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Rectangular array: (row n) = b**c, where b(h) = 2*n-1, c(h) = F(n-1+h), F=A000045 (Fibonacci numbers), n>=1, h>=1, and ** = convolution.
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%I #18 Jul 10 2019 11:16:53

%S 1,4,1,10,5,2,21,14,9,3,40,31,24,14,5,72,61,52,38,23,8,125,112,101,83,

%T 62,37,13,212,197,184,162,135,100,60,21,354,337,322,296,263,218,162,

%U 97,34,585,566,549,519,480,425,353,262,157,55,960,939,920,886

%N Rectangular array: (row n) = b**c, where b(h) = 2*n-1, c(h) = F(n-1+h), F=A000045 (Fibonacci numbers), n>=1, h>=1, and ** = convolution.

%C Principal diagonal: A213766.

%C Antidiagonal sums: A213767.

%C Row 1, (1,3,5,7,9,...)**(1,1,2,3,5,...): A001891.

%C Row 2, (1,3,5,7,9,...)**(1,2,3,5,8,...): A023652.

%C Row 3, (1,3,5,7,9,...)**(2,3,5,8,13,...).

%C For a guide to related arrays, see A213500.

%H Clark Kimberling, <a href="/A213765/b213765.txt">Antidiagonals n = 1..60, flattened</a>

%F T(n,k) = 3*T(n,k-1)-2*T(n,k-2)-T(n,k-3)+T(n,k-4).

%F G.f. for row n: f(x)/g(x), where f(x) = x*(F(n) + F(n+1)*x - F(n-1)*x^2) and g(x) = (1 - x - x^2)(1 - x )^2.

%F T(n,k) = F(n+k+4) - 2*k*F(n+1) - F(n+4), F = A000045. - _Ehren Metcalfe_, Jul 10 2019

%e Northwest corner (the array is read by falling antidiagonals):

%e 1....4....10....21....40....72

%e 1....5....14....31....61....112

%e 2....9....24....52....101...184

%e 3....14...38....83....162...296

%e 5....23...62....135...263...480

%e 8....37...100...218...425...776

%e 13...60...162...353...688...1256

%t b[n_] := 2 n - 1; c[n_] := Fibonacci[n];

%t t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]

%t TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]

%t Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]

%t r[n_] := Table[t[n, k], {k, 1, 60}] (* A213765 *)

%t Table[t[n, n], {n, 1, 40}] (* A213766 *)

%t s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]

%t Table[s[n], {n, 1, 50}] (* A213767 *)

%Y Cf. A213500.

%K nonn,tabl,easy

%O 1,2

%A _Clark Kimberling_, Jun 21 2012