A073254 -id:A073254 - cp's OEIS frontend

A132111 Triangle read by rows: T(n,k) = n^2 + k*n + k^2, 0 <= k <= n.

0, 1, 3, 4, 7, 12, 9, 13, 19, 27, 16, 21, 28, 37, 48, 25, 31, 39, 49, 61, 75, 36, 43, 52, 63, 76, 91, 108, 49, 57, 67, 79, 93, 109, 127, 147, 64, 73, 84, 97, 112, 129, 148, 169, 192, 81, 91, 103, 117, 133, 151, 171, 193, 217, 243, 100, 111, 124, 139, 156, 175, 196, 219

Offset: 0

Reinhard Zumkeller, Aug 10 2007

Permutation of A003136, the Loeschian numbers. [This is false - some terms are repeated, the first being 49. - Joerg Arndt, Dec 18 2015]
Row sums give A132112.
Central terms give A033582.
T(n,k+1) = T(n,k) + n + 2*k + 1;
T(n+1,k) = T(n,k) + 2*n + k + 1;
T(n+1,k+1) = T(n,k) + 3*(n+k+1);
T(n,0) = A000290(n);
T(n,1) = A002061(n+1) for n>0;
T(n,2) = A117950(n+1) for n>1;
T(n,n-2) = A056107(n-1) for n>1;
T(n,n-1) = A003215(n-1) for n>0;
T(n,n) = A033428(n).
T(n,k) is the norm N(alpha) of the integer alpha = n*1 - k*omega, where omega = exp(2*Pi*i/3) = (-1 + i*sqrt(3))/2 in the imaginary quadratic number field Q(sqrt(-3)): N = |alpha|^2 = (n + k/2)^2 + (3/4)*k^2  = n^2 + n*k + k^2 = T(n,k), with n >= 0, and k <= n. See also triangle A073254 for T(n,-k). - Wolfdieter Lang, Jun 13 2021

			From _Philippe Deléham_, Apr 16 2014: (Start)
Triangle begins:
   0;
   1,  3;
   4,  7,  12;
   9, 13,  19,  27;
  16, 21,  28,  37,  48;
  25, 31,  39,  49,  61,  75;
  36, 43,  52,  63,  76,  91, 108;
  49, 57,  67,  79,  93, 109, 127, 147;
  64, 73,  84,  97, 112, 129, 148, 169, 192;
  81, 91, 103, 117, 133, 151, 171, 193, 217, 243;
  ...
(End)

Cf. A073254.

Flatten[Table[n^2+k*n+k^2,{n,0,10},{k,0,n}]] (* Harvey P. Dale, Jun 10 2013 *)

A198063 Triangle read by rows (n >= 0, 0 <= k <= n, m = 3); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

Original entry on oeis.org

0, 1, 1, 8, 4, 8, 27, 15, 15, 27, 64, 40, 32, 40, 64, 125, 85, 65, 65, 85, 125, 216, 156, 120, 108, 120, 156, 216, 343, 259, 203, 175, 175, 203, 259, 343, 512, 400, 320, 272, 256, 272, 320, 400, 512, 729, 585, 477, 405, 369, 369, 405, 477, 585, 729

Offset: 0

Peter Luschny, Oct 26 2011

Read as an infinite symmetric square array, this is the table A(n,k)=(n+k)(n^2+k^2), cf. A321500 for the triangle with k <= n. - M. F. Hasler, Nov 22 2018

			[0]                   0
[1]                  1, 1
[2]                8, 4, 8
[3]             27, 15, 15, 27
[4]           64, 40, 32, 40, 64
[5]        125, 85, 65, 65, 85, 125
[6]   216, 156, 120, 108, 120, 156, 216
[7] 343, 259, 203, 175, 175, 203, 259, 343
From _M. F. Hasler_, Nov 22 2018: (Start)
Can also be seen as the square array A(n,k)=(n+k)*(n^2 + k^2) read by antidiagonals:
n | k: 0   1   2   3 ...
--+----------------------
0 |    0   1   8  27 ...
1 |    1   4  15  40 ...
2 |    8  15  32  65 ...
3 |   27  40  65 108 ...
...      ...     ...
(End)

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A057427, A003056, A073254, A198064, A198065.

[[2*k^2*n-2*k*n^2+n^3: k in [0..n]]: n in [0..12]]; // G. C. Greubel, Nov 23 2018

A198063 := (n,k) -> 2*k^2*n-2*k*n^2+n^3:

t[n_, k_] := 2 k^2*n - 2 k*n^2 + n^3; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Nov 22 2018 *)

A198063(n,k)=2*k^2*n-2*k*n^2+n^3 \\ See also A321500. - M. F. Hasler, Nov 22 2018

[[ 2*k^2*n-2*k*n^2+n^3 for k in range(n+1)] for n in range(12)] # G. C. Greubel, Nov 23 2018

T(n,k) = 2*k^2*n - 2*k*n^2 + n^3.
T(n,0) = T(n,n) = n^m = n^3 = A000578(n).
T(2*n,n) = (m+1)n^m = 4*n^3 = A033430(n).
T(2*n+1,n+1) = (n+1)^(m+1) - n^(m+1) = (n+1)^4 - n^4 = A005917(n).
Sum{k=0..n} T(n,k) = (2*n^4 + 3*n^3 + n^2)/3 = A098077(n).
T(n+1,k+1)*C(n,k)^4/(k+1)^3 = A197653(n,k).

A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)3 and containing (k+1)3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 3, 1, 7, 14, 1, 13, 81, 39, 1, 21, 304, 456, 84, 1, 31, 875, 3000, 1750, 155, 1, 43, 2106, 13875, 18500, 5265, 258, 1, 57, 4459, 50421, 128625, 84035, 13377, 399, 1, 73, 8576, 153664, 669536, 836920, 307328, 30016, 584, 1, 91, 15309, 409536, 2815344, 6001128, 4223016, 955584, 61236, 819, 1

Offset: 0

Susanne Wienand, Oct 10 2011

Definition of a meander:
A binary curve C is a triple (m, S, dir) such that
(a) S is a list with values in {L,R} which starts with an L,
(b) dir is a list of m different values, each value of S being allocated
    a value of dir,
(c) consecutive L's increment the index of dir,
(d) consecutive R's decrement the index of dir,
(e) the integer m > 0 divides the length of S and
(f) C is a meander if each value of dir occurs length(S)/m times.
For this sequence, m = 3.
The values in the triangle are proved by brute force for 0 <= n <= 11. The formulas are not yet proved in general. - Susanne Wienand
Let S(N,n,k) = C(n,k)^(N+1)*Sum_{j=0..N} Sum_{i=0..N} (-1)^(N-j+i)*C(N-i,j)*((n+1)/(k+1))^j. Then S(0,n,k) = A007318(n,k), S(1,n,k) = A103371(n,k), S(2,n,k) = T(n,k), S(3,n,k) = A197653(n,k), S(4,n,k) = A197654(n,k), S(5,n,k) = A197655(n,k). - Peter Luschny, Oct 21 2011
The number triangle can be calculated recursively by the number triangles A103371 and A007318. The first column of the triangle contains the central polygonal numbers A002061. The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A027444. Row sums are in A197657. - Susanne Wienand, Nov 24 2011
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 62. - Susanne Wienand, Jun 24 2015

			For n = 4 and k = 2, T(3,4,2) = 456.
Recursive example:
T(1,4,0) = 1
T(1,4,1) = 4
T(1,4,2) = 6
T(1,4,3) = 4
T(1,4,4) = 1
T(2,4,0) = 5
T(2,4,1) = 40
T(2,4,2) = 60
T(2,4,3) = 20
T(2,4,4) = 1
T(3,4,0) = T(1,4,0)^3 + T(1,4,0)*T(2,4,4-1-0) = 1^3 + 1*20 = 21
T(3,4,1) = T(1,4,1)^3 + T(1,4,1)*T(2,4,4-1-1) = 4^3 + 4*60 = 304
T(3,4,2) = T(1,4,2)^3 + T(1,4,2)*T(2,4,4-1-2) = 6^3 + 6*40 = 456
T(3,4,3) = T(1,4,3)^3 + T(1,4,3)*T(2,4,4-1-3) = 4^3 + 4*5  = 84
T(3,4,4) = 1.
Example for closed formula:
T(4,2) = (C(4,2))^3 + C(4,2) * C(4,3) * C(5,3) = 6^3 + 6 * 4 * 10 = 456.
Some examples of list S and allocated values of dir if n = 4 and k = 2:
Length(S) = (4+1)*3 = 15 and S contains (2+1)*3 = 9 L's.
  S: L,L,L,L,L,L,L,L,L,R,R,R,R,R,R
dir: 1,2,0,1,2,0,1,2,0,0,2,1,0,2,1
  S: L,L,R,L,L,L,L,R,R,L,R,R,L,R,L
dir: 1,2,2,2,0,1,2,2,1,1,1,0,0,0,0
  S: L,R,R,R,L,L,L,L,R,R,L,L,L,R,L
dir: 1,1,0,2,2,0,1,2,2,1,1,2,0,0,0
Each value of dir occurs 15/3 = 5 times.

Susanne Wienand, Table of n, a(n) for n = 0..2015
Peter Luschny, Meanders and walks on the circle.
Susanne Wienand, Animation of a meander.
Susanne Wienand, Example of a meander.

Cf. A103371, A197653, A197654, A197655, A197657, A007318, A002061, A000012, A027444, A073254.

A194595 := (n,k)->binomial(n,k)^3*(k^2+k+1+n^2+n-k*n)/((k+1)^2);
seq(print(seq(A194595(n,k),k=0..n)),n=0..7); # Peter Luschny, Oct 14 2011

T[n_, k_] := Binomial[n, k]^3*(k^2 + k + 1 + n^2 + n - k*n)/((k + 1)^2);
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Peter Luschny *)

A194595(n,k) = {if(n == 1+2*k,3,(1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n))*binomial(n,k)^3} \\ Peter Luschny, Nov 24 2011

Recursive formula (conjectured):
T(n,k) = T(3,n,k) = T(1,n,k)^3 + T(1,n,k)*T(2,n,n-1-k),  0 <= k < n
T(3,n,n) = 1,                                            k = n
T(2,n,k) = T(1,n,k)^2 + T(1,n,k) * T(1,n,n-1-k),         0 <= k < n
T(2,n,n) = 1,                                            k = n
T(2,n,k) = A103371,
T(1,n,k) = A007318 (Pascal's Triangle).
Closed formula (conjectured): T(n,k) = (C(n,k))^3 + C(n,k) * C(n,k+1) * C(n+1,k+1). - Susanne Wienand
Let S(n,k) = binomial(2*n,n)^(k+1)*((n+1)^(k+1)-n^(k+1))/(n+1)^k. Then T(2*n,n) = S(n,2). - Peter Luschny, Oct 20 2011
T(n,k) = A073254(n+1,k+1)C(n,k)^3/(k+1)^2. - Peter Luschny, Oct 29 2011
T(n,k) = h(n,k)*binomial(n,k)^3, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^3)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 3. - Peter Luschny, Nov 24 2011

A198064 Triangle read by rows (n >= 0, 0 <= k <= n, m = 4); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

Original entry on oeis.org

0, 1, 1, 16, 5, 16, 81, 31, 31, 81, 256, 121, 80, 121, 256, 625, 341, 211, 211, 341, 625, 1296, 781, 496, 405, 496, 781, 1296, 2401, 1555, 1031, 781, 781, 1031, 1555, 2401, 4096, 2801, 1936, 1441, 1280, 1441, 1936, 2801, 4096, 6561, 4681, 3355, 2511, 2101

Offset: 0

Peter Luschny, Oct 26 2011

			[0]                      0
[1]                     1, 1
[2]                  16, 5, 16
[3]                81, 31, 31, 81
[4]            256, 121, 80, 121, 256
[5]         625, 341, 211, 211, 341, 625
[6]     1296, 781, 496, 405, 496, 781, 1296
[7] 2401, 1555, 1031, 781, 781, 1031, 1555, 2401

Cf. A057427, A003056, A073254, A198063, A198065.

A198064 := (n,k) -> k^4-2*k^3*n+4*k^2*n^2-3*k*n^3+n^4:

T(n,k) = k^4-2*k^3*n+4*k^2*n^2-3*k*n^3+n^4.
T(n,0) = T(n,n) = n^m = n^4 = A000583(n).
T(2n,n) = (m+1)n^m = 5n^4.
T(2n+1,n+1) = (n+1)^(m+1)-n^(m+1) = (n+1)^5-n^5 = A022521(n).
Sum{k=0..n} T(n,k) = (16n^5+30n^4+15n^3-n)/30.
T(n+1,k+1)C(n,k)^5/(k+1)^4 = A197654(n,k).

A198065 Triangle read by rows (n >= 0, 0 <= k <= n, m = 5); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

Original entry on oeis.org

0, 1, 1, 32, 6, 32, 243, 63, 63, 243, 1024, 364, 192, 364, 1024, 3125, 1365, 665, 665, 1365, 3125, 7776, 3906, 2016, 1458, 2016, 3906, 7776, 16807, 9331, 5187, 3367, 3367, 5187, 9331, 16807, 32768, 19608, 11648, 7448, 6144, 7448, 11648, 19608, 32768, 59049

Offset: 0

Peter Luschny, Oct 26 2011

			[0]                        0
[1]                       1, 1
[2]                    32, 6, 32
[3]                 243, 63, 63, 243
[4]            1024, 364, 192, 364, 1024
[5]         3125, 1365, 665, 665, 1365, 3125
[6]     7776, 3906, 2016, 1458, 2016, 3906, 7776
[7] 16807, 9331, 5187, 3367, 3367, 5187, 9331, 16807

Cf. A057427, A003056, A073254, A198063, A198064.

Cf. A000584, A022522, A197655.

&cat[[n*(k^2-k*n+n^2)*(3*k^2-3*k*n+n^2): k in [0..n]]: n in [0..9]];  // Bruno Berselli, Nov 02 2011

A198065 := (n,k) -> 3*n*k^4-6*k^3*n^2+7*k^2*n^3-4*k*n^4+n^5:

T(n,k) = 3*n*k^4-6*k^3*n^2+7*k^2*n^3-4*k*n^4+n^5.
T(n,0) = T(n,n) = n^m = n^5 = A000584(n).
T(2n,n) = (m+1)n^m = 6n^5.
T(2n+1,n+1) = (n+1)^(m+1)-n^(m+1) = (n+1)^6-n^6 = A022522(n).
Sum{k=0..n} T(n,k) = (13n^6+30n^5+20n^4-3n^2)/30.
T(n+1,k+1)C(n,k)^6/(k+1)^5 = A197655(n,k).

A117066 Partial sums of cupolar numbers (1/3)(n+1)(5n^2+7n+3) (A096000).

Original entry on oeis.org

1, 11, 48, 140, 325, 651, 1176, 1968, 3105, 4675, 6776, 9516, 13013, 17395, 22800, 29376, 37281, 46683, 57760, 70700, 85701, 102971, 122728, 145200, 170625, 199251, 231336, 267148, 306965, 351075, 399776, 453376, 512193, 576555, 646800, 723276, 806341

Offset: 1

Jonathan Vos Post, Apr 17 2006

Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of rectangular prisms with dimensions p, p and q. - Wesley Ivan Hurt, Apr 15 2018

			For n=6, 9*a(6) = 6^3 + 7^3 + 8^3 + 9^3 + 10^3 + 11^3 +12^3 = 9*651. - _Bruno Berselli_, Apr 01 2014

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A005902, A096000.

List([1..40], n-> n^2*(n+1)*(5*n+1)/12); # G. C. Greubel, Jul 05 2019

[n^2*(n+1)*(5*n+1)/12: n in [1..40]]; // Vincenzo Librandi, Apr 16 2018

a:=n->sum ((n+j)^3, j=0..n): seq(a(n)/9, n=1..40);# Zerinvary Lajos, Dec 17 2008

Table[Sum[n i (n + i), {i, 0, n}]/2, {n, 40}] (* Vladimir Joseph Stephan Orlovsky, Jun 03 2011 *)
Accumulate[Table[((n+1)(5n^2+7n+3))/3,{n,0,50}]] (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,11,48,140,325},50] (* Harvey P. Dale, Jan 03 2024 *)

my(x='x+O('x^40)); Vec(x*(3*x^2+6*x+1)/(1-x)^5) \\ G. C. Greubel, Jul 19 2017

a(n) = n^2*(n+1)*(5*n+1)/12; \\ Altug Alkan, Apr 16 2018

[n^2*(n+1)*(5*n+1)/12 for n in (1..40)] # G. C. Greubel, Jul 05 2019

a(n) = Sum_{i=1..n} A096000(i).
a(n) = Sum_{i=1..n} (1/3)*(i+1)*(5*i^2+7*i+3).
a(n) = Sum_{i=1..n} (1/2)*(Q(i) + 3*i^2 + 3*i + 1), where Q(i) are the cuboctahedral numbers (A005902).
a(n) = Sum_{i=0..n} A073254(n,i)*i. - Peter Luschny, Oct 29 2011
G.f.: x*(1+6*x+3*x^2) / (1-x)^5. - Colin Barker, May 08 2013
9*a(n) = Sum_{i=0..n} (n+i)^3, see Maple code by Zerinvary Lajos. - Bruno Berselli, Apr 01 2014
a(n) = n^2*(n+1)*(5*n+1)/12. - Vaclav Kotesovec, Jan 03 2017
E.g.f.: (x/12)*(12 + 54*x + 36*x^2 + 5*x^3)*exp(x). - G. C. Greubel, Jul 19 2017
Another identity: ..., a(4) = (1/2)*(7*(2+4+6+8)+5*(4+6+8)+3*(6+8)+1*(8)) = 140, a(5) = (1/2)*(9*(2+4+6+8+10)+7*(4+6+8+10)+5*(6+8+10)+3*(8+10)+1*(10)) = 325, ... - J. M. Bergot, Aug 25 2022

Terms corrected by Colin Barker, May 08 2013

A198062 Array read by antidiagonals, m>=0, n>=0, k>=0, A(m, n, k) = sum{j=0..m} sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 4, 2, 1, 0, 1, 1, 8, 3, 2, 1, 0, 1, 1, 16, 4, 4, 3, 1, 0, 1, 1, 32, 5, 8, 9, 3, 1, 0, 1, 1, 64, 6, 16, 27, 7, 3, 1, 0, 1, 1, 128, 7, 32, 81, 15, 7, 3, 1, 0, 1, 1, 256, 8, 64, 243, 31, 15, 9, 4, 1, 0, 1, 1, 512, 9

Offset: 0

Peter Luschny, Nov 02 2011

			   [0] [1] [2]  [3] [4]  [5]  [6]  [7]  [8]  [9]
-------------------------------------------------
[0]  1   1   1    1   1    1    1    1    1    1    A000012
[1]  0   1   1    2   2    2    3    3    3    3    A003056
[2]  0   1   1    4   3    4    9    7    7    9    A073254
[3]  0   1   1    8   4    8   27   15   15   27    A198063
[4]  0   1   1   16   5   16   81   31   31   81    A198064
[5]  0   1   1   32   6   32  243   63   63  243    A198065

Vincenzo Librandi, Table of n, a(n) for n = 0..350

Cf. A198060, A198061.

A198062_RowAsTriangle := proc(m) local pow; pow :=(a,b)->`if`(a=0 and b=0,1,a^b): proc(n, k) local i, j; add(add((-1)^(j + i)*binomial(i, j)*pow(n, j)* pow(k, m-j), i=0..m),j=0..m) end: end:
for m from 0 to 2 do seq(print(seq(A198062_RowAsTriangle(m)(n,k),k=0..n)),n=0..5) od;

max = 9; RowAsTriangle[m_][n_, k_] := Module[{pow}, pow[a_, b_] := If[a == 0 && b == 0, 1, a^b]; Module[{i, j}, Sum[Sum[(-1)^(j+i)*Binomial[i, j]*pow[n, j]*pow[k, m-j], {i, 0, m}], {j, 0, m}]]]; t = Flatten /@ Table[RowAsTriangle[m][n, k], {m, 0, max}, {n, 0, max}, {k, 0, n}]; Table[t[[n-k+1, k+1]], {n, 0, max}, {k, 0, n }] // Flatten (* Jean-François Alcover, Feb 25 2014, after Maple *)

A007318(n,k) = A(0,n+1,k+1)*C(n,k)^1/(k+1)^0,
A103371(n,k) = A(1,n+1,k+1)*C(n,k)^2/(k+1)^1,
A194595(n,k) = A(2,n+1,k+1)*C(n,k)^3/(k+1)^2,
A197653(n,k) = A(3,n+1,k+1)*C(n,k)^4/(k+1)^3,
A197654(n,k) = A(4,n+1,k+1)*C(n,k)^5/(k+1)^4,
A197655(n,k) = A(5,n+1,k+1)*C(n,k)^6/(k+1)^5.

A349039 Square array T(n, k) read by antidiagonals, n, k >= 0; T(n, k) = n^2 - n*k + k^2.

Original entry on oeis.org

0, 1, 1, 4, 1, 4, 9, 3, 3, 9, 16, 7, 4, 7, 16, 25, 13, 7, 7, 13, 25, 36, 21, 12, 9, 12, 21, 36, 49, 31, 19, 13, 13, 19, 31, 49, 64, 43, 28, 19, 16, 19, 28, 43, 64, 81, 57, 39, 27, 21, 21, 27, 39, 57, 81, 100, 73, 52, 37, 28, 25, 28, 37, 52, 73, 100, 121, 91, 67, 49, 37, 31, 31, 37, 49, 67, 91, 121

Offset: 0

Rémy Sigrist, Nov 06 2021

T(n, k) is the norm of the Eisenstein integer n + k*w (where w = -1/2 + sqrt(-3)/2 is a primitive cube root of unity).

All terms belong to A003136.

			Array T(n, k) begins:
  n\k|    0   1   2   3   4   5   6   7   8   9   10
  ---+----------------------------------------------
    0|    0   1   4   9  16  25  36  49  64  81  100
    1|    1   1   3   7  13  21  31  43  57  73   91
    2|    4   3   4   7  12  19  28  39  52  67   84
    3|    9   7   7   9  13  19  27  37  49  63   79
    4|   16  13  12  13  16  21  28  37  48  61   76
    5|   25  21  19  19  21  25  31  39  49  61   75
    6|   36  31  28  27  28  31  36  43  52  63   76
    7|   49  43  39  37  37  39  43  49  57  67   79
    8|   64  57  52  49  48  49  52  57  64  73   84
    9|   81  73  67  63  61  61  63  67  73  81   91
   10|  100  91  84  79  76  75  76  79  84  91  100

Rémy Sigrist, Table of n, a(n) for n = 0..10010
Wikipedia, Eisenstein integers: Euclidean domain

Cf. A003136, A004247, A048147, A073254.

T[n_, k_] := n^2 - n*k + k^2; Table[T[k, n - k], {n, 0, 11}, {k, 0, n}] // Flatten (* Amiram Eldar, Nov 08 2021 *)

PARI
```
T(n,k) = n^2 - n*k + k^2
```

T(n, k) = T(k, n).
T(n, 0) = T(n, n) = n^2.
T(n, k) = A048147(n, k) - A004247(n, k).
G.f.: (x - 5*x*y + y*(1 + y) + x^2*(1 + y^2))/((1 - x)^3*(1 - y)^3). - Stefano Spezia, Nov 08 2021

cp's OEIS Frontend

A132111 Triangle read by rows: T(n,k) = n^2 + k*n + k^2, 0 <= k <= n.

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A198063 Triangle read by rows (n >= 0, 0 <= k <= n, m = 3); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j).

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A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*3 and containing (k+1)*3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

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A198064 Triangle read by rows (n >= 0, 0 <= k <= n, m = 4); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j).

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A198065 Triangle read by rows (n >= 0, 0 <= k <= n, m = 5); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j).

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A117066 Partial sums of cupolar numbers (1/3)*(n+1)*(5*n^2+7*n+3) (A096000).

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Extensions

A198062 Array read by antidiagonals, m>=0, n>=0, k>=0, A(m, n, k) = sum{j=0..m} sum{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j).

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A198063 Triangle read by rows (n >= 0, 0 <= k <= n, m = 3); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

A194595 Triangle by rows T(n,k), showing the number of meanders with length (n+1)3 and containing (k+1)3 L's and (n-k)*3 R's, where L's and R's denote arcs of equal length and a central angle of 120 degrees which are positively or negatively oriented.

A198064 Triangle read by rows (n >= 0, 0 <= k <= n, m = 4); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

A198065 Triangle read by rows (n >= 0, 0 <= k <= n, m = 5); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

A117066 Partial sums of cupolar numbers (1/3)(n+1)(5n^2+7n+3) (A096000).

A198062 Array read by antidiagonals, m>=0, n>=0, k>=0, A(m, n, k) = sum{j=0..m} sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).