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

%I A103141 #60 Aug 29 2022 09:46:27
%S A103141 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,
%T A103141 22,116,207,165,66,13,1,1,26,180,441,491,286,91,15,1,1,30,260,840,
%U A103141 1251,996,455,120,17,1,1,34,356,1464,2823,2948,1814,680,153,19,1
%N A103141 Riordan array (1/(1-x), x*(1 + x + x^2 + x^3)/(1-x)).
%C A103141 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.
%H A103141 G. C. Greubel, <a href="/A103141/b103141.txt">Rows n = 0..100 of triangle, flattened</a>
%H A103141 Kuhapatanakul, Kantaphon; Anantakitpaisal, Pornpawee <a href="https://doi.org/10.1080/23311835.2017.1333293">The k-nacci triangle and applications</a>.  Cogent Math. 4, Article ID 1333293, 13 p. (2017).
%F A103141 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.
%F A103141 G.f.: 1/(1-x-x*y*(1+x+x^2+x^3)). - _Vladimir Kruchinin_, Apr 21 2015
%F A103141 From _Werner Schulte_, Dec 07 2018, Dec 12 2018, Dec 13 2018: (Start)
%F A103141 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).
%F A103141 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.
%F A103141 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).
%F A103141 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.
%F A103141 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)
%e A103141 Triangle begins
%e A103141   1;
%e A103141   1,  1;
%e A103141   1,  3,   1;
%e A103141   1,  6,   5,    1;
%e A103141   1, 10,  15,    7,    1;
%e A103141   1, 14,  35,   28,    9,    1;
%e A103141   1, 18,  68,   84,   45,   11,    1;
%e A103141   1, 22, 116,  207,  165,   66,   13,   1;
%e A103141   1, 26, 180,  441,  491,  286,   91,  15,   1;
%e A103141   1, 30, 260,  840, 1251,  996,  455, 120,  17,  1;
%e A103141   1, 34, 356, 1464, 2823, 2948, 1814, 680, 153, 19, 1; ...
%t A103141 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 *)
%o A103141 (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
%o A103141 (Sage)
%o A103141 def A103141Triangle(dim):
%o A103141     def B(n): return n if n < 5 else 4
%o A103141     M = matrix(ZZ, dim, dim)
%o A103141     for k in (0..dim-1): M[k, 0] = 1
%o A103141     for k in (1..dim-1):
%o A103141         for m in (k..dim-1):
%o A103141             M[m, k] = sum(M[j, k-1]*B(m-j) for j in (k-1..m-1))
%o A103141     return M
%o A103141 A103141Triangle(11) # _Peter Luschny_, Dec 22 2018
%Y A103141 Cf. A102036.
%K A103141 easy,nonn,tabl
%O A103141 0,5
%A A103141 _Paul Barry_, Jan 24 2005