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%I #47 Sep 08 2022 08:45:13
%S 3,1,3,1,4,3,1,5,7,3,1,6,12,10,3,1,7,18,22,13,3,1,8,25,40,35,16,3,1,9,
%T 33,65,75,51,19,3,1,10,42,98,140,126,70,22,3,1,11,52,140,238,266,196,
%U 92,25,3,1,12,63,192,378,504,462,288,117,28,3,1,13,75,255,570,882,966,750,405,145,31,3
%N Pascal (1,3) triangle.
%C This is the third member, q=3, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with T(0,0)=2, not 1).
%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} T(n,m)*x^m is G(z,x) = g(z)/(1-x*z*f(z)). Here: g(x) = (3-2*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (3-2*z)/(1-(1+x)*z).
%C The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} T(n-1-k,k) = A000285(n-2), n>=2, with n=1 value 3. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.
%C Central terms: T(2*n,n) = A028329(n) = A100320(n) for n > 0, A028329 are the central terms of triangle A028326. - _Reinhard Zumkeller_, Apr 08 2012
%C Let P be Pascal's triangle, A007318 and R the Riordan array, A097805. Then Pascal triangle (1,q) = ((q-1) * R) + P. Example: Pascal triangle (1,3) = (2 * R) + P. - _Gary W. Adamson_, Sep 12 2015
%H Reinhard Zumkeller, <a href="/A095660/b095660.txt">Rows n=0..150 of triangle, flattened</a>
%H W. Lang, <a href="/A095660/a095660.txt">First 10 rows</a>.
%F Recursion: T(n, m)=0 if m>n, T(0, 0)= 3; T(n, 0)=1 if n>=1; T(n, m) = T(n-1, m) + T(n-1, m-1).
%F G.f. column m (without leading zeros): (3-2*x)/(1-x)^(m+1), m>=0.
%F T(n,k) = (1+2*k/n) * binomial(n,k), for n>0. - _Mircea Merca_, Apr 08 2012
%F Closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - _Boris Putievskiy_, Aug 19 2013
%e Triangle starts:
%e 3;
%e 1, 3;
%e 1, 4, 3;
%e 1, 5, 7, 3;
%e 1, 6, 12, 10, 3;
%e 1, 7, 18, 22, 13, 3;
%e 1, 8, 25, 40, 35, 16, 3;
%e 1, 9, 33, 65, 75, 51, 19, 3;
%e 1, 10, 42, 98, 140, 126, 70, 22, 3;
%e 1, 11, 52, 140, 238, 266, 196, 92, 25, 3;
%e 1, 12, 63, 192, 378, 504, 462, 288, 117, 28, 3;
%e 1, 13, 75, 255, 570, 882, 966, 750, 405, 145, 31, 3;
%p T(n,k):=piecewise(n=0,3,0<n,(1+2*k/n)*binomial(n,k)): # _Mircea Merca_, Apr 08 2012
%t {3}~Join~Table[(1 + 2 k/n) Binomial[n, k], {n, 11}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Sep 14 2015 *)
%o (Haskell)
%o a095660 n k = a095660_tabl !! n !! k
%o a095660_row n = a095660_tabl !! n
%o a095660_tabl = [3] : iterate
%o (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,3]
%o -- _Reinhard Zumkeller_, Apr 08 2012
%o (Magma) A095660:= func< n,k | n eq 0 select 3 else (1+2*k/n)*Binomial(n,k) >;
%o [A095660(n,k): k in [0..n], n in [1..12]]; // _G. C. Greubel_, May 02 2021
%o (Sage)
%o def A095660(n,k): return 3 if n==0 else (1+2*k/n)*binomial(n,k)
%o flatten([[A095660(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, May 02 2021
%Y Row sums: A000079(n+1), n>=1, 3 if n=0. Alternating row sums are [3, -2, followed by 0's].
%Y Column sequences (without leading zeros) give for m=1..9 with n>=0: A000027(n+3), A055998(n+1), A006503(n+1), A095661, A000574, A095662, A095663, A095664, A095665.
%Y Cf. A097805.
%K nonn,easy,tabl
%O 0,1
%A _Wolfdieter Lang_, May 21 2004
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