A152031 -id:A152031 - cp's OEIS frontend

A228275 A(n,k) = Sum_{i=1..k} n^i; square array A(n,k), n>=0, k>=0, read by antidiagonals.

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 6, 3, 0, 0, 4, 14, 12, 4, 0, 0, 5, 30, 39, 20, 5, 0, 0, 6, 62, 120, 84, 30, 6, 0, 0, 7, 126, 363, 340, 155, 42, 7, 0, 0, 8, 254, 1092, 1364, 780, 258, 56, 8, 0, 0, 9, 510, 3279, 5460, 3905, 1554, 399, 72, 9, 0

Offset: 0

Alois P. Heinz, Aug 19 2013

A(n,k) is the total sum of lengths of longest ending contiguous subsequences with the same value over all s in {1,...,n}^k:
A(4,1) = 4 = 1+1+1+1: [1], [2], [3], [4].
A(1,4) = 4: [1,1,1,1].
A(3,2) = 12 = 2+1+1+1+2+1+1+1+2: [1,1], [1,2], [1,3], [2,1], [2,2], [2,3], [3,1], [3,2], [3,3].
A(2,3) = 14 = 3+1+1+2+2+1+1+3: [1,1,1], [1,1,2], [1,2,1], [1,2,2], [2,1,1], [2,1,2], [2,2,1], [2,2,2].

			Square array A(n,k) begins:
  0, 0,  0,   0,    0,     0,      0,      0, ...
  0, 1,  2,   3,    4,     5,      6,      7, ...
  0, 2,  6,  14,   30,    62,    126,    254, ...
  0, 3, 12,  39,  120,   363,   1092,   3279, ...
  0, 4, 20,  84,  340,  1364,   5460,  21844, ...
  0, 5, 30, 155,  780,  3905,  19530,  97655, ...
  0, 6, 42, 258, 1554,  9330,  55986, 335922, ...
  0, 7, 56, 399, 2800, 19607, 137256, 960799, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened

Columns k=0-10 give: A000004, A001477, A002378, A027444, A027445, A152031, A228290, A228291, A228292, A228293, A228294.
Rows n=0-11 give: A000004, A001477, A000918(k+1), A029858(k+1), A080674, A104891, A105281, A104896, A052379(k-1), A052386, A105279, A105280.
Main diagonal gives A031972.
Lower diagonal gives A226238.
Cf. A228250.

A:= (n, k)-> `if`(n=1, k, (n/(n-1))*(n^k-1)):
seq(seq(A(n, d-n), n=0..d), d=0..12);

a[0, 0] = 0; a[1, k_] := k; a[n_, k_] := n*(n^k-1)/(n-1); Table[a[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 16 2013 *)

A(1,k) = k, else A(n,k) = n/(n-1)*(n^k-1).
A(n,k) = Sum_{i=1..k} n^i.
A(n,k) = Sum_{i=1..k+1} binomial(k+1,i)*A(n-i,k)*(-1)^(i+1) for n>k, given values A(0,k), A(1,k),..., A(k,k). - Yosu Yurramendi, Sep 03 2013

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]

A348463 Size of largest bipartite biregular Moore graph of diameter 6 and degrees n and n.

Original entry on oeis.org

12, 126, 728, 2730, 7812

Offset: 2

N. J. A. Sloane, Oct 31 2021

18660 <= a(7) <= 18662, a(8) = 39216, a(9) = 74898, a(10) = 132860.

Table 3 from these references gives the size of the largest bipartite biregular Moore graph of diameter 3 and degrees n and n, and appears to match 2*A002061.

G. Araujo-Pardo, C. Dalfó, M. Á. Fiol and N. López, Bipartite biregular Moore graphs, arXiv:2103.11443 [math.CO], 2021. See Table 2.
G. Araujo-Pardo, C. Dalfó, M. Á. Fiol and N. López, Bipartite biregular Moore graphs, Discrete Math., 334 (2021), # 112582. See Table 3.

Cf. A002061, A053700, A152031, A348461, A348462.

Empirical observation: a(n) = 2*(A152031(n-1) + 1) matches terms a(2)-a(6) and a(8)-a(10). - Hugo Pfoertner, Oct 31 2021

a(n) <= 2*A053700(n-1) = 2*(A152031(n-1) + 1) (the Moore bound). - Pontus von Brömssen, Oct 31 2021

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A228275 A(n,k) = Sum_{i=1..k} n^i; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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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.

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A348463 Size of largest bipartite biregular Moore graph of diameter 6 and degrees n and n.

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