Dennis P. Walsh - cp's OEIS frontend

User: Dennis P. Walsh

Dennis P. Walsh has authored 24 sequences. Here are the ten most recent ones:

A335109 Triangle read by rows: T(n,k) is the number of permutations of length n with each cycle of the permutation containing only elements that are identical (mod k), where 1 <= k <= n.

1, 2, 1, 6, 2, 1, 24, 4, 2, 1, 120, 12, 4, 2, 1, 720, 36, 8, 4, 2, 1, 5040, 144, 24, 8, 4, 2, 1, 40320, 576, 72, 16, 8, 4, 2, 1, 362880, 2880, 216, 48, 16, 8, 4, 2, 1, 3628800, 14400, 864, 144, 32, 16, 8, 4, 2, 1

Offset: 1

Dennis P. Walsh, May 23 2020

Let [n] denote {1,2,...,n} and let [n](j,k) denote the subset of [n] consisting of all elements of [n] that equal j mod k. The cardinality of [n](j,k) equals ceiling(n/k) for j = 1..(n mod k) and equals floor(n/k) for j > (n mod k). Therefore, upon permuting the elements of each [n](j,k) subset, we obtain T(n,k) = (ceiling(n/k)!)^(n mod k)*(floor(n/k)!)^(k-(n mod k)).

			Triangle begins:
    1;
    2  1;
    6  2 1;
   24  4 2 1;
  120 12 4 2 1;
  ...
T(6,3) counts the 8 permutations of [6] where all cycle-mates are identical mod 3, namely, (1 4)(2 5)(3 6), (1 4)(2 5)(3)(6), (1 4)(2)(5)(3 6), (1)(4)(2 5)(3 6), (1 4)(2)(5)(3)(6), (1)(4)(2 5)(3)(6), (1)(4)(2)(5)(3 6) and (1)(2)(3)(4)(5)(6).

Per Alexandersson, Frether Getachew Kebede, Samuel Asefa Fufa, and Dun Qiu, Pattern-Avoidance and Fuss-Catalan Numbers, J. Int. Seq. (2023) Vol. 26, Art. 23.4.2.

Cf. A000142, A010551, A264557, A264635, A264656.

Cf. A275062.

seq(seq((ceil(n/k)!)^(n mod k)*(floor(n/k)!)^(k-(n mod k)), k=1..n), n=1..10);

Table[(Ceiling[n/k]!)^Mod[n, k]*(Floor[n/k]!)^(k - Mod[n, k]), {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Jun 28 2020 *)

T(n,k) = (ceiling(n/k)!)^(n mod k)*(floor(n/k)!)^(k-(n mod k)) for 1 <= k <= n.
T(n,1) = A000142(n).
T(n,2) = A010551(n) for n > 1.
T(n,3) = A264557(n) for n > 2.
T(n,4) = A264635(n) for n > 3.
T(n,5) = A264656(n) for n > 4.
T(n,k) = Product_{i=0..k-1} floor((n+i)/k)!. - Alois P. Heinz, May 23 2020

A335259 Triangle read by rows: T(n,k) = k^ceiling(n/k) for 1 <= k <= n.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 4, 9, 4, 1, 8, 9, 16, 5, 1, 8, 9, 16, 25, 6, 1, 16, 27, 16, 25, 36, 7, 1, 16, 27, 16, 25, 36, 49, 8, 1, 32, 27, 64, 25, 36, 49, 64, 9, 1, 32, 81, 64, 25, 36, 49, 64, 81, 10, 1, 64, 81, 64, 125, 36, 49, 64, 81, 100, 11, 1, 64, 81, 64, 125, 36, 49, 64, 81, 100, 121, 12

Offset: 1

Dennis P. Walsh, May 28 2020

T(n,k) is the number of functions f:[n]->[k] such that f(x)=f(y) whenever i*k-k+1<=x<=y<=i*k where 1<=i<=ceiling(n/k). An example of such a function is f:[8]->[3] defined by f(1)=f(2)=f(3)=2, f(4)=f(5)=f(6)=3, and f(7)=f(8)=2. To count all functions of this type when n=8 and k=3, we note that there are 3 possible values for f(1), f(4), and f(7). Hence T(8,3)=3^3 or, equivalently, 3^ceiling(8/3). A proof of the general result follows the same approach. We also note the following: (i) T(n,1)=1 for all n; (ii) T(n,n)=n for all n; T(n,k)=k^2 when ceiling(n/2)<=k

			Triangle T(n,k):
  1;
  1,  2;
  1,  4,  3;
  1,  4,  9,  4;
  1,  8,  9, 16,  5;
  1,  8,  9, 16, 25,  6;
  1, 16, 27, 16, 25, 36,  7;
  1, 16, 27, 16, 25, 36, 49,  8;
  1, 32, 27, 64, 25, 36, 49, 64,  9;
  1, 32, 81, 64, 25, 36, 49, 64, 81, 10;
...
T(8,3) counts the 27 functions from [8] to [3] where f(1)=f(2)=f(3), f(4)=f(5)=f(6), and f(7)=f(8). Letting f be defined by its vector of images <f(1), ...,f(8)>, the 27 functions are <1,1,1,1,1,1,1,1>, <1,1,1,1,1,1,2,2>, <1,1,1,1,1,1,3,3>, <1,1,1,2,2,2,1,1>, <1,1,1,2,2,2,2,2>, <1,1,1,2,2,2,3,3>, <1,1,1,3,3,3,1,1>, <1,1,1,3,3,3,2,2>, <1,1,1,3,3,3,3,3>, <2,2,2,1,1,1,1,1>, <2,2,2,1,1,1,2,2>, <2,2,2,1,1,1,3,3>, <2,2,2,2,2,2,1,1>, <2,2,2,2,2,2,2,2>, <2,2,2,2,2,2,3,3>, <2,2,2,3,3,3,1,1>, <2,2,2,3,3,3,2,2>, <2,2,2,3,3,3,3,3>, <3,3,3,1,1,1,1,1>, <3,3,3,1,1,1,2,2>, <3,3,3,1,1,1,3,3>, <3,3,3,2,2,2,1,1>, <3,3,3,2,2,2,2,2>, <3,3,3,2,2,2,3,3>, <3,3,3,3,3,3,1,1>, <3,3,3,3,3,3,2,2>, and <3,3,3,3,3,3,3,3>.

Cf. A016116, A127975.

Maple
```
seq(seq(k^ceil(n/k),k=1..n),n=1..20);
```

Table[k^Ceiling[n/k], {n, 12}, {k, n}] // Flatten (* Michael De Vlieger, Jun 28 2020 *)

G.f. for fixed k: k*x^k*(1+k*x+k*x^2+...+k*x^(k-1))/(1-k*x^k).
For n>1, T(n,2) = A016116(n).
For n>2, T(n,3) = A127975(n).

A293211 Triangle T(n,k) is the number of permutations on n elements with at least one k-cycle for 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 4, 3, 2, 15, 9, 8, 6, 76, 45, 40, 30, 24, 455, 285, 200, 180, 144, 120, 3186, 1995, 1400, 1260, 1008, 840, 720, 25487, 15855, 11200, 8820, 8064, 6720, 5760, 5040, 229384, 142695, 103040, 79380, 72576, 60480, 51840, 45360, 40320, 2293839, 1427895, 1030400, 793800, 653184, 604800, 518400, 453600, 403200, 362880

Offset: 1

Dennis P. Walsh, Oct 02 2017

T(n,k) is equivalent to n! minus the number of permutations on n elements with zero k-cycles (sequence A122974).

			T(n,k) (the first 8 rows):
:     1;
:     1,     1;
:     4,     3,     2;
:    15,     9,     8,    6;
:    76,    45,    40,   30,   24;
:   455,   285,   200,  180,  144,  120;
:  3186,  1995,  1400, 1260, 1008,  840,  720;
: 25487, 15855, 11200, 8820, 8064, 6720, 5760, 5040;
  ...
T(4,3)=8 since there are exactly 8 permutations on {1,2,3,4} with at least one 3-cycle: (1)(234), (1)(243), (2)(134), (2)(143), (3)(124), (3)(142), (4)(123), and (4)(132).

Dennis P. Walsh, The number of permutations with no k-cycles

Columns k=1-10 give: A002467, A027616, A027617, A029571, A029572, A029573, A029574, A029575, A029576, A029577.
Row sums give A132961.
T(n,n) gives A000142(n-1) for n>0.
T(2n,n) gives A052145.
Cf. A122974, A126074.

T:=(n,k)->n!*sum((-1)^(j+1)*(1/k)^j/j!,j=1..floor(n/k)); seq(seq(T(n,k),k=1..n),n=1..10);

Table[n!*Sum[(-1)^(j + 1)*(1/k)^j/j!, {j, Floor[n/k]}], {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Oct 02 2017 *)

T(n,k) = n! * Sum_{j=1..floor(n/k)} (-1)^(j+1)*(1/k)^j/j!.
T(n,k) = n! - A122974(n,k).
E.g.f. of column k: (1-exp(-x^k/k))/(1-x). - Alois P. Heinz, Oct 11 2017

A243987 Triangle read by rows: T(n, k) is the number of divisors of n that are less than or equal to k for 1 <= k <= n.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 1, 2, 2, 3, 1, 1, 1, 1, 2, 1, 2, 3, 3, 3, 4, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 3, 3, 3, 3, 4, 1, 1, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 3, 4, 4, 5, 5, 5, 5, 5, 5, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4

Offset: 1

Dennis P. Walsh, Jun 16 2014

This triangular sequence T(n,k) generalizes sequence A000005, the number of divisors of n; in particular, A000005(n) = T(n,n).

Also, for prime p, T(p,k) = 1 when k < p and T(p,p) = 2.

			T(6,4)=3 since there are 3 divisors of 6 that are less than or equal to 4, namely, 1, 2 and 3.
T(n,k) as a triangle, n=1..15:
1,
1, 2,
1, 1, 2,
1, 2, 2, 3,
1, 1, 1, 1, 2,
1, 2, 3, 3, 3, 4,
1, 1, 1, 1, 1, 1, 2,
1, 2, 2, 3, 3, 3, 3, 4,
1, 1, 2, 2, 2, 2, 2, 2, 3,
1, 2, 2, 2, 3, 3, 3, 3, 3, 4
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2,
1, 2, 3, 4, 4, 5, 5, 5, 5, 5, 5, 6,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2,
1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4,
1, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
Dennis P. Walsh, Notes on counting the divisors of n

Cf. A000005 (diagonal), A000012 (first column), A081307 (row sums), A027750 (divisors of n).
Cf. A138553, A051731.
Cf. A340260, A340261.

a243987 n k = a243987_tabl !! (n-1) !! (k-1)
a243987_row n = a243987_tabl !! (n-1)
a243987_tabl = map (scanl1 (+)) a051731_tabl
-- Reinhard Zumkeller, Apr 22 2015

T:=(n,k)->1/n!*eval(diff(sum(x^j/(1-x^j),j=1..k),x$n),x=0):
seq(seq(T(n,k), k=1..n), n=1..10);
# Alternative:
IversonBrackets := expr -> subs(true=1, false=0, evalb(expr)):
T := (n, k) -> add(IversonBrackets(irem(n, j) = 0), j = 1..k):
for n from 1 to 19 do seq(T(n, k), k = 1..n) od; # Peter Luschny, Jan 02 2021

T(n, k) = sumdiv(n, d, d<=k); \\ Michel Marcus, Jun 17 2014

T(n,1) = 1; T(n,n) = A000005(n).
T(n,k) = coefficient of the x^n term in the expansion of Sum(x^j/(1-x^j), j=1..k).
T(n,k) = Sum_{j=1..k} A051731(n,j). - Reinhard Zumkeller, Apr 22 2015

A224711 Number of ballot results from n voters that prompt a run-off election when three candidates vie for two spots on a board.

Original entry on oeis.org

1, 0, 6, 6, 18, 90, 150, 420, 1890, 3570, 10206, 42966, 87318, 252252, 1019304, 2172456, 6319170, 24810786, 54712086, 159906318, 614406078, 1390381278, 4077926034, 15403838346, 35579546262, 104633453340, 389788932240, 915500037120, 2698033909680, 9934966920960

Offset: 0

Dennis P. Walsh, Apr 29 2013

nonn

We assume each of n voters cast two votes, one each for two of three candidates. A run-off election is necessitated if all 3 candidates receive the same number of votes or if there is a tie for the second to most votes. The total number of ballot results is 3^n since each voter must choose two of three candidates. The number of ballot results that necessitate a run-off election is derived in the note "The probability of a run-off election.." cited in the link section below.

The sequence A103221 is used in the derivation. Note that we assign the value 1 to a(0) because if no voters cast ballots on election day another election is needed.

			For n=3, a(3)=6 since a three voter election has 6 possible ballot results that necessitate a run-off. Let A, B, and C denote the three candidates, and, for example, let [AB|AC|BC] denote a ballot result in which voter 1 votes for candidates A and B, voter 2 votes for candidates A and C, and voter 3 votes for candidates B and C. The 6 ballot results that necessitate a run-off election are then given by [AB|AC|BC], [AB|BC|AC], [AC|AB|BC], [AC|BC|AB], [BC|AB|AC], and [BC|AC|AB].

Dennis P. Walsh, The probability of a run-off election when three equally-favored candidates vie for two slots

ind:= n-> piecewise(n mod 3=0, 1, 0):
u:= n-> floor(n/2+1)-floor(n/3+2/3)-1:
a:= n-> 3*add(binomial(n, 2*ceil((n-1)/2)-2*k)*
        binomial(2*ceil((n-1)/2)-2*k, ceil((n-1)/2)-k), k=0..u(n))
        -ind(n)*2*binomial(n, 2*n/3)*binomial(2*n/3, n/3):
seq(a(n), n=0..30);

a(n) = 3*sum(C(n,2*b(k)) *C(2*b(k),b(k)), k=0..u(n)) -2*C(n,2n/3) * C(2n/3,n/3) I[3|n] where b(k) = ceiling((n-1)/2)-k, u(n) = floor((n+2)/2) - floor((n+2)/3)-1 = A103221(n)-1, and I[statement] equals 1 if the statement is true and equals 0 otherwise.

A224542 Number of doubly-surjective functions f:[n]->[4].

Original entry on oeis.org

2520, 30240, 226800, 1367520, 7271880, 35692800, 165957792, 742822080, 3234711480, 13803744864, 58021888080, 241116750624, 993313349544, 4064913201216, 16549636147968, 67112688842496, 271323921459096, 1094303232174240, 4405390451382960, 17709538489849440

Offset: 8

Dennis P. Walsh, Apr 09 2013

Fourth column of A200091: A200091(n,4)=a(n).
Also, a(n) is (i) the number of length-n words on the alphabet A, B, C, and D with each letter of the alphabet occurring at least twice; (ii) the number of ways to distribute n different toys to 4 children so that each child gets at least two toys; (iii) the number of ways to put n numbered balls into 4 labeled boxes so that each box gets at least two balls; (iv) the number of n-digit positive integers consisting of only digits 1,2,3, and 4 with each of these digits appearing at least twice.
A doubly-surjective function f:D->C is such that the pre-image set f^-1(y) has size at least 2 for each y in C.
Triangle A200091 provides the number of doubly-surjective functions f from a set of size n onto a set of size k. Hence a(n) is column 4 of A200091.

			a(9) = 30240 since there are 30240 ways to distribute 9 different toys to 4 children so that each child gets at least 2 toys. One child must get 3 toys and the other children get 2 toys each. There are 4 ways to pick the lucky kid. There are C(9,3) ways to choose the 3 toys for the lucky kid. There are 6!/(2!)^3 ways to distribute the remaining 6 toys among the 3 kids. We obtain 4*C(9,3)*6!/8=30240.

Dennis Walsh, Notes on doubly-surjective finite functions

Cf. A224541, A052515.

seq(eval(diff((exp(x)-x-1)^4,x$n),x=0),n=8..40);

nn=27; Drop[Range[0,nn]! CoefficientList[Series[(Exp[x]-x-1)^4, {x,0,nn}], x], 8] (* Geoffrey Critzer, Sep 28 2013 *)

my(x='x+O('x^66)); Vec(serlaplace((exp(x)-x-1)^4)) /* Joerg Arndt, Apr 10 2013 */

a(n) = 4^n-4*3^n-4*n*3^(n-1)+(9*n+3*n^2)*2^(n-1)+6*2^n-4-8*n-4*n^3;
a(n) = sum(n!/(i!*j!*k!*m!)) over  such that i,j,k, and m are all at least 2 and i+j+k+m=n.
E.g.f.: (exp(x)-x-1)^4.
a(n) = 24*A058844(n). - Alois P. Heinz, Apr 10 2013
G.f.: 24*x^8*(288*x^6-1560*x^5+3500*x^4-4130*x^3+2625*x^2-840*x+105) / ((x-1)^4*(2*x-1)^3*(3*x-1)^2*(4*x-1)). - Colin Barker, Jun 04 2013

A224541 Number of doubly-surjective functions f:[n]->[3].

Original entry on oeis.org

90, 630, 2940, 11508, 40950, 137610, 445896, 1410552, 4390386, 13514046, 41278068, 125405532, 379557198, 1145747538, 3452182656, 10388002848, 31230066186, 93828607686, 281775226860, 845929656900, 2539047258150, 7619759016090, 22864712861880, 68605412870088

Offset: 6

Dennis P. Walsh, Apr 09 2013

Third column of A200091.
Also, a(n) is (i) the number of length-n words on the alphabet A, B, and C with each letter occurring at least twice; (ii) the number of ways to distribute n different toys to 3 different children so that each child gets at least 2 toys; (iii) the number of ways to put n numbered balls into 3 labeled boxes so that each box gets at least 2 balls; (iv) the number of n-digit positive integers consisting only of the digits 1, 2, and 3 with each of these digits appearing at least twice. A doubly-surjective function f has size at least 2 for each pre-image set, that is, |f^-1(y)|>=2 for each element y of the codomain.[Note that a surjective function has |f^-1(y)|>=1.] The triangle A200091 provides the number of doubly-surjective functions f:[n]->[k].  Column 3 of triangle A200091 is a(n).
Sequence A052515 is the number of doubly-surjective functions f:[n]->[2] with exponential generating function (exp(x)-x-1)^2. In general, the number of doubly-surjective functions f:[n]->[k] has exponential generating function (exp(x)-x-1)^k.

			For n=6 we have a(6)=90 since there are 90 six-digit  positive integers using only digits 1, 2, and 3 with each of those digits appearing at least twice. The first 30 of the ninety, namely those with initial digit 1, are given below:
112233, 112323, 112332, 113223, 113232, 113322,
121233, 121323, 121332, 122133, 122313, 122331,
123123, 123132, 123213, 123231, 123312, 123321,
131223, 131232, 131322, 132123, 132132, 132213,
132231, 132312, 132321, 133122, 133212, 133221.

Dennis Walsh, Notes on doubly-surjective finite functions

Cf. A052515, the number of doubly-surjective functions f:[n]->[2].

seq(3^n-3*2^n-3*n*2^(n-1)+3+3*n+3*n^2, n=6..40);

With[{nn=40},Drop[CoefficientList[Series[(Exp[x]-x-1)^3,{x,0,nn}],x] Range[0,nn]!,6]] (* Harvey P. Dale, Oct 01 2015 *)

x='x+O('x^66); Vec(serlaplace((exp(x)-x-1)^3)) \\ Joerg Arndt, Apr 10 2013

a(n) = 3^n-3*2^n-3*n*2^(n-1)+3+3*n+3*n^2.
E.g.f.: (exp(x)-x-1)^3.
From Alois P. Heinz, Apr 10 2013: (Start)
a(n) = 6*A000478(n).
G.f.: -6*(12*x^3-40*x^2+45*x-15)*x^6 / ((3*x-1)*(2*x-1)^2*(x-1)^3).
(End)

A218832 Number of positive integer solutions to the Diophantine equation x + y + 2z = n^2.

Original entry on oeis.org

0, 1, 12, 49, 132, 289, 552, 961, 1560, 2401, 3540, 5041, 6972, 9409, 12432, 16129, 20592, 25921, 32220, 39601, 48180, 58081, 69432, 82369, 97032, 113569, 132132, 152881, 175980, 201601, 229920, 261121, 295392, 332929, 373932, 418609, 467172, 519841, 576840, 638401

Offset: 1

Dennis P. Walsh, Mar 27 2013

The derivation for the number of integer solutions is given in a link below. It is straightforward and uses the fact that the number of positive integer solutions to x + y = n is given by n-1.

			For n=3, a(n)=12 since there are exactly 12 positive integer solutions (x,y,z) to x+y+2z=9, namely, (1,2,3),(1,4,2), (1,6,1), (2,1,3), (2,3,2), (2,5,1), (3,2,2),(3,4,1), (4,1,2), (4,3,1), (5,2,1), and (6,1,1).

Dennis Walsh, The number of positive integer solutions to the Diophantine equation x+y+2z=n^2

Maple
```
seq(floor(n^4/4-n^2+1),n=1..40);
```

a(n) = floor(n^4/4-n^2+1).
Conjectures from Colin Barker, Apr 01 2013: (Start)
a(n) = (7 + (-1)^n - 8*n^2 + 2*n^4)/8.
G.f.: -x^2*(x^4 - 4*x^3 + 6*x^2 + 8*x + 1) / ((x-1)^5*(x+1)). (End)

A182309 Triangle T(n,k) with 2 <= k <= floor(2(n+1)/3) gives the number of length-n binary sequences with exactly k zeros and with length two for the longest run of zeros.

Original entry on oeis.org

1, 2, 3, 2, 4, 6, 1, 5, 12, 6, 6, 20, 18, 3, 7, 30, 40, 16, 1, 8, 42, 75, 50, 10, 9, 56, 126, 120, 45, 4, 10, 72, 196, 245, 140, 30, 1, 11, 90, 288, 448, 350, 126, 15, 12, 110, 405, 756, 756, 392, 90, 5, 13, 132, 550, 1200, 1470, 1008, 357, 50, 1, 14, 156, 726

Offset: 2

Dennis P. Walsh, Apr 23 2012

Triangle T(n,k) captures several well known sequences. In particular, T(n,2)=(n-1), the natural numbers; T(n,3)=(n-2)(n-3)=A002378(n-3), the "oblong" numbers; T(n,4)=(n-3)(n-4)^2/2=A002411(n-4), "pentagonal pyramidal" numbers; and also T(n,5)=(n-4)C(n-4,3)=A004320(n-6). Furthermore, row sums=A000100(n+1).

			For n=6 and k=3, T(6,3)=12 since there are 12 binary sequences of length 6 that contain 3 zeros and that have a maximum run of zeros of length 2, namely, 011100, 101100, 110100, 011001, 101001, 110010, 010011, 100110, 100101, 001110, 001101, and 001011.
Triangle T(n,k) begins
   1,
   2,
   3,   2,
   4,   6,   1,
   5,  12,   6,
   6,  20,  18,    3,
   7,  30,  40,   16,    1,
   8,  42,  75,   50,   10,
   9,  56, 126,  120,   45,    4,
  10,  72, 196,  245,  140,   30,    1,
  11,  90, 288,  448,  350,  126,   15,
  12, 110, 405,  756,  756,  392,   90,    5,
  13, 132, 550, 1200, 1470, 1008,  357,   50,   1,
  14, 156, 726, 1815, 2640, 2268, 1106,  266,  21,
  15, 182, 936, 2640, 4455, 4620, 2898, 1016, 161, 6,

Dennis Walsh, Notes on binary sequences with a maximum run length of two

Row sums of triangle T(n,k)=A000100(n+1);
T(n,3)=A002378(n-3); T(n,4)=A002411(n-4);
T(n,5)=A004320(n-6).

seq(seq(sum(binomial(n-k+1,j)*binomial(n-k+1-j,k-2*j),j=1..floor(k/2)),k=2..floor(2*(n+1)/3)),n=2..20);

t[n_, k_] := Sum[ Binomial[n-k+1, j]*Binomial[n-k-j+1, k-2*j], {j, 1, k/2}]; Table[t[n, k], {n, 2, 15}, {k, 2, 2*(n+1)/3}] // Flatten (* Jean-François Alcover, Jun 06 2013 *)

T(n,k) = Sum_{j=1..k/2} binomial(n-k+1,j)*binomial(n-k-j+1,k-2j) for 2 <= k <= 2(n+1)/3.

A182210 Triangle T(n,k) = floor(k*(n+1)/(k+1)), 1 <= k <= n.

Original entry on oeis.org

1, 1, 2, 2, 2, 3, 2, 3, 3, 4, 3, 4, 4, 4, 5, 3, 4, 5, 5, 5, 6, 4, 5, 6, 6, 6, 6, 7, 4, 6, 6, 7, 7, 7, 7, 8, 5, 6, 7, 8, 8, 8, 8, 8, 9, 5, 7, 8, 8, 9, 9, 9, 9, 9, 10, 6, 8, 9, 9, 10, 10, 10, 10, 10, 10, 11, 6, 8, 9, 10, 10, 11, 11, 11, 11, 11, 11, 12, 7, 9, 10, 11, 11, 12, 12, 12, 12, 12, 12, 12, 13, 7, 10, 11, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 14, 8, 10, 12, 12, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 15

Offset: 1

Dennis P. Walsh, Apr 18 2012

T(n,k) is the maximum number of wins in a sequence of n games in which the longest winning streak is of length k.

T(n,k) generalizes the pattern found in sequence A004523 where A004523(n) = floor(2n/3).

			T(12,4) = 10 since 10 is the maximum number of wins in a 12-game sequence in which the longest winning streak is 4. One such sequence with 10 wins is WWWWLWWWWLWW.
The triangle T(n,k) begins
1,
1, 2,
2, 2, 3,
2, 3, 3,  4,
3, 4, 4,  4,  5,
3, 4, 5,  5,  5,  6,
4, 5, 6,  6,  6,  6,  7,
4, 6, 6,  7,  7,  7,  7,  8,
5, 6, 7,  8,  8,  8,  8,  8,  9,
5, 7, 8,  8,  9,  9,  9,  9,  9, 10,
6, 8, 9,  9, 10, 10, 10, 10, 10, 10, 11,
6, 8, 9, 10, 10, 11, 11, 11, 11, 11, 11, 12,

Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened
Sela Fried and Toufik Mansour, The total number of descents and levels in (cyclic) tensor words, Disc. Math. Lett. (2024) Vol. 13, 44-49. See p. 49.

A004523(n+1) = T(n,2).

a182210 n k = a182210_tabl !! (n-1) !! (k-1)
a182210_tabl = [[k*(n+1) `div` (k+1) | k <- [1..n]] | n <- [1..]]
-- Reinhard Zumkeller, Jul 08 2012

seq(seq(floor(k*(n+1)/(k+1)),k=1..n),n=1..15);

Flatten[Table[Floor[k*(n+1)/(k+1)],{n,0,20},{k,n}]] (* Harvey P. Dale, Jul 21 2015 *)

T(n,k) = floor(k(n+1)/(k+1)).

cp's OEIS Frontend

User: Dennis P. Walsh

A335109 Triangle read by rows: T(n,k) is the number of permutations of length n with each cycle of the permutation containing only elements that are identical (mod k), where 1 <= k <= n.

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A335259 Triangle read by rows: T(n,k) = k^ceiling(n/k) for 1 <= k <= n.

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A293211 Triangle T(n,k) is the number of permutations on n elements with at least one k-cycle for 1 <= k <= n.

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A243987 Triangle read by rows: T(n, k) is the number of divisors of n that are less than or equal to k for 1 <= k <= n.

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A224711 Number of ballot results from n voters that prompt a run-off election when three candidates vie for two spots on a board.

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A224542 Number of doubly-surjective functions f:[n]->[4].

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A224541 Number of doubly-surjective functions f:[n]->[3].

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