A050486 -id:A050486 - cp's OEIS frontend

A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

1, 12, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original name: The first summation of row 4 of Euler's triangle - a row that will recursively accumulate to the power of 4.

D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Dead link]
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (1).

For other sequences based upon MagicNKZ(n,k,z):
..... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7
---------------------------------------------------------------------------
z = 0 | A000007 | A019590 | .......MagicNKZ(n,k,0) = A008292(n,k+1) .......
z = 1 | A000012 | A040000 | A101101 | thisSeq | A101100 | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | A101095 | ....... | .......
z = 3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | .......
z = 4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | .......
z = 5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181
z = 6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177
z = 7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523
z = 8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015
z = 9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541
Cf. A101095 for an expanded table and more about MagicNKZ.

MagicNKZ = Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 1, 1}, {k, 0, 34}]
Join[{1, 12, 23},LinearRecurrence[{1},{24},56]] (* Ray Chandler, Sep 23 2015 *)

a(k) = MagicNKZ(4,k,1) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n (cf. A101095). That is, a(k) = Sum_{j=0..k+1} (-1)^j*binomial(4, j)*(k-j+1)^4.
a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4. - Joerg Arndt, Nov 30 2014
G.f.: x*(1+11*x+11*x^2+x^3)/(1-x). - Colin Barker, Apr 16 2012

New name from Joerg Arndt, Nov 30 2014

Original Formula edited and Crossrefs table added by Danny Rorabaugh, Apr 22 2015

A101095 Fourth difference of fifth powers (A000584).

Original entry on oeis.org

1, 28, 121, 240, 360, 480, 600, 720, 840, 960, 1080, 1200, 1320, 1440, 1560, 1680, 1800, 1920, 2040, 2160, 2280, 2400, 2520, 2640, 2760, 2880, 3000, 3120, 3240, 3360, 3480, 3600, 3720, 3840, 3960, 4080, 4200, 4320, 4440, 4560, 4680, 4800, 4920, 5040, 5160, 5280

Offset: 1

Cecilia Rossiter, Dec 15 2004

Original Name: Shells (nexus numbers) of shells of shells of shells of the power of 5.

The (Worpitzky/Euler/Pascal Cube) "MagicNKZ" algorithm is: MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n, with k>=0, n>=1, z>=0. MagicNKZ is used to generate the n-th accumulation sequence of the z-th row of the Euler Triangle (A008292). For example, MagicNKZ(3,k,0) is the 3rd row of the Euler Triangle (followed by zeros) and MagicNKZ(10,k,1) is the partial sums of the 10th row of the Euler Triangle. This sequence is MagicNKZ(5,k-1,2).

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Archive Machine link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883
Eric Weisstein, Link to section of MathWorld: Eulerian Number
Eric Weisstein, Link to section of MathWorld: Nexus number
Eric Weisstein, Link to section of MathWorld: Finite Differences
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Fourth differences of A000584, third differences of A022521, second differences of A101098, and first differences of A101096.
For other sequences based upon MagicNKZ(n,k,z):
...... |  n = 1  |  n = 2  |  n = 3  |  n = 4  |  n = 5  |  n = 6  |  n = 7  |  n = 8
--------------------------------------------------------------------------------------
z =  0 | A000007 | A019590 | .......  MagicNKZ(n,k,0) = T(n,k+1) from A008292  .......
z =  1 | A000012 | A040000 | A101101 | A101104 | A101100 | ....... | ....... | .......
z =  2 | A000027 | A005408 | A008458 | A101103 | thisSeq | ....... | ....... | .......
z =  3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | ....... | .......
z =  4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | ....... | .......
z =  5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181 | .......
z =  6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177 | A255182
z =  7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523 | A255178
z =  8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015 | A022524
z =  9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541 | A001016
z = 10 | A000582 | A054333 | A254469 | A254681 | A254644 | A254640 | A250212 | A000542
z = 11 | A001287 | A054334 | A254869 | A254470 | A254682 | A254645 | A254641 | A253636
z = 12 | A001288 | A057788 | ....... | A254870 | A254471 | A254683 | A254646 | A254642
z = 13 | A010965 | ....... | ....... | ....... | A254871 | A254472 | A254684 | A254647
z = 14 | A010966 | ....... | ....... | ....... | ....... | A254872 | ....... | .......
--------------------------------------------------------------------------------------
Cf. A047969.

I:=[1,28,121,240,360]; [n le 5 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, May 07 2015

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 5, 5}, {z, 2, 2}, {k, 0, 34}]
CoefficientList[Series[(1 + 26 x + 66 x^2 + 26 x^3 + x^4)/(1 - x)^2, {x, 0, 50}], x] (* Vincenzo Librandi, May 07 2015 *)
Join[{1,28,121,240},Differences[Range[50]^5,4]] (* or *) LinearRecurrence[{2,-1},{1,28,121,240,360},50] (* Harvey P. Dale, Jun 11 2016 *)

a(n)=if(n>3, 120*n-240, 33*n^2-72*n+40) \\ Charles R Greathouse IV, Oct 11 2015

[1,28,121]+[120*(k-2) for k in range(4,36)] # Danny Rorabaugh, Apr 23 2015

a(k+1) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n; n = 5, z = 2.
For k>3, a(k) = Sum_{j=0..4} (-1)^j*binomial(4, j)*(k - j)^5 = 120*(k - 2).
a(n) = 2*a(n-1) - a(n-2), n>5. G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (1-x)^2. - Colin Barker, Mar 01 2012

MagicNKZ material edited, Crossrefs table added, SeriesAtLevelR material removed by Danny Rorabaugh, Apr 23 2015

Name changed and keyword 'uned' removed by Danny Rorabaugh, May 06 2015

A121306 Array read by antidiagonals: a(m,n) = a(m,n-1)+a(m-1,n) but with initialization values a(0,0)=0, a(m>=1,0)=1, a(0,1)=1, a(0,n>1)=0.

Original entry on oeis.org

2, 2, 3, 2, 5, 4, 2, 7, 9, 5, 2, 9, 16, 14, 6, 2, 11, 25, 30, 20, 7, 2, 13, 36, 55, 50, 27, 8, 2, 15, 49, 91, 105, 77, 35, 9, 2, 17, 64, 140, 196, 182, 112, 44, 10, 19, 81, 204, 336, 378, 294, 156, 54, 100, 285, 540, 714, 672, 450, 210, 385, 825, 1254, 1386, 1122

Offset: 0

Thomas Wieder, Aug 04 2006, Aug 06 2006

For a(1,0)=1, a(m>1,0)=0 and a(0,n>=0)=0 one gets Pascal's triangle A007318.

			Array begins
2 2 2 2 2 2 2 2 2 ...
3 5 7 9 11 13 15 17 19 ...
4 9 16 25 36 49 64 81 100 ...
5 14 30 55 91 140 204 285 385 ...
6 20 50 105 196 336 540 825 1210 ...
7 27 77 182 378 714 1254 2079 3289 ...

The first nine rows are: A006527, A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347.
The initial columns are: A000027, A000096, A005581, A005582, A005583, A005584.
Cf. A119800, A007318, A006527, A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347, A000027, A000096, A005581, A005582, A005583, A005584.

=Z(-1)S+ZS(-1). The very first row (not included into the table) contains the initialization values: a(0,1)=1, a(0,n>=2)=0. The very first column (not included into the table) contains the initialization values: a(m>=1,0)=1. The value a(0,0)=0 does not enter into the table.

a(m,n) = a(m,n-1)+a(m-1,n), a(0,0)=0, a(m>=1,0)=1, a(0,1)=1, a(0,n>1)=0.

Edited by N. J. A. Sloane, Sep 15 2006

A191662 a(n) = n! / A000034(n-1).

Original entry on oeis.org

1, 1, 6, 12, 120, 360, 5040, 20160, 362880, 1814400, 39916800, 239500800, 6227020800, 43589145600, 1307674368000, 10461394944000, 355687428096000, 3201186852864000, 121645100408832000, 1216451004088320000, 51090942171709440000, 562000363888803840000

Offset: 1

Paul Curtz, Jun 10 2011

The a(n) are the denominators in the formulas of the k-dimensional square pyramidal numbers:
A005408 = (2*n+1)/1                       = 1, 3,  5,   7,   9, ... (k=1)
A000290 = (n^2)/1                         = 1, 4,  9,  16,  25, ... (k=2)
A000330 = n*(n+1)*(2*n+1)/6               = 1, 5, 14,  30,  55, ... (k=3)
A002415 = (n^2)*(n^2-1)/12                = 1, 6, 20,  50, 105, ... (k=4)
A005585 = n*(n+1)*(n+2)*(n+3)*(2*n+3)/120 = 1, 7, 27,  77, 182, ... (k=5)
A040977 = (n^2)*(n^2-1)*(n^2-4)/360       = 1, 8, 35, 112, 294, ... (k=6)
A050486 (k=7), A053347 (k=8), A054333 (k=9), A054334 (k=10), A057788 (k=11).
The first superdiagonal of this array appears in A029651. - Paul Curtz, Jul 04 2011
The general formula for the k-dimensional square pyramidal numbers is (2*n+k)*binomial(n+k-1,k-1)/k, k >= 1, n >= 0, see A097207. - Johannes W. Meijer, Jun 22 2011

Cf. A000142, A009445, A002674, A064680.
Cf. A000290, A000330, A002415, A005408, A005585, A029651, A040977, A050486, A053347, A054333, A054334, A057788.
Cf. A052612.

A191662:= proc(n): n!/A000034(n-1) end: A000034 := proc(n) op((n mod 2)+1, [1, 2]) ; end proc: seq(A191662(n),n=1..17); # Johannes W. Meijer, Jun 22 2011

Array[If[EvenQ[#],#!/2,#!]&,20] (* Harvey P. Dale, Mar 14 2014 *)

a(2*n-1) = (2*n-1)!, a(2*n) = (2*n)!/2.
a(n+1) = A064680(n+1) * a(n).
From Amiram Eldar, Jul 06 2022: (Start)
Sum_{n>=1} 1/a(n) = sinh(1) + 2*cosh(1) - 2.
Sum_{n>=1} (-1)^(n+1)/a(n) = sinh(1) - 2*cosh(1) + 2. (End)
D-finite with recurrence: a(n) - (n-1)*n*a(n-2) = 0 for n >= 3 with a(1)=a(2)=1. - Georg Fischer, Nov 25 2022
a(n) = A052612(n)/2 for n >= 1. - Alois P. Heinz, Sep 05 2023

More terms from Harvey P. Dale, Mar 14 2014

A208509 Triangle of coefficients of polynomials v(n,x) jointly generated with A208508; see the Formula section.

Original entry on oeis.org

1, 3, 5, 1, 7, 5, 9, 14, 1, 11, 30, 7, 13, 55, 27, 1, 15, 91, 77, 9, 17, 140, 182, 44, 1, 19, 204, 378, 156, 11, 21, 285, 714, 450, 65, 1, 23, 385, 1254, 1122, 275, 13, 25, 506, 2079, 2508, 935, 90, 1, 27, 650, 3289, 5148, 2717, 442, 15, 29, 819, 5005, 9867

Offset: 1

Clark Kimberling, Feb 27 2012

			First five rows:
  1
  3
  5    1
  7    5
  9   14   1
First five polynomials v(n,x):
  1
  3
  5 +   x
  7 +  5x
  9 + 14x + x^2

Columns 0 to 5: A005408, A000330, A005585, A050486, A054333, A057788.
Row sums, v(n,1): A003948.
Alternating row sums, v(n,-1): A090131.
Cf. A208508.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A208508 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A208509 *)

u(n,x) = u(n-1,x) + x*v(n-1,x), v(n,x) = u(n-1,x) + v(n-1,x) + 1, with u(1,x)=1, v(1,x)=1.

Conjecture: T(n,k) = binomial(n-1,2*k+1) + binomial(n,2*k+1). - Knud Werner, Jan 11 2022

A038164 Expansion of 1/((1-x)*(1-x^2))^4.

Original entry on oeis.org

1, 4, 14, 36, 85, 176, 344, 624, 1086, 1800, 2892, 4488, 6798, 10032, 14520, 20592, 28743, 39468, 53482, 71500, 94523, 123552, 159952, 205088, 260780, 328848, 411672, 511632, 631788, 775200, 945744, 1147296, 1384701, 1662804

Offset: 0

N. J. A. Sloane

nonn

From Gary W. Adamson, Mar 02 2010: (Start)
Given the tetrahedral numbers, A000292, shift the offset to 0; then
(1 + 4x + 10x^2 + 20x^3 + ...)*(1 + 4x^2 + 10x^4 + 20x^6 + ...) =
(1 + 4x^2 + 14x^3 + 36x^4 + ...) (End)

M. R. Bremner, Free associative algebras, noncommutative Grobner bases, and universal associative envelopes for nonassociative structures, arXiv preprint arXiv:1303.0920 [math.RA], 2013.
Index entries for linear recurrences with constant coefficients, signature (4,-2,-12,17,8,-28,8,17,-12,-2,4,-1).

Cf. A008619, A006918, A038163.

Cf. A000292. - Gary W. Adamson, Mar 02 2010

CoefficientList[Series[1/((1-x)(1-x^2))^4,{x,0,40}],x] (* or *) LinearRecurrence[ {4,-2,-12,17,8,-28,8,17,-12,-2,4,-1},{1,4,14,36,85,176,344,624,1086,1800,2892,4488},40] (* Harvey P. Dale, Jul 02 2011 *)

Vec(1/((1-x)*(1-x^2))^4 + O(x^40)) \\ Michel Marcus, Jan 13 2024

a(2*k) = (4*k^2 + 24*k + 21)*binomial(k + 5, 5)/21 = A059600(k); a(2*k + 1) = 4*binomial(k + 6, 6)*(7 + 2*k)/7 = 4*A050486(k), k >= 0.

a(0)=1, a(1)=4, a(2)=14, a(3)=36, a(4)=85, a(5)=176, a(6)=344, a(7)=624, a(8)=1086, a(9)=1800, a(10)=2892, a(11)=4488, a(n)=4*a(n-1)-2*a(n-2)- 12*a(n-3)+17*a(n-4)+8*a(n-5)-28*a(n-6)+8*a(n-7)+17*a(n-8)-12*a(n-9)- 2*a(n-10)+4*a(n-11)-a(n-12). - Harvey P. Dale, Jul 02 2011

A207543 Triangle read by rows, expansion of (1+yx)/(1-2y*x+y(y-1)x^2).

Original entry on oeis.org

1, 0, 3, 0, 1, 5, 0, 0, 5, 7, 0, 0, 1, 14, 9, 0, 0, 0, 7, 30, 11, 0, 0, 0, 1, 27, 55, 13, 0, 0, 0, 0, 9, 77, 91, 15, 0, 0, 0, 0, 1, 44, 182, 140, 17, 0, 0, 0, 0, 0, 11, 156, 378, 204, 19, 0, 0, 0, 0, 0, 1, 65, 450, 714, 285, 21, 0

Offset: 0

Philippe Deléham, Feb 24 2012

Previous name was: "A scaled version of triangle A082985."

Triangle, read by rows, given by (0, 1/3, -1/3, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (3, -4/3, 1/3, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

			Triangle begins :
1
0, 3
0, 1, 5
0, 0, 5, 7
0, 0, 1, 14, 9
0, 0, 0, 7, 30, 11
0, 0, 0, 1, 27, 55, 13
0, 0, 0, 0, 9, 77, 91, 15
0, 0, 0, 0, 1, 44, 182, 140, 17
0, 0, 0, 0, 0, 11, 156, 378, 204, 19
0, 0, 0, 0, 0, 1, 65, 450, 714, 285, 21
0, 0, 0, 0, 0, 0, 13, 275, 1122, 1254, 385, 23

R. Zielinski, Induction and Analogy in a Problem of Finite Sums arXiv:1608.04006 [math.GM], 2016.

Cf. A082985 which is another version of this triangle.

Cf. Diagonals : A005408, A000330, A005585, A050486, A054333, A057788. Cf. A119900.

s := (1+y*x)/(1-2*y*x+y*(y-1)*x^2): t := series(s,x,12):
seq(print(seq(coeff(coeff(t,x,n),y,m),m=0..n)),n=0..11); # Peter Luschny, Aug 17 2016

T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2), T(0,0) = 1, T(1,0) = 0, T(1,1) = 3.
G.f.: (1+y*x)/(1-2*y*x+y*(y-1)*x^2).
Sum_{i, i>=0} T(n+i,n) = A000204(2*n+1).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A078069(n), A000007(n), A003945(n), A111566(n) for x = -1, 0, 1, 2 respectively.
Sum_{k, 0<=k<=n} T(n,k)*x^(n-k) = A090131(n), A005408(n), A003945(n), A078057(n), A028859(n), A000244(n), A063782(n), A180168(n) for x = -1, 0, 1, 2, 3, 4, 5, 6 respectively.
From Peter Bala, Aug 17 2016: (Start)
Let S(k,n) = Sum_{i = 1..n} i^k. Calculations in Zielinski 2016 suggest the following identity holds involving the p-th row elements of this triangle:
Sum_{k = 0..p} T(p,k)*S(2*k,n) = 1/2*(2*n + 1)*(n*(n + 1))^p.
For example, for row 6 we find S(6,n) + 27*S(8,n) + 55*S(10,n) + 13*S(12,n) = 1/2*(2*n + 1)*(n*(n + 1))^6.
There appears to be a similar result for the odd power sums S(2*k + 1,n) involving A119900. (End)

New name using a formula of the author from Peter Luschny, Aug 17 2016

A129710 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 01 subwords (0 <= k <= floor(n/2)). A Fibonacci binary word is a binary word having no 00 subword.

Original entry on oeis.org

1, 2, 2, 1, 2, 3, 2, 5, 1, 2, 7, 4, 2, 9, 9, 1, 2, 11, 16, 5, 2, 13, 25, 14, 1, 2, 15, 36, 30, 6, 2, 17, 49, 55, 20, 1, 2, 19, 64, 91, 50, 7, 2, 21, 81, 140, 105, 27, 1, 2, 23, 100, 204, 196, 77, 8, 2, 25, 121, 285, 336, 182, 35, 1, 2, 27, 144, 385, 540, 378, 112, 9, 2, 29, 169, 506

Offset: 0

Emeric Deutsch, May 12 2007

Also number of Fibonacci binary words of length n and having k 10 subwords.
Row n has 1+floor(n/2) terms.
Row sums are the Fibonacci numbers (A000045).
T(n,0)=2 for n >= 1.
Sum_{k>=0} k*T(n,k) = A023610(n-2).
Triangle, with zeros omitted, given by (2, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Jan 14 2012
Riordan array ((1+x)/(1-x), x^2/(1-x)), zeros omitted. - Philippe Deléham, Jan 14 2012

			T(5,2)=4 because we have 10101, 01101, 01010 and 01011.
Triangle starts:
  1;
  2;
  2, 1;
  2, 3;
  2, 5, 1;
  2, 7, 4;
  2, 9, 9, 1;
Triangle (2, -1, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, ...) begins:
  1;
  2, 0;
  2, 1, 0;
  2, 3, 0, 0;
  2, 5, 1, 0, 0;
  2, 7, 4, 0, 0, 0;
  2, 9, 9, 1, 0, 0, 0;

Michael De Vlieger, Table of n, a(n) for n = 0..10200 (rows 0 <= n <= 200, flattened.)
Jean-Luc Baril and José Luis Ramírez, Descent distribution on Catalan words avoiding ordered pairs of Relations, arXiv:2302.12741 [math.CO], 2023.
Thomas Grubb and Frederick Rajasekaran, Set Partition Patterns and the Dimension Index, arXiv:2009.00650 [math.CO], 2020. Mentions this sequence.

Cf. A000045, A023610.
Cf. A029635, A029653.
Columns: A040000, A005408, A000290, A000330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.

T:=proc(n,k) if n=0 and k=0 then 1 elif k<=floor(n/2) then binomial(n-k,k)+binomial(n-k-1,k) else 0 fi end: for n from 0 to 18 do seq(T(n,k),k=0..floor(n/2)) od; # yields sequence in triangular form

MapAt[# - 1 &, #, 1] &@ Table[Binomial[n - k, k] + Binomial[n - k - 1, k], {n, 0, 16}, {k, 0, Floor[n/2]}] // Flatten (* Michael De Vlieger, Nov 15 2019 *)

T(n,k) = binomial(n-k,k) + binomial(n-k-1,k) for n >= 1 and 0 <= k <= floor(n/2).
G.f. = G(t,z) = (1+z)/(1-z-tz^2).
Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A078050(n), A057079(n), A040000(n), A000045(n+2), A000079(n), A006138(n), A026597(n), A133407(n), A133467(n), A133469(n), A133479(n), A133558(n), A133577(n), A063092(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively. - Philippe Deléham, Jan 14 2012
T(n,k) = T(n-1,k) + T(n-2,k-1) with T(0,0)=1, T(1,0)=2, T(1,1)=0 and T(n,k) = 0 if k > n or if k < 0. - Philippe Deléham, Jan 14 2012

A259775 Stepped path in P(k,n) array of k-th partial sums of squares (A000290).

Original entry on oeis.org

1, 5, 6, 20, 27, 77, 112, 294, 450, 1122, 1782, 4290, 7007, 16445, 27456, 63206, 107406, 243542, 419900, 940576, 1641486, 3640210, 6418656, 14115100, 25110020, 54826020, 98285670, 213286590, 384942375

Offset: 1

Luciano Ancora, Jul 05 2015

The term "stepped path" in the name field is the same used in A001405.

Interleaving of terms of the sequences A220101 and A129869. - Michel Marcus, Jul 05 2015

			The array of k-th partial sums of squares begins:
[1], [5],  14,   30,    55,     91,  ...  A000330
1,   [6], [20],  50,   105,    196,  ...  A002415
1,    7,  [27], [77],  182,    378,  ...  A005585
1,    8,   35, [112], [294],   672,  ...  A040977
1,    9,   44,  156,  [450], [1122], ...  A050486
1,   10,   54,  210,   660,  [1782], ...  A053347
This is essentially A110813 without its first two columns.

Cf. A000330, A002415, A005585, A040977, A050486, A053347.

Table[DifferenceRoot[Function[{a, n}, {(-9168 - 14432*n - 8412*n^2 - 2152*n^3 - 204*n^4)*a[n] +(-1332 - 1902*n - 792*n^2 - 102*n^3)*a[1 + n] + (2100 + 3884*n + 2493*n^2 + 640*n^3 + 51*n^4)*a[2 + n] == 0, a[1] == 1 , a[2] == 5}]][n], {n, 29}]

Conjecture: -(n+5)*(13*n-11)*a(n) +(8*n^2+39*n-35)*a(n-1) +2*(26*n^2+48*n+25)*a(n-2) -4*(8*n+5)*(n-1)*a(n-3)=0. - R. J. Mathar, Jul 16 2015

A266561 12-dimensional square numbers.

Original entry on oeis.org

1, 14, 104, 546, 2275, 8008, 24752, 68952, 176358, 419900, 940576, 1998724, 4056234, 7904456, 14858000, 27041560, 47805615, 82317690, 138389160, 227613750, 366913365, 580610160, 903171360, 1382805840, 2086129500, 3104160696, 4559958144, 6618272584

Offset: 0

Antal Pinter, Dec 31 2015

2*a(n) is number of ways to place 11 queens on an (n+11) X (n+11) chessboard so that they diagonally attack each other exactly 55 times. The maximal possible attack number, p=binomial(k,2)=55 for k=11 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form the corresponding complete graph.

Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 15.
OEIS Wiki, Square hyperpyramidal numbers, line d=12 of first table.

Cf. A000330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.

[Binomial(n+11,11)*(n+6)/6: n in [0..40]]; // Vincenzo Librandi, Jan 01 2016

CoefficientList[Series[(1 + x)/(1 - x)^13, {x, 0, 33}], x] (* Vincenzo Librandi, Jan 01 2016 *)

a(n) = binomial(n+11,11)*(n+6)/6.
a(n) = 2*binomial(n+12,12) - binomial(n+11,11).
a(n) = binomial(n+11,11) + 2*binomial(n+11,12) for n>0.
G.f.: (1+x)/(1-x)^13. - Vincenzo Librandi, Jan 01 2016

cp's OEIS Frontend

A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

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A101095 Fourth difference of fifth powers (A000584).

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A121306 Array read by antidiagonals: a(m,n) = a(m,n-1)+a(m-1,n) but with initialization values a(0,0)=0, a(m>=1,0)=1, a(0,1)=1, a(0,n>1)=0.

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A191662 a(n) = n! / A000034(n-1).

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A208509 Triangle of coefficients of polynomials v(n,x) jointly generated with A208508; see the Formula section.

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A038164 Expansion of 1/((1-x)*(1-x^2))^4.

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A207543 Triangle read by rows, expansion of (1+y*x)/(1-2*y*x+y*(y-1)*x^2).

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A207543 Triangle read by rows, expansion of (1+yx)/(1-2y*x+y(y-1)x^2).