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A096940 Pascal (1,5) triangle.

%I A096940 #41 Apr 08 2025 08:46:02
%S A096940 5,1,5,1,6,5,1,7,11,5,1,8,18,16,5,1,9,26,34,21,5,1,10,35,60,55,26,5,1,
%T A096940 11,45,95,115,81,31,5,1,12,56,140,210,196,112,36,5,1,13,68,196,350,
%U A096940 406,308,148,41,5,1,14,81,264,546,756,714,456,189,46,5,1,15,95,345,810,1302
%N A096940 Pascal (1,5) triangle.
%C A096940 This is the fifth member, q=5, 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), A095666 (q=4), A096956 (q=6).
%C A096940 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)=(5-4*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(5-4*z)/(1-(1+x)*z).
%C A096940 The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022096(n-2), n>=2, with n=1 value 5. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.
%H A096940 David A. Corneth, <a href="/A096940/b096940.txt">Table of n, a(n) for n = 0..9999</a>
%H A096940 Wolfdieter Lang, <a href="/A096940/a096940.txt">First 10 rows</a>.
%F A096940 Recursion: a(n, m)=0 if m>n, a(0, 0)= 5; a(n, 0)=1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
%F A096940 G.f. column m (without leading zeros): (5-4*x)/(1-x)^(m+1), m>=0.
%F A096940 a(n,k) = (1+4*k/n)*binomial(n,k), for n>0. - _Mircea Merca_, Apr 08 2012
%e A096940 Triangle begins:
%e A096940   5;
%e A096940   1,  5;
%e A096940   1,  6,  5;
%e A096940   1,  7, 11,   5;
%e A096940   1,  8, 18,  16,   5;
%e A096940   1,  9, 26,  34,  21,   5;
%e A096940   1, 10, 35,  60,  55,  26,   5;
%e A096940   1, 11, 45,  95, 115,  81,  31,   5;
%e A096940   1, 12, 56, 140, 210, 196, 112,  36,   5;
%e A096940   1, 13, 68, 196, 350, 406, 308, 148,  41,  5;
%e A096940   1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5; etc.
%p A096940 a(n,k):=piecewise(n=0,5,0<n,(1+4*k/n)*binomial(n,k)) # _Mircea Merca_, Apr 08 2012
%o A096940 (PARI) a(n) = {if(n <= 1, return(5 - 4*(n==1))); my(m = (sqrtint(8*n + 1) - 1)\2, t = n - binomial(m + 1, 2)); (1+4*t/m)*binomial(m,t)} \\ _David A. Corneth_, Aug 28 2019
%Y A096940 Row sums: A007283(n-1), n>=1, 5 if n=0; g.f.: (5-4*x)/(1-2*x). Alternating row sums are [5, -4, followed by 0's].
%Y A096940 Column sequences (without leading zeros) give for m=1..9, with n>=0: A000027(n+5), A056000(n-1), A096941-7.
%K A096940 nonn,easy,tabl
%O A096940 0,1
%A A096940 _Wolfdieter Lang_, Jul 16 2004