A273975 -id:A273975 - cp's OEIS frontend

A077042 Square array read by falling antidiagonals of central polynomial coefficients: largest coefficient in expansion of (1 + x + x^2 + ... + x^(n-1))^k = ((1-x^n)/(1-x))^k, i.e., the coefficient of x^floor(k(n-1)/2) and of x^ceiling(k(n-1)/2); also number of compositions of floor(k*(n+1)/2) into exactly k positive integers each no more than n.

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 3, 1, 1, 0, 1, 6, 7, 4, 1, 1, 0, 1, 10, 19, 12, 5, 1, 1, 0, 1, 20, 51, 44, 19, 6, 1, 1, 0, 1, 35, 141, 155, 85, 27, 7, 1, 1, 0, 1, 70, 393, 580, 381, 146, 37, 8, 1, 1, 0, 1, 126, 1107, 2128, 1751, 780, 231, 48, 9, 1, 1, 0, 1, 252, 3139

Offset: 0

Henry Bottomley, Oct 22 2002

From Michel Marcus, Dec 01 2012: (Start)
A pair of numbers written in base n are said to be comparable if all digits of the first number are at least as big as the corresponding digit of the second number, or vice versa. Otherwise, this pair will be defined as uncomparable. A set of pairwise uncomparable integers will be called anti-hierarchic.
T(n,k) is the size of the maximal anti-hierarchic set of integers written with k digits in base n.
For example, for base n=2 and k=4 digits:
- 0 (0000) and 15 (1111) are comparable, while 6 (0110) and 9 (1001) are uncomparable,
- the maximal antihierarchic set is {3 (0011), 5 (0101), 6 (0110), 9 (1001), 10 (1010), 12 (1100)} with 6 elements that are all pairwise uncomparable. (End)

			Rows of square array start:
  1,    0,    0,    0,    0,    0,    0, ...
  1,    1,    1,    1,    1,    1,    1, ...
  1,    1,    2,    3,    6,   10,   20, ...
  1,    1,    3,    7,   19,   51,  141, ...
  1,    1,    4,   12,   44,  155,  580, ...
  1,    1,    5,   19,   85,  381, 1751, ...
  ...
Read by antidiagonals:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 1, 1;
  0, 1, 2, 1, 1;
  0, 1, 3, 3, 1, 1;
  0, 1, 6, 7, 4, 1, 1;
  ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Denis Bouyssou, Thierry Marchant, Marc Pirlot, The size of the largest antichains in products of linear orders, arXiv:1903.07569 [math.CO], 2019.
J. W. Sander, On maximal antihierarchic sets of integers, Discrete Mathematics, Volume 113, Issues 1-3, 5 April 1993, Pages 179-189.
Index entries for sequences related to compositions

Rows include A000007, A000012, A001405, A002426, A005190, A005191, A018901, A025012, A025013, A025014, A025015, A201549, A225779, A201550. Columns include A000012, A000012, A001477, A077043, A005900, A077044, A071816, A133458.
Central diagonal is A077045, with A077046 and A077047 either side. Cf. A067059, A270918, A201552.
Cf. A273975.

t[n_, k_] := Max[ CoefficientList[ Series[ ((1-x^n) / (1-x))^k, {x, 0, k*(n-1)}], x]]; t[0, 0] = 1; t[0, ] = 0; Flatten[ Table[ t[n-k, k], {n, 0, 12}, {k, n, 0, -1}]] (* _Jean-François Alcover, Nov 04 2011 *)

T(n,k)=if(n<1 || k<1,k==0,vecmax(Vec(((1-x^n)/(1-x))^k)))

By the central limit theorem, T(n,k) is roughly n^(k-1)*sqrt(6/(Pi*k)).

T(n,k) = Sum{j=0,h/n} (-1)^j*binomial(k,j)*binomial(k-1+h-n*j,k-1) with h=floor(k*(n-1)/2), k>0. - Michel Marcus, Dec 01 2012

A277949 Triangle read by rows, in which row n gives coefficients in expansion of ((x^n - 1)/(x - 1))^4.

Original entry on oeis.org

1, 1, 4, 6, 4, 1, 1, 4, 10, 16, 19, 16, 10, 4, 1, 1, 4, 10, 20, 31, 40, 44, 40, 31, 20, 10, 4, 1, 1, 4, 10, 20, 35, 52, 68, 80, 85, 80, 68, 52, 35, 20, 10, 4, 1, 1, 4, 10, 20, 35, 56, 80, 104, 125, 140, 146, 140, 125, 104, 80, 56, 35, 20, 10, 4, 1

Offset: 1

Juan Pablo Herrera P., Nov 05 2016

Sum of n-th row is n^4. The n-th row contains 4n-3 entries. Largest coefficients of each row are listed in A005900.
The n-th row is the fourth row of the n-nomial triangle. For example, row 2 (1,4,6,4,1) is the fourth row in the binomial triangle.
T(n,k) gives the number of possible ways of randomly selecting k cards from n-1 sets, each with four different playing cards. It is also the number of lattice paths from (0,0) to (4,k) using steps (1,0), (1,1), (1,2), ..., (1,n-1).

			Triangle starts:
1;
1, 4, 6, 4, 1;
1, 4, 10, 16, 19, 16, 10, 4, 1;
1, 4, 10, 20, 31, 40, 44, 40, 31, 20, 10, 4, 1;
1, 4, 10, 20, 35, 52, 68, 80, 85, 80, 68, 52, 35, 20, 10, 4, 1;
1, 4, 10, 20, 35, 56, 80, 104, 125, 140, 146, 140, 125, 104, 80, 56, 35, 20, 10, 4, 1.
...
There are T(3,2) = 10 ways to select 2 cards from two sets of four playing cards ABCD, namely, {AA}, {AB}, {AC}, {AD}, {BB}, {BC}, {BD}, {CC}, {CD}, and {DD}.

Juan Pablo Herrera P., Rows n=1..60 of the triangle, flattened

Cf. A000292, A004737, A005900, A109439, A210440, A277950, A277951.

Mentioned in: A273975

Table[CoefficientList[Series[((x^n - 1)/(x - 1))^4, {x, 0, 4 n}], x], {n, 6}] // Flatten (* Michael De Vlieger, Nov 10 2016 *)

row(n) = Vec(((1 - x^n)/(1 - x))^4);
tabf(nn) = for (n=1, nn, print(row(n)));

T(n,k) = Sum_{i=k-n+1..k} A109439(T(n,i)).
T(n,k) = A000292(k+1) = (k+3)!/(k!*6) if 0 =< k < n,
T(n,k) = ((k+3)*(k+2)*(k+1)-4*(k-n+3)*(k-n+2)*(k-n+1))/6 if n =< k < 2*n,
T(n,k) = ((4*n-1-k)*(4*n-2-k)*(4*n-3-k)-4*(3*n-1-k)*(3*n-2-k)*(3*n-3-k))/6 if 2*n-3 =< k < 3*n-3,
T(n,k) = A000292(4*n-3-k) = (4*n-1-k)!/((4*n-4-k)!*6) if 3*n-3  =< k < 4n-3.

A277950 Triangle read by rows, in which row n gives coefficients in expansion of ((x^n - 1)/(x - 1))^5.

Original entry on oeis.org

1, 1, 5, 10, 10, 5, 1, 1, 5, 15, 30, 45, 51, 45, 30, 15, 5, 1, 1, 5, 15, 35, 65, 101, 135, 155, 155, 135, 101, 65, 35, 15, 5, 1, 1, 5, 15, 35, 70, 121, 185, 255, 320, 365, 381, 365, 320, 255, 185, 121, 70, 35, 15, 5, 1

Offset: 1

Juan Pablo Herrera P., Nov 05 2016

Sum of n-th row is n^5. The n-th row contains 5n-4 entries. Largest coefficients of each row are listed in A077044.
The n-th row is the fifth row of the n-nomial triangle.  For example, row 2 (1,5,10,10,5,1) is the fifth row in the binomial triangle.
T(n,k) gives the number of possible ways of randomly selecting k cards from n-1 sets, each with five different playing cards. It is also the number of lattice paths from (0,0) to (5,k) using steps (1,0), (1,1), (1,2), ..., (1,n-1).

			Triangle starts:
1;
1, 5, 10, 10, 5, 1;
1, 5, 15, 30, 45, 51, 45, 30, 15, 5, 1;
1, 5, 15, 35, 65, 101, 135, 155, 155, 135, 101, 65, 35, 15, 5, 1;
1, 5, 15, 35, 70, 121, 185, 255, 320, 365, 381, 365, 320, 255, 185, 121, 70, 35, 15, 5, 1;

Juan Pablo Herrera P., Rows n=1..60 of the triangle, flattened

Cf. A000332, A004737, A077044, A109439, A154286, A277949, A277951.

Mentioned in: A273975.

Table[CoefficientList[Series[((x^n - 1)/(x - 1))^5, {x, 0, 5 n}], x], {n, 10}] // Flatten

row(n) = Vec(((1 - x^n)/(1 - x))^5); tabf(nn) = for (n=1, nn, print(row(n)));

T(n,k) = Sum_{i=k-n+1..k} A277949(T(n,i)).
From Juan Pablo Herrera P., Dec 20 2016: (Start)
T(n,k) = A000332(k+4) = (k+4)!/(k!*24) if 0 =< k < n.
T(n,k) = ((k+4)!/k!-5*(k-n+4)!/(k-n)!)/24 if n =< k < 2*n.
T(n,k) = ((k+4)!/k!-5*(k-n+4)!/(k-n)!+10*(k-2*n+4)!/(k-2*n)!)/24 if 2*n =< k < 3*n.
T(n,k) = ((5*n-k-1)!/(5*n-k-5)!-5*(4*n-k-1)!/(4*n-k-5)!)/24 if 3*n-4 =< k < 4*n-4.
T(n,k) = A000332(5*n-k-1) = (5*n-k-1)!/(5*n-k-5)!*24 4*n-4 =< k < 5*n-4. (End)

A277951 Triangle read by rows, in which row n gives coefficients in expansion of ((x^n - 1)/(x - 1))^6.

Original entry on oeis.org

1, 1, 6, 15, 20, 15, 6, 1, 1, 6, 21, 50, 90, 126, 141, 126, 90, 50, 21, 6, 1, 1, 6, 21, 56, 120, 216, 336, 456, 546, 580, 546, 456, 336, 216, 120, 56, 21, 6, 1, 1, 6, 21, 56, 126, 246, 426, 666, 951, 1246, 1506, 1686, 1751, 1686, 1506, 1246, 951, 666, 426, 246, 126, 56, 21, 6, 1

Offset: 1

Juan Pablo Herrera P., Nov 18 2016

Sum of n-th row is n^6. The n-th row contains 6n-5 entries. Largest coefficients of each row are listed in A071816.
The n-th row is the sixth row of the n-nomial triangle.  For example, row 2 (1,6,15,20,15,6,1) is the sixth row in the binomial triangle
T(n,k) gives the number of possible ways of randomly selecting k cards from n-1 sets, each with six different playing cards. It is also the number of lattice paths from (0,0) to (6,k) using steps (1,0), (1,1), (1,2), ..., (1,n-1).

			Triangle starts:
1;
1, 6, 15, 20, 15, 6, 1;
1, 6, 21, 50, 90, 126, 141, 126, 90, 50, 21, 6, 1;
1, 6, 21, 56, 120, 216, 336, 456, 546, 580, 546, 456, 336, 216, 120, 56, 21, 6, 1.

Juan Pablo Herrera P., Rows n=1..50 of the triangle, flattened

Cf. A000332, A004737, A071816, A109439, A154286, A277949, A277950.

Mentioned in: A273975

Table[CoefficientList[Series[((x^n - 1)/(x - 1))^6, {x, 0, 6 n}], x], {n, 10}] // Flatten

row(n) = Vec(((1 - x^n)/(1 - x))^6);
tabf(nn) = for (n=1, nn, print(row(n)));

T(n,k) = Sum_{i=k-n+1..k} A277950(T(n,i))

A163269 T(n,k) = largest coefficient in the expansion of (1 + ... + x^(n-1))^(2*k).

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 20, 19, 4, 1, 70, 141, 44, 5, 1, 252, 1107, 580, 85, 6, 1, 924, 8953, 8092, 1751, 146, 7, 1, 3432, 73789, 116304, 38165, 4332, 231, 8, 1, 12870, 616227, 1703636, 856945, 135954, 9331, 344, 9, 1, 48620, 5196627, 25288120, 19611175, 4395456

Offset: 1

R. H. Hardin, Jul 24 2009

T(n,k) = number of ways the sums of all components of two 1..n k-vectors can be equal.
T(n,k) is an odd polynomial in n of order 2*k-1.
Examples:
T(n,1) = n.
T(n,2) = (2/3)*n^3 + (1/3)*n.
T(n,3) = (11/20)*n^5 + (1/4)*n^3 + (1/5)*n.
T(n,4) = (151/315)*n^7 + (2/9)*n^5 + (7/45)*n^3 + (1/7)*n.
Table starts:
  1  1    1     1      1 ...
  2  6   20    70    252 ...
  3 19  141  1107   8953 ...
  4 44  580  8092 116304 ...
  5 85 1751 38165 856945 ...
  ...

			For n = 3 and k = 2, (1 + x + x^2)^(2*2) = x^8 + 4*x^7 + 10*x^6 + 16*x^5 + 19*x^4 + 16*x^3 + 10*x^2 + 4*x + 1, whose largest coefficient is T(3,2) = 19.

R. H. Hardin, Table of n, a(n) for n = 1..3240

Removing the leftmost column of A349933 generates this sequence.

Cf. A273975.

T(n,k) = polcoef(sum(i=0, n-1, x^i)^(2*k), k*(n-1)); \\ Michel Marcus, Jan 23 2024

T(n,k) = A273975(2*k, n, (n-1)*k). - Andrey Zabolotskiy, Jan 23 2024

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A277949 Triangle read by rows, in which row n gives coefficients in expansion of ((x^n - 1)/(x - 1))^4.

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A277950 Triangle read by rows, in which row n gives coefficients in expansion of ((x^n - 1)/(x - 1))^5.

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A277951 Triangle read by rows, in which row n gives coefficients in expansion of ((x^n - 1)/(x - 1))^6.

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A163269 T(n,k) = largest coefficient in the expansion of (1 + ... + x^(n-1))^(2*k).

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