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 1-6 of 6 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

A213888 Triangle of coefficients of representations of columns of A213744 in binomial basis.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 0, 5, 10, 10, 5, 1, 0, 0, 4, 15, 20, 15, 6, 1, 0, 0, 3, 18, 35, 35, 21, 7, 1, 0, 0, 2, 19, 52, 70, 56, 28, 8, 1, 0, 0, 1, 18, 68, 121, 126, 84, 36, 9, 1, 0
Offset: 0

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Author

Keywords

Comments

This triangle is the fourth array in the sequence of arrays A026729, A071675, A213887,..., such that the first two arrays are 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 A213744. For example, s_1(n)=binomial(n,1)=n is the first column of A213744 for n>1, s_2(n)=binomial(n,1)+binomial(n,2)is the second column of A213744 for n>1, etc. In particular (see comment inA213744), in cases k=7,8,9 s_k(n) is A063262(n+2), A063263(n+2), A063264(n+2) respectively.

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....1....4....6....4....1
.6..|..0....0....5...10...10....5....1
.7..|..0....0....4...15...20...15....6....1
.8..|..0....0....3...18...35...35...21....7....1
.9..|..0....0....2...19...52...70...56...28....8....1
		

Crossrefs

Cf. A026729, A071675, A213887, A030528 (parts <=2), A078803 (parts <=3), A213887 (parts <=4).

Programs

  • Maple
    pts := 5; # A213888
    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

A061676 Triangle T(n,k) of number of ways of throwing k standard dice to produce a total of n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 0, 6, 15, 20, 15, 6, 1, 0, 5, 21, 35, 35, 21, 7, 1, 0, 4, 25, 56, 70, 56, 28, 8, 1, 0, 3, 27, 80, 126, 126, 84, 36, 9, 1, 0, 2, 27, 104, 205, 252, 210, 120, 45, 10, 1, 0, 1, 25, 125, 305, 456, 462, 330, 165, 55, 11, 1
Offset: 1

Views

Author

Henry Bottomley, Apr 01 2002

Keywords

Examples

			Rows start:
1;
1,1;
1,2,1;
1,3,3,1;
1,4,6,4,1;
1,5,10,10,5,1;
0,6,15,20,15,6,1;
0,5,21,35,35,21,7,1;
etc.
T(8,2)=5 since 8 =2+6 =3+5 =4+4 =5+3 =6+2.
		

Crossrefs

First 21 terms as A007318 (see formula). Cf. A001592, A069713.
Cf. A030528 (2-sided dice), A078803 (3-sided), A213887 (4-sided), A213888 (5-sided).

Programs

  • Maple
    pts := 6; # A213889 and A061676
    g := 1/(1-t*z*add(z^i,i=0..pts-1)) ;
    for n from 1 to 13 do
        for k from 1 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

Formula

T(n, k)=T(n-1, k-1)+T(n-2, k-1)+T(n-3, k-1)+T(n-4, k-1)+T(n-5, k-1)+T(n-6, k-1) starting with T(0, 0)=1. T(n, k)=T(7k-n, k); if n>6k or n6k-6, T(n, k)=C(7k-n-1, k-1); T([7k/2], k)=A018901(k).

A213889 Triangle of coefficients of representations of columns of A213745 in binomial basis.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 0, 1, 5, 10, 10, 5, 1, 0, 0, 6, 15, 20, 15, 6, 1, 0, 0, 5, 21, 35, 35, 21, 7, 1, 0, 0, 4, 25, 56, 70, 56, 28, 8, 1, 0, 0, 3, 27, 80, 126, 126, 84, 36, 9, 1
Offset: 0

Views

Author

Keywords

Comments

This array is the fifth array in the sequence of arrays A026729, A071675, A213887, A213888,..., such that the first two arrays are 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 A213745. For example, s_1(n)=binomial(n,1)=n is the first column of A213745 for n>1, s_2(n)=binomial(n,1)+binomial(n,2)is the second column of A213745 for n>1, etc. In particular (see comment in A213745), in cases k=8,9 s_k(n) is A063417(n+2), A063418(n+2) respectively.

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....1....4....6....4....1
.6..|..0....1....5...10...10....5....1
.7..|..0....0....6...15...20...15....6....1
.8..|..0....0....5...21...35...35...21....7....1
.9..|..0....0....4...25...56...70...56...28....8....1
		

Crossrefs

Cf. A026729, A071675, A078803 (parts <=3), A213887 (parts <=4), A213888 (parts <=5).
Essentially the same as A061676.

Programs

  • Maple
    pts := 6; # A213889 and A061676
    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

A078802 Triangular array T given by T(n,k) = number of 01-words of length n containing k 1's, no three of which are consecutive.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 0, 1, 4, 6, 2, 0, 1, 5, 10, 7, 1, 0, 1, 6, 15, 16, 6, 0, 0, 1, 7, 21, 30, 19, 3, 0, 0, 1, 8, 28, 50, 45, 16, 1, 0, 0, 1, 9, 36, 77, 90, 51, 10, 0, 0, 0, 1, 10, 45, 112, 161, 126, 45, 4, 0, 0, 0, 1, 11, 55, 156, 266, 266, 141, 30, 1, 0, 0, 0, 1, 12, 66, 210, 414
Offset: 0

Views

Author

Clark Kimberling, Dec 06 2002

Keywords

Comments

The rows of T are essentially the antidiagonals of A027907 (trinomial coefficients). Reversing the rows produces A078803. Row sums: A000073.
Also, the diagonals of T are essentially the rows of A027907, so diagonal sums = 3^n. Antidiagonal sums are essentially A060961 (number of ordered partitions of n into 1's, 3's and 5's). - Gerald McGarvey, May 13 2005

Examples

			T(4,3) = 2 counts 1+0+1+1 and 1+1+0+1. Top of triangle T:
  1;
  1, 1;
  1, 2, 1;
  1, 3, 3, 0;
  1, 4, 6, 2, 0;
		

References

  • Clark Kimberling, Binary words with restricted repetitions and associated compositions of integers, in Applications of Fibonacci Numbers, vol. 10, Proceedings of the Eleventh International Conference on Fibonacci Numbers and Their Applications, William Webb, editor, Congressus Numerantium, Winnipeg, Manitoba 194 (2009) 141-151.

Crossrefs

Cf. A027907, A078803. See A082601 for another version.

Programs

  • Maple
    seq(seq(sum(binomial(n+1-k,k-j)*binomial(k-j,j),j=0..ceil((k-1)/2)),k=0..n),n=0..20); # Dennis P. Walsh, Apr 04 2012
  • Mathematica
    nn=15; a=1+y x+y^2 x^2;f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[a/(1-x a),{x,0,nn}],{x,y}]]//Grid (* Geoffrey Critzer, Sep 15 2012 *)

Formula

T(n, k) = T(n-1, k) + T(n-2, k-1) + T(n-3, k-2) with initial values as in first 3 rows.
T(n,k) = Sum_{j=0..ceiling((k-1)/2)} C(n+1-k, k-j)*C(k-j, j). - Dennis P. Walsh, Apr 04 2012
G.f.: (1 + y*x + y^2*x^2)/(1 - (x*(1 + y*x + y^2*x^2))). - Geoffrey Critzer, Sep 15 2012

A339884 Triangle read by rows: T(n, m) gives the number of partitions of n with m parts and parts from {1, 2, 3}.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 2, 1, 1, 0, 0, 2, 2, 2, 1, 1, 0, 0, 1, 3, 2, 2, 1, 1, 0, 0, 1, 2, 3, 2, 2, 1, 1, 0, 0, 0, 2, 3, 3, 2, 2, 1, 1, 0, 0, 0, 1, 3, 3, 3, 2, 2, 1, 1, 0, 0, 0, 1, 2, 4, 3, 3, 2, 2, 1, 1
Offset: 1

Views

Author

Wolfdieter Lang, Jan 31 2021

Keywords

Comments

Row sums give A001399(n), for n >= 1.
One could add the column [1,repeat 0] for m = 0 starting with n >= 0.

Examples

			The triangle T(n,m) begins:
  n\m  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ...
  1:   1
  2:   1 1
  3:   1 1 1
  4:   0 2 1 1
  5:   0 1 2 1 1
  6:   0 1 2 2 1 1
  7:   0 0 2 2 2 1 1
  8:   0 0 1 3 2 2 1 1
  9:   0 0 1 2 3 2 2 1 1
  10:  0 0 0 2 3 3 2 2 1  1
  11:  0 0 0 1 3 3 3 2 2  1  1
  12:  0 0 0 1 2 4 3 3 2  2  1  1
  13:  0 0 0 0 2 3 4 3 3  2  2  1  1
  14:  0 0 0 0 1 3 4 4 3  3  2  2  1  1
  15:  0 0 0 0 1 2 4 4 4  3  3  2  2  1  1
  16:  0 0 0 0 0 2 3 5 4  4  3  3  2  2  1  1
  17:  0 0 0 0 0 1 3 4 5  4  4  3  3  2  2  1  1
  18:  0 0 0 0 0 1 2 4 5  5  4  4  3  3  2  2  1  1
  19:  0 0 0 0 0 0 2 3 5  5  5  4  4  3  3  2  2  1  1
  20:  0 0 0 0 0 0 1 3 4  6  5  5  4  4  3  3  2  2  1  1
  ...
Row n = 6: the partitions of 6 with number of parts m = 1,2, ...., 6, and parts from {1,2,3} are (in Abramowitz-Stegun order): [] | [],[],[3,3] | [],[1,2,3],[2^3] | [1^3,3],[1^2,2^2] | [1^4,2] | 1^6, giving 0, 1, 2, 2, 1, 1.
		

Crossrefs

Cf. A001399, A008284 (all parts), A145362 (parts 1, 2), A232539 (parts <=4), A291983.
Compositions: A007818, A030528 (parts 1, 2), A078803 (parts 1, 2, 3).

Formula

Sum_{k=0..n} (-1)^k * T(n,k) = A291983(n). - Alois P. Heinz, Feb 01 2021
G.f.: 1/((1-u*t)*(1-u*t^2)*(1-u*t^3)). [Comtet page 97 [2c]]. - R. J. Mathar, May 27 2025
Showing 1-6 of 6 results.