A198064 -id:A198064 - 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).

A197654 Triangle by rows T(n,k), showing the number of meanders with length 5(n+1) and containing 5(k+1) L's and 5(n-k) R's, where L's and R's denote arcs of equal length and a central angle of 72 degrees which are positively or negatively oriented.

Original entry on oeis.org

1, 5, 1, 31, 62, 1, 121, 1215, 363, 1, 341, 13504, 20256, 1364, 1, 781, 96875, 500000, 193750, 3905, 1, 1555, 501066, 7321875, 9762500, 1252665, 9330, 1, 2801, 2033647, 72656661, 262609375, 121094435, 6100941, 19607, 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 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.
Then C is a meander if each value of dir occurs length(S)/m times.
Let T(m,n,k) = number of meanders (m, S, dir) in which S contains m(k+1) L's and m(n-k) R's, so that length(S) = m(n+1).
For this sequence, m = 5, T(n,k) = T(5,n,k).
The values in the triangle were proved by brute force for 0 <= n <= 6. The formulas have not yet been proved in general.
The number triangle can be calculated recursively by the number triangles and A007318, A103371, A194595 and A197653. The first column seems to be A053699.  The diagonal right hand is A000012. The diagonal with k = n-1 seems to be A152031 and to start with the second number of A152031. Row sums are in A198257.
The conjectured formulas are confirmed by dynamic programming for 0 <= n <= 29. - Susanne Wienand, Jul 01 2015

			For n = 5 and k = 2, T(n,k) = 500000
Example for recursive formula:
T(1,5,2) = 10
T(4,5,5-1-2) = T(4,5,2) = 40000
T(5,5,2) = 10^5 + 10*40000 = 500000
Example for closed formula:
T(5,2) = A + B + C + D + E
A = 10^5
B = 10^4 * 10
C = 10^3 * 10^2
D = 10^2 * 10^3
E = 10   * 10^4
T(5,2) = 5 * 10^5 = 500000
Some examples of list S and allocated values of dir if n = 5 and k = 2:
Length(S) = (5+1)*5 = 30 and S contains (2+1)*5 = 15 Ls.
  S: 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,R,R,R
dir: 1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,0,4,3,2,1,0,4,3,2,1,0,4,3,2,1
  S: L,L,L,L,L,L,L,R,R,L,L,R,R,R,L,R,R,R,L,R,L,L,L,R,R,L,R,R,R,R
dir: 1,2,3,4,0,1,2,2,1,1,2,2,1,0,0,0,4,3,3,3,3,4,0,0,4,4,4,3,2,1
  S: L,L,L,L,L,R,L,L,L,L,L,R,L,R,R,R,R,R,R,R,R,R,R,L,L,L,L,R,R,R
dir: 1,2,3,4,0,0,0,1,2,3,4,4,4,4,3,2,1,0,4,3,2,1,0,0,1,2,3,3,2,1
Each value of dir occurs 30/5 = 6 times.
The triangle begins:
1,
5, 1,
31, 62, 1,
121, 1215, 363, 1,
341, 13504, 20256, 1364, 1,
781, 96875, 500000, 193750, 3905, 1,
...

Alois P. Heinz, Rows n = 0..140, flattened
Peter Luschny, Meanders and walks on the circle.
Susanne Wienand, Example of a meander counted by A197654

Cf. A000012, A007318, A053699, A103371, A152031, A194595, A197653, A197655, A198257.

A197654 := (n,k)->(k^4+2*k^3*(1-n)+2*k^2*(2+n+2*n^2)+k*(3+n-n^2-3*n^3)+ n^4+n^3+n^2+n+1)*binomial(n,k)^5/(1+k)^4;
seq(print(seq(A197654(n, k), k=0..n)), n=0..7);  # Peter Luschny, Oct 20 2011

T[n_, k_] := (k^4 + 2*k^3*(1-n) + 2*k^2*(2+n+2*n^2) + k*(3+n-n^2-3*n^3) + n^4+n^3+n^2+n+1)*Binomial[n, k]^5/(1+k)^4; Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 20 2017, after Peter Luschny *)

A197654(n,k) = {if(n ==1+2*k,5,(1+k)*(1-((n-k)/(1+k))^5)/(1+2*k-n))*binomial(n,k)^5} \\ 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 A197654(n,k) : return S(4,n,k)
for n in (0..5) : print([A197654(n,k) for k in (0..n)])  # Peter Luschny, Oct 24 2011

Recursive formula (conjectured):
T(n,k) = 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(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,              k < n
                     A = (C(n,k))^5
                     B = (C(n,k))^4 * C(n,n-1-k)
                     C = (C(n,k))^3 *(C(n,n-1-k))^2
                     D = (C(n,k))^2 *(C(n,n-1-k))^3
                     E =  C(n,k)    *(C(n,n-1-k))^4     [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,4). [Peter Luschny, Oct 20 2011]
T(n,k) = A198064(n+1,k+1)C(n,k)^5/(k+1)^4. [Peter Luschny, Oct 29 2011]
T(n,k) = h(n,k)*binomial(n,k)^5, where h(n,k) = (1+k)*(1-((n-k)/(1+k))^5)/(1+2*k-n) if 1+2*k-n <> 0 else h(n,k) = 5. [Peter Luschny, Nov 24 2011]

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).

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.

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

A197654 Triangle by rows T(n,k), showing the number of meanders with length 5(n+1) and containing 5(k+1) L's and 5(n-k) R's, where L's and R's denote arcs of equal length and a central angle of 72 degrees which are positively or negatively oriented.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Sage

Formula

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).

Views

Author

Keywords

Examples

Crossrefs

Programs

Magma

Maple

Formula

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).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

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).

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).

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).