A057884 -id:A057884 - cp's OEIS frontend

A058187 Expansion of (1+x)/(1-x^2)^4: duplicated tetrahedral numbers.

1, 1, 4, 4, 10, 10, 20, 20, 35, 35, 56, 56, 84, 84, 120, 120, 165, 165, 220, 220, 286, 286, 364, 364, 455, 455, 560, 560, 680, 680, 816, 816, 969, 969, 1140, 1140, 1330, 1330, 1540, 1540, 1771, 1771, 2024, 2024, 2300, 2300, 2600, 2600, 2925, 2925, 3276, 3276

Offset: 0

Henry Bottomley, Nov 20 2000

For n >= i, i = 6,7, a(n - i) is the number of incongruent two-color bracelets of n beads, i of which are black (cf. A005513, A032280), having a diameter of symmetry. The latter means the following: if we imagine (0,1)-beads as points (with the corresponding labels) dividing a circumference of a bracelet into n identical parts, then a diameter of symmetry is a diameter (connecting two beads or not) such that a 180-degree turn of one of two sets of points around it (obtained by splitting the circumference by this diameter) leads to the coincidence of the two sets (including their labels). - Vladimir Shevelev, May 03 2011
From Johannes W. Meijer, May 20 2011: (Start)
The Kn11, Kn12, Kn13, Fi1 and Ze1 triangle sums, see A180662 for their definitions, of the Connell-Pol triangle A159797 are linear sums of shifted versions of the duplicated tetrahedral numbers, e.g., Fi1(n) = a(n-1) + 5*a(n-2) + a(n-3) + 5*a(n-4).
The Kn11, Kn12, Kn13, Kn21, Kn22, Kn23, Fi1, Fi2, Ze1 and Ze2 triangle sums of the Connell sequence A001614 as a triangle are also linear sums of shifted versions of the sequence given above. (End)
The number of quadruples of integers [x, u, v, w] that satisfy x > u > v > w >= 0, n + 5 = x + u. - Michael Somos, Feb 09 2015
Also, this sequence is the fourth column in the triangle of the coefficients of the sum of two consecutive Fibonacci polynomials F(n+1, x) and F(n, x) (n>=0) in ascending powers of x. - Mohammad K. Azarian, Jul 18 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..880
Hansraj Gupta, Enumeration of incongruent cyclic k-gons, Indian J. Pure and Appl. Math., Vol. 10, No. 8 (1979), 964-999.
Vladimir Shevelev, A problem of enumeration of two-color bracelets with several variations, arXiv:0710.1370 [math.CO], 2007-2011.
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Cf. A057884. Sum of 2 consecutive terms gives A006918, whose sum of 2 consecutive terms gives A002623, whose sum of 2 consecutive terms gives A000292, which is this sequence without the duplication. Continuing to sum 2 consecutive terms gives A000330, A005900, A001845, A008412 successively.
Cf. A096338, A005513, A032280
Cf. A000292, A190717, A190718. - Johannes W. Meijer, May 20 2011

a058187 n = a058187_list !! n
a058187_list = 1 : f 1 1 [1] where
   f x y zs = z : f (x + y) (1 - y) (z:zs) where
     z = sum $ zipWith (*) [1..x] [x,x-1..1]
-- Reinhard Zumkeller, Dec 21 2011

A058187:= proc(n) option remember; A058187(n):= binomial(floor(n/2)+3, 3) end: seq(A058187(n), n=0..51); # Johannes W. Meijer, May 20 2011

a[n_]:= Length @ FindInstance[{x>u, u>v, v>w, w>=0, x+u==n+5}, {x, u, v, w}, Integers, 10^9]; (* Michael Somos, Feb 09 2015 *)
With[{tetra=Binomial[Range[30]+2,3]},Riffle[tetra,tetra]] (* Harvey P. Dale, Mar 22 2015 *)

{a(n) = binomial(n\2+3, 3)}; /* Michael Somos, Jun 07 2005 */

[binomial((n//2)+3, 3) for n in (0..60)] # G. C. Greubel, Feb 18 2022

a(n) = A006918(n+1) - a(n-1).
a(2*n) = a(2*n+1) = A000292(n) = (n+1)*(n+2)*(n+3)/6.
a(n) = (2*n^3 + 21*n^2 + 67*n + 63)/96 + (n^2 + 7*n + 11)(-1)^n/32. - Paul Barry, Aug 19 2003
a(n) = A108299(n-3,n)*(-1)^floor(n/2) for n > 2. - Reinhard Zumkeller, Jun 01 2005
Euler transform of finite sequence [1, 3]. - Michael Somos, Jun 07 2005
G.f.: 1 / ((1 - x) * (1 - x^2)^3) = 1 / ((1 + x)^3 * (1 - x)^4). a(n) = -a(-7-n) for all n in Z.
a(n) = binomial(floor(n/2) + 3, 3). - Vladimir Shevelev, May 03 2011
a(-n) = -a(n-7); a(n) = A000292(A008619(n)). - Guenther Schrack, Sep 13 2018
Sum_{n>=0} 1/a(n) = 3. - Amiram Eldar, Aug 18 2022

A058393 A square array based on 1^n (A000012) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 1, 2, 1, 1, 1, 2, 3, 1, 0, 1, 2, 4, 4, 1, 1, 1, 2, 4, 7, 5, 1, 0, 1, 2, 4, 8, 11, 6, 1, 1, 1, 2, 4, 8, 15, 16, 7, 1, 0, 1, 2, 4, 8, 16, 26, 22, 8, 1, 1, 1, 2, 4, 8, 16, 31, 42, 29, 9, 1, 0, 1, 2, 4, 8, 16, 32, 57, 64, 37, 10, 1, 1, 1, 2, 4, 8, 16, 32, 63, 99, 93, 46, 11, 1, 0

Offset: 0

Henry Bottomley, Nov 24 2000

Changing the formula by replacing T(0,2n)=T(1,n) by T(0,2n)=T(m,n) for some other value of m, would make the generating function change to coefficient of x^n in expansion of (1+x)^k/(1-x^2)^m. This would produce A058394, A058395, A057884, (and effectively A007318).

			Rows are (1,0,1,0,1,0,1,...), (1,1,1,1,1,1,...), (1,2,2,2,2,2,...), (1,3,4,4,4,...) etc.

Rows are A000035 (A000012 with zeros), A000012, A040000 etc. Columns are A000012, A001477, A000124, A000125, A000127, A006261, A008859, A008860, A008861, A008862, A008863 etc. Diagonals include A000079, A000225, A000295, A002662, A002663, A002664, A035038, A035039, A035040, A035041, etc. The triangles A008949, A054143 and A055248 also appear in the half of the array which is not powers of 2.

T(n, k)=T(n-1, k-1)+T(n, k-1) with T(0, k)=1, T(1, 1)=1, T(0, 2n)=T(1, n) and T(0, 2n+1)=0. Coefficient of x^n in expansion of (1+x)^k/(1-x^2).

A058395 Square array read by antidiagonals. Based on triangular numbers (A000217) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 3, 1, 1, 0, 3, 2, 1, 6, 3, 4, 3, 1, 0, 6, 6, 6, 4, 1, 10, 6, 9, 10, 9, 5, 1, 0, 10, 12, 15, 16, 13, 6, 1, 15, 10, 16, 21, 25, 25, 18, 7, 1, 0, 15, 20, 28, 36, 41, 38, 24, 8, 1, 21, 15, 25, 36, 49, 61, 66, 56, 31, 9, 1, 0, 21, 30, 45, 64, 85, 102, 104, 80, 39, 10, 1, 28, 21, 36, 55, 81, 113, 146, 168, 160, 111, 48, 11, 1

Offset: 0

Henry Bottomley, Nov 24 2000

Changing the formula by replacing T(2n, 0) = T(n, 3) with T(2n, 0) = T(n, m) for some other value of m would change the generating function to the coefficient of x^n in expansion of (1 + x)^k / (1 - x^2)^m. This would produce A058393, A058394, A057884 (and effectively A007318).

			The array T(n, k) starts:
[0] 1, 0,  3,   0,   6,   0,  10,    0,   15,    0, ...
[1] 1, 1,  3,   3,   6,   6,  10,   10,   15,   15, ...
[2] 1, 2,  4,   6,   9,  12,  16,   20,   25,   30, ...
[3] 1, 3,  6,  10,  15,  21,  28,   36,   45,   55, ...
[4] 1, 4,  9,  16,  25,  36,  49,   64,   81,  100, ...
[5] 1, 5, 13,  25,  41,  61,  85,  113,  145,  181, ...
[6] 1, 6, 18,  38,  66, 102, 146,  198,  258,  326, ...
[7] 1, 7, 24,  56, 104, 168, 248,  344,  456,  584, ...
[8] 1, 8, 31,  80, 160, 272, 416,  592,  800, 1040, ...
[9] 1, 9, 39, 111, 240, 432, 688, 1008, 1392, 1840, ...

Rows are A000217 with zeros, A008805, A002620, A000217, A000290, A001844, A005899.
Columns are A000012, A001477, A016028.
Diagonals include A058396, A049611, A001793, A001788, A055580, A055581, A055582.
The triangle A055252 also appears in half of the array.

gf := n -> (1 + x)^n / (1 - x^2)^3: ser := n -> series(gf(n), x, 20):
seq(lprint([n], seq(coeff(ser(n), x, k), k = 0..9)), n = 0..9); # Peter Luschny, Apr 12 2023

T[0, k_] := If[OddQ[k], 0, (k+2)(k+4)/8];
T[n_, k_] := T[n, k] = If[k == 0, 1, T[n-1, k-1] + T[n-1, k]];
Table[T[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Apr 13 2023 *)

T(n, k) = T(n-1, k-1) + T(n, k-1) with T(0, k) = 1, T(2*n, 0) = T(n, 3) and T(2*n + 1, 0) = 0. Coefficient of x^n in expansion of (1 + x)^k / (1 - x^2)^3.

A058394 A square array based on natural numbers (A000027) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 2, 1, 1, 0, 2, 2, 1, 3, 2, 3, 3, 1, 0, 3, 4, 5, 4, 1, 4, 3, 5, 7, 8, 5, 1, 0, 4, 6, 9, 12, 12, 6, 1, 5, 4, 7, 11, 16, 20, 17, 7, 1, 0, 5, 8, 13, 20, 28, 32, 23, 8, 1, 6, 5, 9, 15, 24, 36, 48, 49, 30, 9, 1, 0, 6, 10, 17, 28, 44, 64, 80, 72, 38, 10, 1, 7, 6, 11, 19, 32, 52, 80, 112, 129

Offset: 0

Henry Bottomley, Nov 24 2000

Changing the formula by replacing T(2n,0)=T(n,2) by T(2n,0)=T(n,m) for some other value of m, would make the generating function change to coefficient of x^n in expansion of (1+x)^k/(1-x^2)^m. This would produce A058393, A058395, A057884 (and effectively A007318).

			Rows are (1,0,2,0,3,0,4,...), (1,1,2,2,3,3,...), (1,2,3,4,5,6,...), (1,3,5,7,9,11,...), etc.

Rows are A027656 (A000027 with zeros), A008619, A000027, A005408, A008574 etc. Columns are A000012, A001477, A022856 etc. Diagonals include A034007, A045891, A045623, A001792, A001787, A000337, A045618, A045889, A034009, A055250, A055251 etc. The triangle A055249 also appears in half of the array.

T(n, k)=T(n-1, k-1)+T(n, k-1) with T(0, k)=1, T(2n, 0)=T(n, 2) and T(2n+1, 0)=0. Coefficient of x^n in expansion of (1+x)^k/(1-x^2)^2.

cp's OEIS Frontend

A058187 Expansion of (1+x)/(1-x^2)^4: duplicated tetrahedral numbers.

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A058393 A square array based on 1^n (A000012) with each term being the sum of 2 consecutive terms in the previous row.

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A058395 Square array read by antidiagonals. Based on triangular numbers (A000217) with each term being the sum of 2 consecutive terms in the previous row.

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Programs

Maple

Mathematica

Formula

A058394 A square array based on natural numbers (A000027) with each term being the sum of 2 consecutive terms in the previous row.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula