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%I #16 Jun 01 2023 18:23:04
%S 1,1,0,1,1,0,1,2,1,0,1,3,2,1,0,1,4,5,3,1,0,1,5,9,5,4,1,0,1,6,14,14,9,
%T 5,1,0,1,7,20,28,14,14,6,1,0,1,8,27,48,42,28,20,7,1,0,1,9,35,75,90,42,
%U 48,27,8,1,0,1,10,44,110,165,132,90,75,35,9,1,0,1,11,54,154,275,297,132,165,110,44,10,1,0
%N Pendular Catalan triangle, read by rows, where row n is formed from row n-1 by the recurrence: if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else T(n,k) = T(n,n-1-k) - T(n-1,k) - T(n-1,k+1), for n>=k>=0, with T(n,0)=1 and T(n,n)=0^n.
%C See A118340 for definition of pendular triangles and pendular sums.
%H G. C. Greubel, <a href="/A118344/b118344.txt">Rows n = 0..50 of the triangle, flattened</a>
%F T(2*n+m, n) = [A000108^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A000108.
%F Sum_{k=0..n} T(n,k) = (1/2)*[n=0] + A026010(n-1) = (1/2)*[n=0] + (1/2)^((5 + (-1)^n)/2)*(6*n + 1 + 3*(-1)^n)*Catalan((2*n - 1 + (-1)^n)/4). - _G. C. Greubel_, Mar 17 2021
%e Row 6 equals the pendular sums of row 5:
%e [1, 4, 5, 3, 1, 0], where the sums proceed as follows:
%e [1, __, __, __, __, __]: T(6,0) = T(5,0) = 1;
%e [1, __, __, __, __, 1]: T(6,5) = T(6,0) - T(5,5) = 1 - 0 = 1;
%e [1, 5, __, __, __, 1]: T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;
%e [1, 5, __, __, 4, 1]: T(6,4) = T(6,1) - T(5,4) - T(5,5) = 5-1-0 = 4;
%e [1, 5, 9, __, 4, 1]: T(6,2) = T(6,4) + T(5,2) = 4 + 5 = 9;
%e [1, 5, 9, 5, 4, 1]: T(6,3) = T(6,2) - T(5,3) - T(5,4) = 9-3-1 = 5;
%e [1, 5, 9, 5, 4, 1, 0] finally, append a zero to obtain row 6.
%e Triangle begins:
%e 1;
%e 1, 0;
%e 1, 1, 0;
%e 1, 2, 1, 0;
%e 1, 3, 2, 1, 0;
%e 1, 4, 5, 3, 1, 0;
%e 1, 5, 9, 5, 4, 1, 0;
%e 1, 6, 14, 14, 9, 5, 1, 0;
%e 1, 7, 20, 28, 14, 14, 6, 1, 0;
%e 1, 8, 27, 48, 42, 28, 20, 7, 1, 0;
%e 1, 9, 35, 75, 90, 42, 48, 27, 8, 1, 0;
%e 1, 10, 44, 110, 165, 132, 90, 75, 35, 9, 1, 0;
%e 1, 11, 54, 154, 275, 297, 132, 165, 110, 44, 10, 1, 0;
%e Central terms are Catalan numbers T(2*n,n) = A000108(n);
%e semi-diagonals form successive self-convolutions of the central terms:
%e T(2*n+1,n) = [A000108^2](n),
%e T(2*n+2,n) = [A000108^3](n).
%p T:= proc(n, k) option remember;
%p if k<0 or k>n then 0;
%p elif k=0 then 1;
%p elif k=n then 0;
%p elif n>2*k then T(n, n-k) +T(n-1, k);
%p else T(n, n-k-1) -T(n-1, k) -T(n-1, k+1);
%p fi; end:
%p seq(seq(T(n, k), k=0..n), n=0..12); # _G. C. Greubel_, Mar 17 2021
%t T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, 1, If[k==n, 0, If[n>2*k, T[n, n-k] +T[n-1, k], T[n, n-k-1] -T[n-1, k] -T[n-1, k+1] ]]]];
%t Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* _G. C. Greubel_, Mar 17 2021 *)
%o (PARI) T(n,k)=if(n<k || k<0,0,if(k==0,1,if(n==k,0, if(n>2*k,T(n,n-k)+T(n-1,k),T(n,n-1-k)-T(n-1,k)-if(n-1>k,T(n-1,k+1)) ))))
%o (Sage)
%o @CachedFunction
%o def T(n, k):
%o if (k<0 or k>n): return 0
%o elif (k==0): return 1
%o elif (k==n): return 0
%o elif (n>2*k): return T(n, n-k) +T(n-1, k)
%o else: return T(n, n-k-1) -T(n-1, k) -T(n-1, k+1)
%o flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Mar 17 2021
%Y Cf. A000108, A033184, A118340, A026010 (row sums shift left).
%K nonn,tabl
%O 0,8
%A _Paul D. Hanna_, Apr 26 2006
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