A198065 -id:A198065 - cp's OEIS frontend

A073254 Array read by antidiagonals, A(n,k) = n^2 + n*k + k^2.

0, 1, 1, 4, 3, 4, 9, 7, 7, 9, 16, 13, 12, 13, 16, 25, 21, 19, 19, 21, 25, 36, 31, 28, 27, 28, 31, 36, 49, 43, 39, 37, 37, 39, 43, 49, 64, 57, 52, 49, 48, 49, 52, 57, 64, 81, 73, 67, 63, 61, 61, 63, 67, 73, 81, 100, 91, 84, 79, 76, 75, 76, 79, 84, 91, 100, 121, 111, 103, 97

Offset: 0

Michael Somos, Jul 23 2002

Norm of elements in planar hexagonal lattice A_2.

Only numbers which appear in A003136 (Loeschian numbers) can appear in this array. - Peter Luschny, Nov 10 2021

			Triangle T(n, k) starts:
[0]               0
[1]              1, 1
[2]            4, 3, 4
[3]           9, 7, 7, 9
[4]       16, 13, 12, 13, 16
[5]     25, 21, 19, 19, 21, 25
[6]   36, 31, 28, 27, 28, 31, 36
[7] 49, 43, 39, 37, 37, 39, 43, 49

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (triangular rows 0 <= n <= 150, flattened)
G. Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2.
Index entries for sequences related to A2 = hexagonal = triangular lattice

A033994 gives antidiagonal sums.
Cf. A003136, A000290, A033428, A003215, A033994, A194595.
Cf. A004016, A057427, A003056, A132111.
Cf. A198063 (m=3), A198064 (m=4), A198065 (m=5).

# Using the triangle formula:
A073254 := (n,k) -> k^2 - k*n + n^2: # Peter Luschny, Oct 26 2011

(* Using the array formula: *)
A[n_, k_] := n^2 + n k + k^2;
Table[A[n - k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 22 2018 *)

PARI
```
{A(n, k) = n^2 + n*k + k^2}
```

From Peter Luschny, Oct 26 2011: (Start)
Let m = 2, for the cases m = 3, 4, and 5 see the cross-references.
T(n,k) = k^2 - k*n + n^2 = A(n-k,k).
T(n,k) = Sum_{j=0..m} Sum_{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j) for m = 2.
T(n,0) = T(n,n) = n^m = n^2 = A000290(n).
T(2n,n) = (m+1)*n^m = 3*n^2 = A033428(n).
T(2n+1,n+1) = (n+1)^(m+1) - n^(m+1) = (n+1)^3 - n^3 = A003215(n).
Sum_{k=0..n} T(n,k) = (5*n^3 + 6*n^2 + n)/6 = A033994(n).
T(n+1, k+1)*binomial(n, k)^3/(k+1)^2 = A194595(n,k). (End)

Edited by Peter Luschny, Nov 10 2021

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).

A197655 Triangle by rows T(n,k), showing the number of meanders with length (n+1)6 and containing (k+1)6 Ls and (n-k)*6 Rs, where Ls and Rs denote arcs of equal length and a central angle of 60 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 6, 1, 63, 126, 1, 364, 4374, 1092, 1, 1365, 85120, 127680, 5460, 1, 3906, 984375, 6000000, 1968750, 19530, 1, 9331, 7562646, 157828125, 210437500, 18906615, 55986, 1, 19608, 42824236, 2628749256, 11029593750, 4381248760, 128472708, 137256, 1

Offset: 0

Susanne Wienand, Oct 19 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 Ls increment the index of dir,
(d) consecutive Rs 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 = 6.
The values in the triangle are proved by brute force for 0 <= n <= 5. The formulas are not yet proved in general.
The number triangle can be calculated recursively by the number triangles and A007318, A103371, A194595, A197653 and A197654. For n > 0, the first column seems to be A053700. The diagonal right hand is A000012. Row sums are in A198258.
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 19. - Susanne Wienand, Jul 04 2015

			For n = 4 and k = 2, T(n,k) = 127680
Example for recursive formula:
T(1,4,2) = 6
T(5,4,4-1-2) = T(5,4,1) = 13504
T(6,4,2) = 6^6 + 6*13504 = 127680
Example for closed formula:
T(4,2) = A + B + C + D + E + F
A = 6^6       =  46656
B = 6^5 * 4   =  31104
C = 6^4 * 4^2 =  20736
D = 6^3 * 4^3 =  13824
E = 6^2 * 4^4 =   9216
F = 6   * 4^5 =   6144
T(4,2)        = 127680
Some examples of list S and allocated values of dir if n = 4 and k = 2:
Length(S) = (4+1)*6 = 30 and S contains (2+1)*6 = 18 Ls.
  S: L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,L,R,R,R,R,R,R,R,R,R,R,R,R
dir: 1,2,3,4,5,0,1,2,3,4,5,0,1,2,3,4,5,0,0,5,4,3,2,1,0,5,4,3,2,1
  S: L,L,L,L,L,L,L,L,L,L,R,L,L,R,L,R,R,R,L,R,R,L,R,R,R,L,L,R,R,L
dir: 1,2,3,4,5,0,1,2,3,4,4,4,5,5,5,5,4,3,3,3,2,2,2,1,0,0,1,1,0,0
  S: L,L,L,L,R,L,L,R,L,L,R,L,R,R,L,L,L,L,L,R,R,L,L,L,R,R,R,R,L,R
dir: 1,2,3,4,4,4,5,5,5,0,0,0,0,5,5,0,1,2,3,3,2,2,3,4,4,3,2,1,1,1
Each value of dir occurs 30/6 = 5 times.

Susanne Wienand, Table of n, a(n) for n = 0..209
Peter Luschny, Meanders and walks on the circle.

Cf. A000012, A007318, A053700, A103371, A194595, A197653, A197654, A198258.

A197655 := (n,k) -> (1+n)*(1+3*k+3*k^2-n-3*k*n+n^2)*(1+k+k^2+n-k*n+n^2)* binomial(n,k)^6/(1+k)^5; seq(print(seq(A197655(n, k), k=0..n)), n=0..7); # Peter Luschny, Oct 21 2011

T[n_, k_] := (1 + n)(1 + 3k + 3k^2 - n - 3k*n + n^2)(1 + k + k^2 + n - k*n + n^2) Binomial[n, k]^6/(1 + k)^5;
Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Peter Luschny *)

A197655(n,k) = {if(n==1+2*k,6,(1+k)*(1-((n-k)/(1+k))^6)/(1+2*k-n))*binomial(n,k)^6} \\ Peter Luschny, Nov 24 2011

def S(N,n,k) : return binomial(n,k)^(N+1)*sum(sum((-1)^(N-j+i)*binomial(N-i,j)*((n+1)/(k+1))^j for i in (0..N) for j in (0..N)))
def A197655(n,k) : return S(5,n,k)
for n in (0..5) : print([A197655(n,k) for k in (0..n)])  # Peter Luschny, Oct 24 2011

recursive formula (conjectured):
T(n,k) = T(6,n,k)
T(6,n,k) = T(1,n,k)^6 + T(1,n,k)*T(5,n,n-1-k), 0 <= k < n
T(6,n,n) = 1                                        k = n
T(5,n,k) = T(1,n,k)^5 + T(1,n,k)*T(4,n,n-1-k), 0 <= k < n
T(5,n,n) = 1                                        k = n
T(4,n,k) = T(1,n,k)^4 + T(1,n,k)*T(3,n,n-1-k), 0 <= k < n
T(4,n,n) = 1                                        k = n
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(5,n,k) = A197654
T(4,n,k) = A197653
T(3,n,k) = A194595
T(2,n,k) = A103371
T(1,n,k) = A007318 (Pascal's Triangle)
closed formula (conjectured): T(n,n) = 1,                         k = n
                T(n,k) = A + B + C + D + E + F,     k < n
                     A = (C(n,k))^6
                     B = (C(n,k))^5 * C(n,n-1-k)
                     C = (C(n,k))^4 *(C(n,n-1-k))^2
                     D = (C(n,k))^3 *(C(n,n-1-k))^3
                     E = (C(n,k))^2 *(C(n,n-1-k))^4
                     F =  C(n,k)    *(C(n,n-1-k))^5
[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,5). (Cf. A103371, A194595, A197653). [Peter Luschny, Oct 21 2011]
T(n,k) = A198065(n+1,k+1)C(n,k)^6/(k+1)^5. [Peter Luschny, Oct 29 2011]
T(n,k) = h(n,k)*binomial(n,k)^6, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^6)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 6. [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).

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.

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A073254 Array read by antidiagonals, A(n,k) = n^2 + n*k + k^2.

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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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A197655 Triangle by rows T(n,k), showing the number of meanders with length (n+1)*6 and containing (k+1)*6 Ls and (n-k)*6 Rs, where Ls and Rs denote arcs of equal length and a central angle of 60 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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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).

A197655 Triangle by rows T(n,k), showing the number of meanders with length (n+1)6 and containing (k+1)6 Ls and (n-k)*6 Rs, where Ls and Rs denote arcs of equal length and a central angle of 60 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).

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).