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Pascal (1,4) triangle.
17

%I #27 Aug 28 2019 16:04:35

%S 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 10,39,80,95,66,25,4,1,11,49,119,175,161,91,29,4,1,12,60,168,294,336,

%U 252,120,33,4,1,13,72,228,462,630,588,372,153,37,4,1,14,85,300,690,1092

%N Pascal (1,4) triangle.

%C 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.

%C 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 nr. 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 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 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 Reinhard Zumkeller, <a href="/A095666/b095666.txt">Rows n=0..150 of triangle, flattened</a>

%H W. Lang, <a href="/A095666/a095666.txt">First 10 rows</a>.

%F 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 G.f. column m (without leading zeros): (4-3*x)/(1-x)^(m+1), m >= 0.

%F a(n,k) = (1 + 3*k/n)*binomial(n,k). [_Mircea Merca_, Apr 08 2012]

%e [4];

%e [1,4];

%e [1,5,4];

%e [1,6,9,4];

%e [1,7,15,13,4];

%e ...

%p a(n,k):=(1+3*k/n)*binomial(n,k) # _Mircea Merca_, Apr 08 2012

%o (Haskell)

%o a095666 n k = a095666_tabl !! n !! k

%o a095666_row n = a095666_tabl !! n

%o a095666_tabl = [4] : iterate

%o (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,4]

%o -- _Reinhard Zumkeller_, Apr 08 2012

%Y Row sums: A020714(n-1), n >= 1, 4 if n=0.

%Y Alternating row sums are [4, -3, followed by 0's].

%Y 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 nonn,easy,tabl

%O 0,1

%A _Wolfdieter Lang_, Jun 11 2004