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A114197 A Pascal-Fibonacci triangle.

%I A114197 #18 Sep 20 2019 07:33:55
%S A114197 1,1,1,1,2,1,1,3,3,1,1,4,7,4,1,1,5,13,13,5,1,1,6,21,31,21,6,1,1,7,31,
%T A114197 61,61,31,7,1,1,8,43,106,142,106,43,8,1,1,9,57,169,286,286,169,57,9,1,
%U A114197 1,10,73,253,520,659,520,253,73,10,1
%N A114197 A Pascal-Fibonacci triangle.
%C A114197 T(2n,n) is A114198. Row sums are A114199. Row sums of inverse are 0^n.
%H A114197 Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL9/Barry/barry91.html">On Integer-Sequence-Based Constructions of Generalized Pascal Triangles</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
%F A114197 As a number triangle, T(n,k) = Sum_{j=0..n-k} C(n-k, j)C(k, j)F(j);
%F A114197 As a number triangle, T(n,k) = Sum_{j=0..n} C(n-k, n-j)C(k, j-k)F(j-k);
%F A114197 As a number triangle, T(n,k) = Sum_{j=0..n} C(k, j)C(n-k, n-j)F(k-j) if k <= n, 0 otherwise.
%F A114197 As a square array, T(n,k) = Sum_{j=0..n} C(n, j)C(k, j)F(j);
%F A114197 As a square array, T(n,k) = Sum_{j=0..n+k} C(n, n+k-j)C(k, j-k)F(j-k);
%F A114197 Column k has g.f.: (Sum_{j=0..k} C(k, j)F(j+1)(x/(1-x))^j)*x^k/(1-x);
%F A114197 G.f.: -((x^2-x)*y-x+1)/((x^4+x^3-x^2)*y^2+(x^3-3*x^2+2*x)*y-x^2+2*x-1). - _Vladimir Kruchinin_, Jan 15 2018
%e A114197 Triangle begins
%e A114197   1;
%e A114197   1,   1;
%e A114197   1,   2,   1;
%e A114197   1,   3,   3,   1;
%e A114197   1,   4,   7,   4,   1;
%e A114197   1,   5,  13,  13,   5,   1;
%e A114197   1,   6,  21,  31,  21,   6,   1;
%e A114197   1,   7,  31,  61,  61,  31,   7,   1;
%e A114197   1,   8,  43, 106, 142, 106,  43,   8,   1;
%Y A114197 Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A114197, A162741, A228074.
%K A114197 easy,nonn,tabl
%O A114197 0,5
%A A114197 _Paul Barry_, Nov 16 2005