A103141 Riordan array (1/(1-x), x*(1 + x + x^2 + x^3)/(1-x)).

1, 1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 10, 15, 7, 1, 1, 14, 35, 28, 9, 1, 1, 18, 68, 84, 45, 11, 1, 1, 22, 116, 207, 165, 66, 13, 1, 1, 26, 180, 441, 491, 286, 91, 15, 1, 1, 30, 260, 840, 1251, 996, 455, 120, 17, 1, 1, 34, 356, 1464, 2823, 2948, 1814, 680, 153, 19, 1

Offset

0,5

Comments

Generalized Pascal matrix: row sums are generalized Pell numbers A103142 and diagonal sums are the Pentanacci numbers A001591(n+4). One of a family of generalized Pascal triangles given by the Riordan arrays (1/(1-x), x*Sum_{j=0..k} x^k/(1-x)). This array has the 'k+2-nacci' numbers as diagonal sums and generalized Pell numbers b(n) = 2b(n-1) + Sum_{j=1..k} b(n-1-j) as row sums. The first two arrays of the family are Pascal's triangle and the Delannoy number triangle.

Links

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Kuhapatanakul, Kantaphon; Anantakitpaisal, Pornpawee The k-nacci triangle and applications. Cogent Math. 4, Article ID 1333293, 13 p. (2017).

Formula

Triangle, read by rows, where the terms are generated by the rule: T(n, k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k-1) + T(n-3, k-1) + T(n-4, k-1), with T(0, 0)=1.

G.f.: 1/(1-x-x*y*(1+x+x^2+x^3)). - Vladimir Kruchinin, Apr 21 2015

From Werner Schulte, Dec 07 2018, Dec 12 2018, Dec 13 2018: (Start)

G.f. of column k: Sum_{n>=0} T(n+k,k) * x^n = (1+x+x^2+x^3)^k / (1-x)^(k+1) = (1-x^4)^k / (1-x)^(2*k+1).

Let k >= 0 be some fixed integer and a_k(n) be multiplicative with a_k(p^e) = T(e+k,k) for prime p and e >= 0. Then we have the Dirichlet g.f.: Sum{n>0} a_k(n) / n^s = (zeta(s))^(2*k+1) / (zeta(4*s))^k.

T(n,k) = Sum_{i=0..n-k} binomial(n-i,k) * (Sum_{j=0..i} binomial(k,j) * binomial(3*k-2*j,i-j) * (-2)^j) for 0 <= k <= n (conjectured).

T(n,k) = Sum_{i=0..n-k} binomial(n-i,k) * (Sum_{j=0..floor(i/4)} (-1)^j * binomial(k,j) * binomial(k-1+i-4*j,i-4*j)) for 0 <= k <= n.

T(n,k) = Sum_{i=0..n-k} binomial(n-i,k) * (Sum_{j=0..floor(i/2)} binomial(k,j) * binomial(k,i-2*j)) for 0 <= k <= n. (End)

Example

Triangle begins
  1;
  1,  1;
  1,  3,   1;
  1,  6,   5,    1;
  1, 10,  15,    7,    1;
  1, 14,  35,   28,    9,    1;
  1, 18,  68,   84,   45,   11,    1;
  1, 22, 116,  207,  165,   66,   13,   1;
  1, 26, 180,  441,  491,  286,   91,  15,   1;
  1, 30, 260,  840, 1251,  996,  455, 120,  17,  1;
  1, 34, 356, 1464, 2823, 2948, 1814, 680, 153, 19, 1; ...

Mathematica

T[_?Positive, 0] = 1; T[n_, n_] = 1; T[n_, k_] /; 0<k<n := T[n, k] = T[n-1, k] + T[n-1, k-1] + T[n-2, k-1] + T[n-3, k-1] + T[n-4, k-1]; T[_, _] = 0; Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 24 2017 *)

Prog

(PARI) T(n, k)=polcoef(polcoef(1/(1-x-x*y*(1+x+x^2+x^3)) + O(x*x^n), n), k) \\ Andrew Howroyd, Dec 12 2018
(Sage)
def A103141Triangle(dim):
    def B(n): return n if n < 5 else 4
    M = matrix(ZZ, dim, dim)
    for k in (0..dim-1): M[k, 0] = 1
    for k in (1..dim-1):
        for m in (k..dim-1):
            M[m, k] = sum(M[j, k-1]*B(m-j) for j in (k-1..m-1))
    return M
A103141Triangle(11) # Peter Luschny, Dec 22 2018

Crossrefs

Cf. A102036.

Sequence in context: A102036 A121524 A361682 * A085478 A129818 A123970

Adjacent sequences: A103138 A103139 A103140 * A103142 A103143 A103144

Keyword

easy,nonn,tabl

Author

Paul Barry, Jan 24 2005

Status

approved