A109439 -id:A109439 - cp's OEIS frontend

A008867 Triangle of truncated triangular numbers: k-th term in n-th row is number of dots in hexagon of sides k, n-k, k, n-k, k, n-k.

1, 3, 3, 6, 7, 6, 10, 12, 12, 10, 15, 18, 19, 18, 15, 21, 25, 27, 27, 25, 21, 28, 33, 36, 37, 36, 33, 28, 36, 42, 46, 48, 48, 46, 42, 36, 45, 52, 57, 60, 61, 60, 57, 52, 45, 55, 63, 69, 73, 75, 75, 73, 69, 63, 55, 66, 75, 82, 87, 90, 91, 90, 87, 82, 75, 66, 78, 88, 96, 102, 106, 108, 108, 106, 102, 96, 88, 78

Offset: 2

N. J. A. Sloane, J. H. Conway and R. K. Guy

Closely related to A109439. The current sequence is made of truncated triangular numbers, the latter gives the full description. Both can help to build a cube with layers perpendicular to the great diagonal. E.g.: 15,18,19,18,15 in A008867 is a truncation of the lesser triangular numbers of 1,3,6,10,15,18,19,18,15,10,6,3,1 in A109439. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 02 2005

The sequence is a triangle read by rows where the n-th row is obtained by multiplying by (1/3)*(n+1)*(2*(n+1)^2+1) the first row of the limit as k approaches infinity of P(n)^k where P(n) is the stochastic matrix associated with a variant of the Ehrenfest model using n balls. The elements of the stochastic matrix P(n) we have considered are given by P(n)[i,j] = n+1-(max(i,j)-min(i,j)), where each row must be normalized using the L1 norm and where i,j belong to the set {0,1,2,...,n}. They are defined as the probabilities of arriving in a state j given the previous state i. In particular the sum of every row of a stochastic matrix must be 1, and so the sum of the terms of the n-th row of this triangle is (1/3)*(n+1)*(2*(n+1)^2+1) (since the limit of a stochastic matrix is again a stochastic matrix). Furthermore, by the properties of Markov chains, we can interpret P(n)^k as the k-step transition matrix of this variant of the Ehrenfest model using n balls. It is important to note that the rows of the limit of the stochastic matrix are identical and since we know the first we know all the others. - Luca Onnis, Oct 29 2023

			Triangle begins:
n = 0:  1;
n = 1:  3,  3;
n = 2:  6,  7,  6;
n = 3: 10, 12, 12, 10;
n = 4: 15, 18, 19, 18, 15;
n = 5: 21, 25, 27, 27, 25, 21;
n = 6: 28, 33, 36, 37, 36, 33, 28;

Paul and Tatjana Ehrenfest, Über zwei bekannte Einwände gegen das Boltzmannsche H-Theorem, Physikalische Zeitschrift, vol. 8 (1907), pp. 311-314.

Hugo Pfoertner, Table of n, a(n) for n = 2..1226, rows 1..50, flattened.
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
Luca Onnis, Animation of this variant of the Ehrenfest model using n = 15 balls.
Wikipedia, Ehrenfest model.

Row sums are A005900(n-1).

Cf. A109439.

T:= (n, k)-> n*(n-3)/2 - k^2+k*n+1:
seq(seq(T(n,k), k=1..n-1), n=2..14);

T[n_,k_] := n*(n-3)/2 - k^2 + k*n + 1; Table[T[n,k], {n,3,20}, {k,n,2,-1}] // Flatten (* Amiram Eldar, Dec 12 2018 *)

T(n,k) = n*(n-3)/2 - k^2 + k*n + 1.

A273975 Three-dimensional array written by antidiagonals in k,n: T(k,n,h) with k >= 1, n >= 0, 0 <= h <= n*(k-1) is the coefficient of x^h in the polynomial (1 + x + ... + x^(k-1))^n = ((x^k-1)/(x-1))^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 3, 2, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 6, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 4, 3, 2, 1, 1, 3, 6, 10, 12, 12, 10, 6, 3, 1, 1, 4, 10

Offset: 1

Andrey Zabolotskiy, Nov 10 2016

Equivalently, T(k,n,h) is the number of ordered sets of n nonnegative integers < k with the sum equal to h.
From Juan Pablo Herrera P., Nov 21 2016: (Start)
T(k,n,h) is the number of possible ways of randomly selecting h cards from k-1 sets, each with n different playing cards. It is also the number of lattice paths from (0,0) to (n,h) using steps (1,0), (1,1), (1,2), ..., (1,k-1).
Shallow diagonal sums of each triangle with fixed k give the k-bonacci numbers. (End)
T(k,n,h) is the number of n-dimensional grid points of a k X k X ... X k grid, which are lying in the (n-1)-dimensional hyperplane which is at an L1 distance of h from one of the grid's corners, and normal to the corresponding main diagonal of the grid. - Eitan Y. Levine, Apr 23 2023

			For first few k and for first few n, the rows with h = 0..n*(k-1) are given:
k=1:  1;  1;  1;  1;  1; ...
k=2:  1;  1, 1;  1, 2, 1;  1, 3, 3, 1;  1, 4, 6, 4, 1; ...
k=3:  1;  1, 1, 1;  1, 2, 3, 2, 1;  1, 3, 6, 7, 6, 3, 1; ...
k=4:  1;  1, 1, 1, 1;  1, 2, 3, 4, 3, 2, 1; ...
For example, (1 + x + x^2)^3 = 1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6, hence T(3,3,2) = T(3,3,4) = 6.
From _Eitan Y. Levine_, Apr 23 2023: (Start)
Example for the repeated cumulative sum formula, for (k,n)=(3,3) (each line is the cumulative sum of the previous line, and the first line is the padded, alternating 3rd row from Pascal's triangle):
  1  0  0 -3  0  0  3  0  0 -1
  1  1  1 -2 -2 -2  1  1  1
  1  2  3  1 -1 -3 -2 -1
  1  3  6  7  6  3  1
which is T(3,3,h). (End)

Andrey Zabolotskiy, slices with k+n = 1..20, flattened
Florentin Smarandache, K-Nomial Coefficients, arXiv:math/0612062 [math.GM], 2006 (originally published in French in: F. Smarandache, Généralisations et Généralités, Ed. Nouvelle, 1984, pp. 24-26).

k-nomial arrays for fixed k=1..10: A000012, A007318, A027907, A008287, A035343, A063260, A063265, A171890, A213652, A213651.
Arrays for fixed n=0..6: A000012, A000012, A004737, A109439, A277949, A277950, A277951.
Central n-nomial coefficients for n=1..9, i.e., sequences with h=floor(n*(k-1)/2) and fixed n: A000012, A000984 (A001405), A002426, A005721 (A005190), A005191, A063419 (A018901), A025012, (A025013), A025014, A174061 (A025015), A201549, (A225779), A201550. Arrays: A201552, A077042, see also cfs. therein.
Triangle n=k-1: A181567. Triangle n=k: A163181.

a = Table[CoefficientList[Sum[x^(h-1),{h,k}]^n,x],{k,10},{n,0,9}];
Flatten@Table[a[[s-n,n+1]],{s,10},{n,0,s-1}]
(* alternate program *)
row[k_, n_] := Nest[Accumulate,Upsample[Table[((-1)^j)*Binomial[n,j],{j,0,n}],k],n][[;;n*(k-1)+1]] (* Eitan Y. Levine, Apr 23 2023 *)

T(k,n,h) = Sum_{i = 0..floor(h/k)} (-1)^i*binomial(n,i)*binomial(n+h-1-k*i,n-1). [Corrected by Eitan Y. Levine, Apr 23 2023]
From Eitan Y. Levine, Apr 23 2023: (Start)
(T(k,n,h))_{h=0..n*(k-1)} = f(f(...f(g(P))...)), where:
(x_i)_{i=0..m} denotes a tuple (in particular, the LHS contains the values for 0 <= h <= n*(k-1)),
f repeats n times,
f((x_i){i=0..m}) = (Sum{j=0..i} x_j)_{i=0..m} is the cumulative sum function,
g((x_i){i=0..m}) = (x(i/k) if k|i, otherwise 0)_{i=0..m*k} is adding k-1 zeros between adjacent elements,
and P=((-1)^i*binomial(n,i))_{i=0..n} is the n-th row of Pascal's triangle, with alternating signs. (End)
From Eitan Y. Levine, Jul 27 2023: (Start)
Recurrence relations, the first follows from the sequence's defining polynomial as mentioned in the Smarandache link:
T(k,n+1,h) = Sum_{i = 0..s-1} T(k,n,h-i)
T(k+1,n,h) = Sum_{i = 0..n} binomial(n,i)*T(k,n-i,h-i*k) (End)

A156039 Number of compositions (ordered partitions) of n into 4 parts, where the first is at least as great as each of the others.

Original entry on oeis.org

1, 1, 4, 7, 11, 17, 26, 35, 48, 63, 81, 102, 127, 154, 187, 223, 263, 308, 359, 413, 474, 540, 612, 690, 775, 865, 964, 1069, 1181, 1301, 1430, 1565, 1710, 1863, 2025, 2196, 2377, 2566, 2767, 2977, 3197, 3428, 3671, 3923, 4188, 4464, 4752, 5052, 5365, 5689

Offset: 0

Jack W Grahl, Feb 02 2009, Feb 11 2009

nonn

For n=1,2 these are just the tetrahedral numbers. a(n) is always at least 1/4 of the corresponding tetrahedral number, since each partition of this type gives up to four ordered partitions with the same cyclical order.

Diagonal sums of the irregular triangle A109439, for example a(0)=1, a(1)=1, a(2)=1+3, a(3)=1+3+3, a(4)=1+3+6+1. - Bob Selcoe, Feb 09 2014

			For n = 3 the a(3) = 7 compositions are: (3 0 0 0) (2 1 0 0) (2 0 1 0) (2 0 0 1) (1 1 1 0) (1 1 0 1) (1 0 1 1).

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2, -1, 1, -1, -1, 1, -1, 2, -1).

For partitions into 3 summands see A156040; also see A156041 and A156042.

a:= proc(n) local m, r; m:= iquo(n, 12, 'r'); r:= r+1; (9 +(27 +72*m +18*r)*m +((9 +3*r) *r-12) /2)*m +[1, 1, 4, 7, 11, 17, 26, 35, 48, 63, 81, 102][r] end: seq(a(n), n=0..60); # Alois P. Heinz, Jun 14 2009

LinearRecurrence[{2, -1, 1, -1, -1, 1, -1, 2, -1}, {1, 1, 4, 7, 11, 17, 26, 35, 48}, 60] (* Jean-François Alcover, May 17 2018 *)

G.f.: ( 1-x+3*x^2-x^3+x^4 ) / ( (1+x)*(1+x^2)*(1+x+x^2)*(x-1)^4 ). - Alois P. Heinz, Jun 14 2009

More terms from Alois P. Heinz, Jun 14 2009

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))

cp's OEIS Frontend

A008867 Triangle of truncated triangular numbers: k-th term in n-th row is number of dots in hexagon of sides k, n-k, k, n-k, k, n-k.

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A273975 Three-dimensional array written by antidiagonals in k,n: T(k,n,h) with k >= 1, n >= 0, 0 <= h <= n*(k-1) is the coefficient of x^h in the polynomial (1 + x + ... + x^(k-1))^n = ((x^k-1)/(x-1))^n.

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A156039 Number of compositions (ordered partitions) of n into 4 parts, where the first is at least as great as each of the others.

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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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