n) = 2*n + ((-1)^(n-1) - 1)/2. - Gary Detlefs, Oct 29 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 - log(2)/4. - Amiram Eldar, Dec 05 2021
E.g.f.: 1 + ((4*x - 1)*exp(x) - exp(-x))/2. - David Lovler, Aug 08 2022

Extensions

Edited by N. J. A. Sloane, Jun 30 2008 at the suggestion of R. J. Mathar
Corrected by Jaroslav Krizek, Dec 18 2009

A103881 Square array T(n,k) (n >= 1, k >= 0) read by antidiagonals: coordination sequence for root lattice A_n.

Original entry on oeis.org

1, 1, 2, 1, 6, 2, 1, 12, 12, 2, 1, 20, 42, 18, 2, 1, 30, 110, 92, 24, 2, 1, 42, 240, 340, 162, 30, 2, 1, 56, 462, 1010, 780, 252, 36, 2, 1, 72, 812, 2562, 2970, 1500, 362, 42, 2, 1, 90, 1332, 5768, 9492, 7002, 2570, 492, 48, 2, 1, 110, 2070, 11832, 26474, 27174, 14240, 4060, 642, 54, 2, 1, 132, 3080, 22530, 66222, 91112, 65226, 26070, 6040, 812, 60, 2
Offset: 1

Views

Author

Ralf Stephan, Feb 20 2005

Keywords

Comments

T(n,k) is the number of integer sequences of length n+1 with sum zero and sum of absolute values 2k. - R. H. Hardin, Feb 23 2009

Examples

			Array begins:
  1,   2,     2,      2,       2,        2,         2,          2, ... A040000;
  1,   6,    12,     18,      24,       30,        36,         42, ... A008458;
  1,  12,    42,     92,     162,      252,       362,        492, ... A005901;
  1,  20,   110,    340,     780,     1500,      2570,       4060, ... A008383;
  1,  30,   240,   1010,    2970,     7002,     14240,      26070, ... A008385;
  1,  42,   462,   2562,    9492,    27174,     65226,     137886, ... A008387;
  1,  56,   812,   5768,   26474,    91112,    256508,     623576, ... A008389;
  1,  72,  1332,  11832,   66222,   271224,    889716,    2476296, ... A008391;
  1,  90,  2070,  22530,  151560,   731502,   2777370,    8809110, ... A008393;
  1, 110,  3080,  40370,  322190,  1815506,   7925720,   28512110, ... A008395;
  1, 132,  4422,  68772,  643632,  4197468,  20934474,   85014204, ... A035837;
  1, 156,  6162, 112268, 1219374,  9129276,  51697802,  235895244, ... A035838;
  1, 182,  8372, 176722, 2206932, 18827718, 120353324,  614266354, ... A035839;
  1, 210, 11130, 269570, 3838590, 37060506, 265953170, 1511679210, ... A035840;
  ...
Antidiagonals:
  1;
  1,  2;
  1,  6,    2;
  1, 12,   12,    2;
  1, 20,   42,   18,    2;
  1, 30,  110,   92,   24,    2;
  1, 42,  240,  340,  162,   30,    2;
  1, 56,  462, 1010,  780,  252,   36,   2;
  1, 72,  812, 2562, 2970, 1500,  362,  42,  2;
  1, 90, 1332, 5768, 9492, 7002, 2570, 492, 48,  2;
		

Crossrefs

Programs

  • GAP
    T:=Flat(List([1..12],n->Concatenation([1],List([1..n-1],k->Sum([1..n],i->Binomial(n-k+1,i)*Binomial(k-1,i-1)*Binomial(n-i,k)))))); # Muniru A Asiru, Oct 14 2018
    
  • Magma
    A103881:= func< n,k | k le 0 select 1 else (&+[Binomial(n-k+1, j)*Binomial(k-1, j-1)*Binomial(n-j, k): j in [1..n-k]]) >;
    [A103881(n,k): k in [0..n-1], n in [1..15]]; // G. C. Greubel, Oct 16 2018; May 24 2023
    
  • Maple
    T:=proc(n,k) option remember; local i;
    if k=0 then 1 else
    add( binomial(n+1,i)*binomial(k-1,i-1)*binomial(n-i+k,k),i=1..n); fi;
    end:
    g:=n->[seq(T(n-i,i),i=0..n-1)]:
    for n from 1 to 14 do lprint(op(g(n))); od:
  • Mathematica
    T[n_, k_]:= (n+1)*(n+k-1)!*HypergeometricPFQ[{1-k,1-n,-n}, {2,-n-k+1}, 1]/(k!*(n-1)!); T[, 0]=1; Flatten[Table[T[n-k, k], {n,12}, {k,0,n-1}]] (* _Jean-François Alcover, Dec 27 2012 *)
  • PARI
    A103881(n,k) = if(k==0, 1, sum(j=1, n-k, binomial(n-k+1, j)*binomial(k-1, j-1)*binomial(n-j, k)));
    for(n=1, 15, for(k=0, n-1, print1(A103881(n,k), ", "))) \\ G. C. Greubel, Oct 16 2018; May 24 2023
    
  • SageMath
    def A103881(n,k): return 1 if k==0 else (n-k+1)*binomial(n-1,k)*hypergeometric([k-n,1+k-n,1-k], [2,1-n], 1).simplify()
    flatten([[A103881(n,k) for k in range(n)] for n in range(1,16)]) # G. C. Greubel, May 24 2023

Formula

T(n,k) = Sum_{i=1..n} C(n+1, i)*C(k-1, i-1)*C(n-i+k, k), T(n,0)=1.
G.f. of n-th row: (Sum_{i=0..n} C(n, i)^2*x^i)/(1-x)^n.
From G. C. Greubel, May 24 2023: (Start)
T(n, k) = Sum_{j=0..n} binomial(n,j)^2 * binomial(n+k-j-1, n-1) (array).
T(n, k) = (n+1)*binomial(n+k-1,k)*hypergeometric([-n,1-n,1-k], [2,1-n-k], 1), with T(n, k) = 1 (array).
t(n, k) = (n-k+1)*binomial(n-1,k)*hypergeometric([k-n,1+k-n,1-k], [2,1-n], 1), with t(n, 0) = 1 (antidiagonals).
Sum_{k=0..n-1} t(n, k) = A047085(n). (End)
From Peter Bala, Jul 09 2023: (Start)
T(n,k) = [x^k] Legendre_P(n, (1 + x)/(1 - x)).
(n+1)*T(n+1,k) = (n+1)*T(n+1,k-1) + (2*n+1)*(T(n,k) + T(n,k-1)) - n*(T(n-1,k) - T(n-1,k-1)). (End)

Extensions

Corrected by N. J. A. Sloane, Dec 15 2012, at the suggestion of Manuel Blum

A103451 Triangular array T read by rows: T(n, 0) = T(n, n) = 1, T(n, k) = 0 for 0 < k < n.

Original entry on oeis.org

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

Views

Author

Paul Barry, Feb 06 2005

Keywords

Comments

Equals Pascal's triangle (A007318) where all elements > 1 are replaced with zero. Therefore it might be called "binomial skeleton".
Row sums are in A040000, antidiagonal sums are in A040001. When construed as a lower triangular matrix, the matrix inverse is A103452.

Examples

			First few rows are:
  1;
  1, 1;
  1, 0, 1;
  1, 0, 0, 1;
  1, 0, 0, 0, 1;
  1, 0, 0, 0, 0, 1;
  ...
		

Crossrefs

Programs

  • Magma
    r:=14; T:=ScalarMatrix(r, 1); for n in [1..r] do T[n, 1]:=1; end for; &cat[ [ T[n, k]: k in [1..n] ]: n in [1..r] ];
    
  • Magma
    /* As triangle */ [[Binomial(n, k-n)+Binomial(n, -k)-Binomial(0, n+k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 20 2016
    
  • Mathematica
    Table[Boole[n == 0 || Mod[k, n] == 0], {n, 0, 14}, {k, 0, n}] (* or *)
    Table[Binomial[n, k - n] + Binomial[n, -k] - Binomial[0, n + k], {n, 0, 14}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 19 2016 *)
  • PARI
    for(n=0,15, for(k=0,n, print1(if(k==0||k==n, 1, 0), ", "))) \\ G. C. Greubel, Dec 08 2018
    
  • Python
    from math import isqrt, comb
    def A103451(n):
        if n==0: return 1
        a = (m:=isqrt(k:=n+1<<1))-(k<=m*(m+1))
        return int(not (n-comb(a+1,2))%a) # Chai Wah Wu, Jun 24 2025
  • Sage
    def A103451(n,k): return 1 if (k==0 or k==n) else 0
    flatten([[A103451(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Feb 14 2021
    

Formula

a(n) = A097806(n-1) for n > 0. - Philippe Deléham, Oct 16 2007
T(n,k) = C(n,k-n) + C(n,-k) - C(0,n+k), 0 <= k <= n. - Eric Werley, Jul 01 2011
From Stefano Spezia, Jul 04 2024: (Start)
G.f.: (1 - x^2*y)/((1 - x)*(1 - x*y)).
E.g.f.: BesselI(0, 2*sqrt(x*y)) + exp(x) - 1. (End)

Extensions

Edited by Klaus Brockhaus, Jan 26 2011
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