cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A213887 Triangle of coefficients of representations of columns of A213743 in binomial basis.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 0, 4, 6, 4, 1, 0, 0, 3, 10, 10, 5, 1, 0, 0, 2, 12, 20, 15, 6, 1, 0, 0, 1, 12, 31, 35, 21, 7, 1, 0, 0, 0, 10, 40, 65, 56, 28, 8, 1, 0, 0, 0, 6, 44, 101, 120, 84, 36, 9, 1, 0
Offset: 0

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Author

Keywords

Comments

This triangle is the third array in the sequence of arrays A026729, A071675 considered as triangles.
Let {a_(k,i)}, k>=1, i=0,...,k, be the k-th row of the triangle. 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 A213743. For example, s_1(n)=binomial(n,1)=n is the first column of A213743 for n>1, s_2(n)=binomial(n,1)+binomial(n,2)is the second column of A213743 for n>1, etc. In particular (see comment in A213743), in cases k=6,7,8,9 s_k(n) is A064056(n+2), A064057(n+2), A064058(n+2), A000575(n+3) respectively.
Riordan array (1,x+x^2+x^3+x^4). A186332 with additional 0 column. - Ralf Stephan, Dec 31 2013

Examples

			As a triangle, this begins
n/k.|..0....1....2....3....4....5....6....7....8....9
=====================================================
.0..|..1
.1..|..0....1
.2..|..0....1....1
.3..|..0....1....2....1
.4..|..0....1....3....3....1
.5..|..0....0....4....6....4....1
.6..|..0....0....3...10...10....5....1
.7..|..0....0....2...12...20...15....6....1
.8..|..0....0....1...12...31...35...21....7....1
.9..|..0....0....0...10...40...65...56...28....8....1
		

Crossrefs

Cf. A026729, A071675, A030528 (parts <=2), A078803 (parts <=3), A213888 (parts <=5), A061676 and A213889 (parts <=6).

Programs

  • Maple
    pts := 4; # A213887
    g := 1/(1-t*z*add(z^i,i=0..pts-1)) ;
    for n from 0 to 13 do
        for k from 0 to n do
            coeftayl(g,z=0,n) ;
            coeftayl(%,t=0,k) ;
            printf("%d ",%) ;
        end do:
        printf("\n") ;
    end do: # R. J. Mathar, May 28 2025

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

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Comments

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.

Examples

			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
		

Crossrefs

Programs

  • Mathematica
    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 *)

Formula

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

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Author

Keywords

Comments

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.

Examples

			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
		

Crossrefs

Programs

  • Mathematica
    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 *)

Formula

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

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Author

Keywords

Comments

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

Examples

			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
		

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    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

Formula

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.
Showing 1-4 of 4 results.