A186332 Riordan array (1, x + x^2 + x^3 + x^4) without 0-column.

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 0, 4, 6, 4, 1, 0, 3, 10, 10, 5, 1, 0, 2, 12, 20, 15, 6, 1, 0, 1, 12, 31, 35, 21, 7, 1, 0, 0, 10, 40, 65, 56, 28, 8, 1, 0, 0, 6, 44, 101, 120, 84, 36, 9, 1, 0, 0, 3, 40, 135, 216, 203, 120, 45, 10, 1, 0, 0, 1, 31, 155, 336, 413, 322, 165, 55, 11, 1

Offset: 1

Author

Vladimir Kruchinin, Feb 17 2011

Keywords

• nonn
• easy
• tabl

Comments

Columns k >= 1 contain the expansion coefficients T(n,k) = [x^(n-k)] (x + x^2 + x^3 + x^4)^k.

Number of lattice paths from (0,0) to (n,k) using steps (1,1), (2,1), (3,1), (4,1). - Joerg Arndt, Jul 05 2011

Examples

			Array begins
  1;
  1, 1;
  1, 2,  1;
  1, 3,  3,  1;
  0, 4,  6,  4,   1;
  0, 3, 10, 10,   5,   1;
  0, 2, 12, 20,  15,   6,   1;
  0, 1, 12, 31,  35,  21,   7,   1;
  0, 0, 10, 40,  65,  56,  28,   8,   1;
  0, 0,  6, 44, 101, 120,  84,  36,   9,  1;
  0, 0,  3, 40, 135, 216, 203, 120,  45, 10,  1;
  0, 0,  1, 31, 155, 336, 413, 322, 165, 55, 11, 1;
  ...
		

Links

• Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.

Programs

Mathematica

T[n_, k_] := Sum[(-1)^m*Binomial[k, k - m]*Binomial[n - 4*m - 1, k - 1], {m, 0, (n - k)/4}];
Table[T[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 22 2018, from 2nd formula *)

Formula

T(n,k) = Sum_{j=0..k} binomial(k,j) * Sum_{i=0..n-k} binomial(j,i)*binomial(k-j,n-3*k+2*j-i), n>0, n>=k.

T(n,k) = Sum_{m=0..floor((n-k)/4)} (-1)^m*binomial(k,k-m)*binomial(n-4*m-1,k-1), n>0, n>=k.

O.g.f. of row polynomials R(n, x). I.e., o.g.f. of triangle (Riordan): G(z,x) = 1/(1 - x*z*(1+z)*(1+z^2)) - 1 (without column k=0). - Wolfdieter Lang, Jan 29 2021