A213742 -id:A213742 - cp's OEIS frontend

A111808 Left half of trinomial triangle (A027907), triangle read by rows.

1, 1, 1, 1, 2, 3, 1, 3, 6, 7, 1, 4, 10, 16, 19, 1, 5, 15, 30, 45, 51, 1, 6, 21, 50, 90, 126, 141, 1, 7, 28, 77, 161, 266, 357, 393, 1, 8, 36, 112, 266, 504, 784, 1016, 1107, 1, 9, 45, 156, 414, 882, 1554, 2304, 2907, 3139, 1, 10, 55, 210, 615, 1452, 2850, 4740, 6765, 8350

Offset: 1

Reinhard Zumkeller, Aug 17 2005

Consider a doubly infinite chessboard with squares labeled (n,k), ranks or rows n in Z, files or columns k in Z (Z denotes ...,-2,-1,0,1,2,... ); number of king-paths of length n from (0,0) to (n,k), 0 <= k <= n, is T(n,n-k). - Harrie Grondijs, May 27 2005. Cf. A026300, A114929, A114972.

Triangle of numbers C^(2)(n-1,k), n>=1, of combinations with repetitions from elements {1,2,...,n} over k, such that every element i, i=1,...,n, appears in a k-combination either 0 or 1 or 2 times (cf. also A213742-A213745). - Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

Harrie Grondijs, Neverending Quest of Type C, Volume B - the endgame study-as-struggle.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Eric Weisstein's World of Mathematics, Trinomial Triangle
Eric Weisstein's World of Mathematics, Trinomial Coefficient

Row sums give A027914; central terms give A027908;
T(n, 0) = 0;
T(n, 1) = n for n>1;
T(n, 2) = A000217(n) for n>1;
T(n, 3) = A005581(n) for n>2;
T(n, 4) = A005712(n) for n>3;
T(n, 5) = A000574(n) for n>4;
T(n, 6) = A005714(n) for n>5;
T(n, 7) = A005715(n) for n>6;
T(n, 8) = A005716(n) for n>7;
T(n, 9) = A064054(n-5) for n>8;
T(n, n-5) = A098470(n) for n>4;
T(n, n-4) = A014533(n-3) for n>3;
T(n, n-3) = A014532(n-2) for n>2;
T(n, n-2) = A014531(n-1) for n>1;
T(n, n-1) = A005717(n) for n>0;
T(n, n) = central terms of A027907 = A002426(n).

T := (n,k) -> simplify(GegenbauerC(k, -n, -1/2)):
for n from 0 to 9 do seq(T(n,k), k=0..n) od; # Peter Luschny, May 09 2016

Table[GegenbauerC[k, -n, -1/2], {n,0,10}, {k,0,n}] // Flatten (* G. C. Greubel, Feb 28 2017 *)

(1 + x + x^2)^n = Sum(T(n,k)*x^k: 0<=k<=n) + Sum(T(n,k)*x^(2*n-k): 0<=k

T(n, k) = A027907(n, k) = Sum_{i=0,..,(k/2)} binomial(n, n-k+2*i) * binomial(n-k+2*i, i), 0<=k<=n.

T(n, k) = GegenbauerC(k, -n, -1/2). - Peter Luschny, May 09 2016

Corrected and edited by Johannes W. Meijer, Oct 05 2010

A071675 Array read by antidiagonals of trinomial coefficients.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 0, 3, 3, 1, 0, 0, 2, 6, 4, 1, 0, 0, 1, 7, 10, 5, 1, 0, 0, 0, 6, 16, 15, 6, 1, 0, 0, 0, 3, 19, 30, 21, 7, 1, 0, 0, 0, 1, 16, 45, 50, 28, 8, 1, 0, 0, 0, 0, 10, 51, 90, 77, 36, 9, 1, 0, 0, 0, 0, 4, 45, 126, 161, 112, 45, 10, 1, 0, 0, 0, 0, 1, 30, 141, 266, 266

Offset: 0

Henry Bottomley, May 30 2002

Read as a number triangle, this is the Riordan array (1, x(1+x+x^2)) with T(n,k) = Sum_{i=0..floor((n+k)/2)} C(k,2i+2k-n)*C(2i+2k-n,i). Rows start {1}, {0,1}, {0,1,1}, {0,1,2,1}, {0,0,3,3,1},... Row sums are then the trinomial numbers A000073(n+2). Diagonal sums are A013979. - Paul Barry, Feb 15 2005

Let {a_(k,i)}, k>=1, i=0,...,k, be the k-th antidiagonal of the array. Then s_k(n)=sum{i=0,...,k}a_(k,i)* binomial(n,k) is the n-th element of the k-th column of A213742. For example, s_1(n)=binomial(n,1)=n is the first column of A213742 for n>1, s_2(n)=binomial(n,1)+binomial(n,2)is the second column of A213742 for n>1, etc. In particular (see comment in A213742) in cases k=4,5,6,7,8, s_k(n) is A005718(n+2), A005719(n), A005720(n), A001919(n), A064055(n+3), respectively. - Vladimir Shevelev and Peter J. C. Moses, Jun 22 2012

			Rows start
1, 0,  0,  0,  0,  0, ...;
1, 1,  1,  0,  0,  0,  0, ...;
1, 2,  3,  2,  1,  0,  0, ...;
1, 3,  6,  7,  6,  3,  1, 0, ...;
1, 4, 10, 16, 19, 16, 10, 4, 1, ...; etc.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Milan Janjic, Binomial Coefficients and Enumeration of Restricted Words, Journal of Integer Sequences, 2016, Vol 19, #16.7.3

Visible version of A027907. Row sums are 3^n, i.e. A000244. Central diagonal is A002426. Cf. A071676 for a slight variation.

T[n_, k_] := Sum[Binomial[n - k - j, j]*Binomial[k, n - k - j], {j, 0,
Floor[(n - k)/2]}]; Table[T[n, k], {n,0,10}, {k,0,n}] // Flatten (* G. C. Greubel, Feb 28 2017 *)

T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-1, k-2) starting with T(0, 0)=1. See A027907 for more.

As a number triangle, T(n, k) = Sum_{i=0..floor((n-k)/2)} C(n-k-i, i) * C(k, n-k-i). - Paul Barry, Apr 26 2005

A213743 Triangle T(n,k), read by rows, of numbers T(n,k)=C^(4)(n,k) of combinations with repetitions from n different elements over k for each of them not more than four appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 121, 1, 6, 21, 56, 126, 246, 426, 1, 7, 28, 84, 210, 455, 875, 1520, 1, 8, 36, 120, 330, 784, 1652, 3144, 5475, 1, 9, 45, 165, 495, 1278, 2922, 6030, 11385, 19855, 1, 10, 55, 220, 715, 1992, 4905, 10890, 22110, 41470, 72403

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

The left side of triangle consists of 1's, while the right side is formed by A187925. Further, T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A000332(n) for n>=7, T(n,5)=A027659(n) for n>=3, T(n,6)=A064056(n) for n>=4, T(n,7)=A064057(n) for n>=5, T(n,8)=A064058(n) for n>=6, T(n,9)=A000575(n) for n>=6.

			Triangle begins
  n/k.|..0.....1.....2.....3.....4.....5.....6.....7
  ==================================================
  .0..|..1
  .1..|..1.....1
  .2..|..1.....2.....3
  .3..|..1.....3.....6....10
  .4..|..1.....4....10....20....35
  .5..|..1.....5....15....35....70....121
  .6..|..1.....6....21....56...126....246...426
  .7..|..1.....7....28....84...210....455...875....1520
T(4,2)=C^(4)(4,2): From 4 elements {1,2,3,4}, we have the following 10 allowed combinations of 2 elements: {1,1}, {1,2}, {1,3}, {1,4}, {2,2}, {2,3}, {2,4}, {3,3}, {3,4}, {4,4}.

Peter J. C. Moses, Rows n = 0..50 of triangle, flattened

Cf. A007318, A005725, A111808, A187925, A213742, A000217, A000292, A000332, A027659, A064056, A064057, A064058, A000575.

Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[5] (* Peter J. C. Moses, Apr 16 2013 *)

C^(4)(n,k) = Sum_{r=0...floor(k/5)} (-1)^r*C(n,r)*C(n-5*r+k-1, n-1).

A213744 Triangle of numbers C^(5)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 5 appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 456, 1, 7, 28, 84, 210, 462, 917, 1667, 1, 8, 36, 120, 330, 792, 1708, 3368, 6147, 1, 9, 45, 165, 495, 1287, 2994, 6354, 12465, 22825, 1, 10

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

For k<=4, the triangle coincides with triangle A213743.

We have over columns of the triangle: T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A000332(n) for n>=7, T(n,5)=A000389(n) for n>=9, T(n,6)=A062989(n) for n>=5, T(n,7)=A063262 for n>=5, T(n,8)=A063263 for n>=6, T(n,9)=A063264 for n>=7.

			Triangle begins
n/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1.....2.....3
.3..|..1.....3.....6....10
.4..|..1.....4....10....20....35
.5..|..1.....5....15....35....70....126
.6..|..1.....6....21....56...126....252...456
.7..|..1.....7....28....84...210....462...917....1667

Peter J. C. Moses, Rows n = 0..50 of triangle, flattened

Cf. A007318, A005725, A111808, A187925, A213742, A213743, A000217, A000292, A000332, A000389, A062989, A063262, A063263, A063264.

Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[6] (* Peter J. C. Moses, Apr 16 2013 *)

C^(5)(n,k)=sum{r=0,...,floor(k/6)}(-1)^r*C(n,r)*C(n-6*r+k-1, n-1)

A213745 Triangle of numbers C^(6)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 6 appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1709, 1, 8, 36, 120, 330, 792, 1716, 3424, 6371, 1, 9, 45, 165, 495, 1287, 3003, 6426, 12789, 23905, 1, 10

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

For k<=5, the triangle coincides with triangle A213744.

We have over columns of the triangle: T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A000332(n) for n>=7, T(n,5)=A000389(n) for n>=9, T(n,6)=A000579(n) for n>=11, T(n,7)=A063267 for n>=5, T(n,8)=A063417 for n>=6, T(n,9)=A063418 for n>=7.

			Triangle begins
n/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1.....2.....3
.3..|..1.....3.....6....10
.4..|..1.....4....10....20....35
.5..|..1.....5....15....35....70....126
.6..|..1.....6....21....56...126....252...462
.7..|..1.....7....28....84...210....462...924....1709

Peter J. C. Moses, Rows n = 0..50 of triangle, flattened

Cf. A007318, A005725, A059481, A111808, A187925, A213742, A213743, A213744, A000217, A000292, A000332, A000389, A000579, A063267, A063417, A063418.

Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[7] (* Peter J. C. Moses, Apr 16 2013 *)

C^(6)(n,k)=sum{r=0,...,floor(k/7)}(-1)^r*C(n,r)*C(n-7*r+k-1, n-1).
A generalization. The numbers C^(t)(n,k) of combinations with repetitions from n different elements over k, for each of them not more than t>=1 appearances allowed, are enumerated by the formula:
C^(t)(n,k)=sum{r=0,...,floor(k/(t+1))}(-1)^r*C(n,r)*C(n-(t+1)*r+k-1, n-1).
In case t=1, it is binomial coefficient C^(t)(n,k)=C(n,k), and we have the combinatorial identity: sum{r=0,...,floor(k/2)}(-1)^r*C(n,r)*C(n-2*r+k-1, n-1)=C(n,k). On the other hand, if t=n, then r=0, and for the corresponding numbers of combinations with repetitions without a restriction on appearances of elements we obtain a well known formula C(n+k-1, n-1) (cf. triangle A059481).
In addition, note that, if k<=t, then C^(t)(n,k)=C(n+k-1, n-1). Therefore, triangle {C^(t+1)(n,k)} coincides with the previous triangle {C^(t)(n,k)} for k<=t.

A213808 Triangle of numbers C^(7)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 7 appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1716, 1, 8, 36, 120, 330, 792, 1716, 3432, 6427, 1, 9, 45, 165, 495, 1287, 3003, 6435, 12861, 24229, 1, 10, 55, 220, 715, 2002, 5005, 11440, 24300, 48520, 91828

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jun 20 2012

For k <= 6, the triangle coincides with triangle A213745.

			Triangle begins
n/k |  0     1     2     3     4     5     6     7     8
----+---------------------------------------------------
  0 |  1
  1 |  1     1
  2 |  1     2     3
  3 |  1     3     6    10
  4 |  1     4    10    20    35
  5 |  1     5    15    35    70   126
  6 |  1     6    21    56   126   252   462
  7 |  1     7    28    84   210   462   924  1716
  8 |  1     8    36   120   330   792  1716  3432  6427

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A007318, A005725, A059481, A111808, A187925, A213742, A213743, A213744, A000217, A000292, A000332, A000389, A000579, A000580.

Table[Sum[(-1)^r*Binomial[n, r]*Binomial[n - 8*r + k - 1, n - 1], {r, 0, Floor[k/8]}], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Nov 25 2017 *)

for(n=0,10, for(k=0,n, print1(if(n==0 && k==0, 1, sum(r=0, floor(k/8), (-1)^r*binomial(n,r)*binomial(n-8*r + k-1,n-1))), ", "))) \\ G. C. Greubel, Nov 25 2017

T(n,k) = Sum_{r=0..floor(k/8)} (-1)^r*C(n,r)*C(n-8*r+k-1, n-1).

T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n > 1, T(n,3)=A000292(n) for n >= 3, T(n,4)=A000332(n) for n >= 7, T(n,5)=A000389(n) for n >= 9, T(n,6)=A000579(n) for n >= 11, T(n,7)=A000580(n) for n >= 13.

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cp's OEIS Frontend

A111808 Left half of trinomial triangle (A027907), triangle read by rows.

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A071675 Array read by antidiagonals of trinomial coefficients.

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A213743 Triangle T(n,k), read by rows, of numbers T(n,k)=C^(4)(n,k) of combinations with repetitions from n different elements over k for each of them not more than four appearances allowed.

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A213744 Triangle of numbers C^(5)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 5 appearances allowed.

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A213745 Triangle of numbers C^(6)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 6 appearances allowed.

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A213808 Triangle of numbers C^(7)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 7 appearances allowed.

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