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A095666 Pascal (1,4) triangle.

%I A095666 #44 Apr 14 2025 11:34:27
%S A095666 4,1,4,1,5,4,1,6,9,4,1,7,15,13,4,1,8,22,28,17,4,1,9,30,50,45,21,4,1,
%T A095666 10,39,80,95,66,25,4,1,11,49,119,175,161,91,29,4,1,12,60,168,294,336,
%U A095666 252,120,33,4,1,13,72,228,462,630,588,372,153,37,4,1,14,85,300,690,1092
%N A095666 Pascal (1,4) triangle.
%C A095666 This is the fourth member, q=4, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A096940 (q=5), A096956 (q=6).
%C A095666 This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) := Sum_{m=0..n} a(n,m)*x^m is G(z,x) = g(z)/(1 - x*z*f(z)). Here: g(x) = (4-3*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (4-3*z)/(1-(1+x)*z).
%C A095666 The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022095(n-2), n >= 2, with n=1 value 4. [Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.]
%C A095666 T(2*n,n) = A029609(n) for n > 0, A029609 are the central terms of the Pascal (2,3) triangle A029600. - _Reinhard Zumkeller_, Apr 08 2012
%H A095666 Reinhard Zumkeller, <a href="/A095666/b095666.txt">Rows n=0..150 of triangle, flattened</a>
%H A095666 Wolfdieter Lang, <a href="/A095666/a095666.txt">First 10 rows</a>.
%F A095666 Recursion: a(n, m) = 0 if m > n, a(0, 0) = 4; a(n, 0) = 1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
%F A095666 G.f. column m (without leading zeros): (4-3*x)/(1-x)^(m+1), m >= 0.
%F A095666 a(n,k) = (1 + 3*k/n)*binomial(n,k). - _Mircea Merca_, Apr 08 2012
%e A095666 Triangle begins:
%e A095666   [4];
%e A095666   [1,4];
%e A095666   [1,5,4];
%e A095666   [1,6,9,4];
%e A095666   [1,7,15,13,4];
%e A095666   ...
%p A095666 a(n,k):=(1+3*k/n)*binomial(n,k) # _Mircea Merca_, Apr 08 2012
%t A095666 A095666[n_, k_] := If[n == k,  4, (3*k/n + 1)*Binomial[n, k]];
%t A095666 Table[A095666[n, k], {n, 0, 12}, {k, 0, n}] (* _Paolo Xausa_, Apr 14 2025 *)
%o A095666 (Haskell)
%o A095666 a095666 n k = a095666_tabl !! n !! k
%o A095666 a095666_row n = a095666_tabl !! n
%o A095666 a095666_tabl = [4] : iterate
%o A095666    (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,4]
%o A095666 -- _Reinhard Zumkeller_, Apr 08 2012
%Y A095666 Row sums: A020714(n-1), n >= 1, 4 if n=0.
%Y A095666 Alternating row sums are [4, -3, followed by 0's].
%Y A095666 Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+4), A055999(n+1), A060488(n+3), A095667-71, A095819.
%K A095666 nonn,easy,tabl
%O A095666 0,1
%A A095666 _Wolfdieter Lang_, Jun 11 2004