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 1-3 of 3 results.

A187889 Riordan matrix (1/(1-x-x^2-x^3),(x+x^2+x^3)/(1-x-x^2-x^3)).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 4, 8, 5, 1, 7, 19, 18, 7, 1, 13, 43, 54, 32, 9, 1, 24, 94, 147, 117, 50, 11, 1, 44, 200, 375, 375, 216, 72, 13, 1, 81, 418, 913, 1100, 799, 359, 98, 15, 1, 149, 861, 2147, 3027, 2657, 1507, 554, 128, 17, 1, 274, 1753, 4914, 7937, 8174, 5610, 2603, 809, 162, 19, 1
Offset: 0

Views

Author

Emanuele Munarini, Mar 15 2011

Keywords

Examples

			Triangle begins:
1
1,1
2,3,1
4,8,5,1
7,19,18,7,1
13,43,54,32,9,1
24,94,147,117,50,11,1
44,200,375,375,216,72,13,1
81,418,913,1100,799,359,98,15,1
		

Crossrefs

Programs

  • Mathematica
    (* Function RiordanSquare defined in A321620. *)
    RiordanSquare[1/(1 - x - x^2- x^3), 11] // Flatten (* Peter Luschny, Nov 27 2018 *)
  • Maxima
    trinomial(n,k):=coeff(expand((1+x+x^2)^n),x,k);
    create_list(sum(binomial(i+k,k)*trinomial(i+k,n-k-i),i,0,n-k),n,0,8,k,0,n);

Formula

a(n,k) = Sum_{i=0..n-k} binomial(i+k,k)*trinomial(i+k,n-k-i), where trinomial(n,k) are the trinomial coefficients (A027907).
Recurrence: a(n+3,k+1) = a(n+2,k+1) + a(n+2,k) + a(n+1,k+1) + a(n+1,k) + a(n,k+1) + a(n,k)

A202193 Triangle read by rows: T(n,m) = coefficient of x^n in expansion of (x/(1 - x - x^2 - x^3 - x^4))^m.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 5, 3, 1, 8, 12, 9, 4, 1, 15, 28, 25, 14, 5, 1, 29, 62, 66, 44, 20, 6, 1, 56, 136, 165, 129, 70, 27, 7, 1, 108, 294, 401, 356, 225, 104, 35, 8, 1, 208, 628, 951, 944, 676, 363, 147, 44, 9, 1, 401, 1328, 2211, 2424, 1935, 1176, 553, 200, 54, 10, 1
Offset: 1

Views

Author

Vladimir Kruchinin, Dec 14 2011

Keywords

Comments

From Philippe Deléham, Feb 16 2014: (Start)
As a Riordan array, this is (1/(1 - x - x^2 - x^3 - x^4), x/(1 - x - x^2 - x^3 - x^4)).
T(n,0) = A000078(n+3); T(n+1,1) = A118898(n+4).
Row sums are A103142(n).
Diagonal sums are A077926(n)*(-1)^n.
Tetranacci convolution triangle. (End)

Examples

			Triangle begins:
   1;
   1,  1;
   2,  2,  1;
   4,  5,  3,  1;
   8, 12,  9,  4,  1;
  15, 28, 25, 14,  5,  1;
  29, 62, 66, 44, 20,  6,  1;
		

Crossrefs

Similar sequences : A037027 (Fibonacci convolution triangle), A104580 (tribonacci convolution triangle). - Philippe Deléham, Feb 16 2014

Programs

  • Maxima
    T(n,m):=if n=m then 1 else sum(sum((-1)^i*binomial(k,k-i)*binomial(n-m-4*i-1,k-1),i,0,(n-m-k)/4)*binomial(k+m-1,m-1),k,1,n-m);

Formula

T(n,m) = Sum_{k=1..n-m} (Sum_{i=0..floor((n-m-k)/4)} (-1)^i*binomial(k,k-i)*binomial(n-m-4*i-1,k-1))*binomial(k+m-1,m-1), n > m, T(n,n)=1.
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-3,k) + T(n-4,k), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Feb 16 2014
G.f. for column m: (x/(1 - x - x^2 - x^3 - x^4))^m. - Jason Yuen, Feb 17 2025

A189187 Riordan matrix (1/(1-x-x^2-x^3),(x+x^2)/(1-x-x^2-x^3)).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 4, 7, 5, 1, 7, 17, 16, 7, 1, 13, 38, 46, 29, 9, 1, 24, 82, 122, 99, 46, 11, 1, 44, 174, 304, 303, 184, 67, 13, 1, 81, 362, 728, 857, 641, 309, 92, 15, 1, 149, 743, 1690, 2291, 2031, 1212, 482, 121, 17, 1, 274, 1509, 3827, 5869, 6004, 4260, 2108, 711, 154, 19, 1
Offset: 0

Views

Author

Emanuele Munarini, Apr 18 2011

Keywords

Comments

Row sums are A077936, diagonal sums are A077946

Examples

			Triangle begins:
1
1,1
2,3,1
4,7,5,1
7,17,16,7,1
13,38,46,29,9,1
24,82,122,99,46,11,1
44,174,304,303,184,67,13,1
81,362,728,857,641,309,92,15,1
		

Crossrefs

Programs

  • Mathematica
    Flatten[Table[Sum[Binomial[i+k,k]Sum[Binomial[i+k,j]Binomial[n-i-j,i+k],{j,0,n-k-2i}],{i,0,n}],{n,0,20},{k,0,n}]]
  • Maxima
    create_list(sum(binomial(i+k,k)*sum(binomial(i+k,j)*binomial(n-i-j,i+k),j,0,n-k-2*i),i,0,n),n,0,8,k,0,n);

Formula

T(n,k) = [x^n](x+x^2)^k/(1-x-x^2-x^3)^(k+1).
T(n,k) = sum(binomial(i+k,k)*sum(binomial(i+k,j)*binomial(n-i-j,i+k),j=0..n-k-2*i),i=0..n).
T(n,k) = sum(binomial(k,i)*(-1)^(k-i)*sum(binomial(j+k,k)*trinomial(i+j,n-3*k+2*i-j),j=0..n-k),i=0..k)
Recurrence: T(n+3,k+1) = T(n+2,k+1) + T(n+2,k) + T(n+1,k+1) + T(n+1,k) + T(n,k+1)

Extensions

a(23) and a(40) corrected by Georg Fischer, Feb 20 2021 and Apr 29 2022
Showing 1-3 of 3 results.