A213774 Rectangular array: (row n) = b**c, where b(h) = F(h+1), c(h) = 2*n-3+2*h, F=A000045 (Fibonacci numbers), n>=1, h>=1, and ** = convolution.
1, 5, 3, 14, 11, 5, 31, 26, 17, 7, 61, 53, 38, 23, 9, 112, 99, 75, 50, 29, 11, 197, 176, 137, 97, 62, 35, 13, 337, 303, 240, 175, 119, 74, 41, 15, 566, 511, 409, 304, 213, 141, 86, 47, 17, 939, 850, 685, 515, 368, 251, 163, 98, 53, 19, 1545, 1401, 1134
Offset: 1
Examples
Northwest corner (the array is read by falling antidiagonals): 1....5....14...31....61....112 3....11...26...53....99....176 5....17...38...75....137...240 7....23...50...97....175...304 9....29...62...119...213...368 11...35...74...141...251...432
Links
- Clark Kimberling, Antidiagonals n=1..60, flattened
Programs
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Mathematica
b[n_] := Fibonacci[n + 1]; c[n_] := 2 n - 1; t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}] TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]] Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]] r[n_] := Table[t[n, k], {k, 1, 60}] (* A213774 *) Table[t[n, n], {n, 1, 40}] (* A213775 *) s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}] Table[s[n], {n, 1, 50}] (* A213776 *)
Formula
T(n,k) = 3*T(n,k-1)-2*T(n,k-2)-T(n,k-3)+T(n,k-4).
G.f. for row n: f(x)/g(x), where f(x) = x*(2*n - 1 + 2*x - (2*n - 3)*x^2) and g(x) = (1 - x - x^2)*(1 - x )^2.
T(n,k) = 2*n*Fibonacci(k+3) + Lucas(k+3) - 4*(k+n+1). - Ehren Metcalfe, Jul 08 2019
Comments