A213768 Rectangular array: (row n) = b**c, where b(h) = F(h), c(h) = 2*n-3+2*h, F=A000045 (Fibonacci numbers), n>=1, h>=1, and ** = convolution.
1, 4, 3, 10, 8, 5, 21, 18, 12, 7, 40, 35, 26, 16, 9, 72, 64, 49, 34, 20, 11, 125, 112, 88, 63, 42, 24, 13, 212, 191, 152, 112, 77, 50, 28, 15, 354, 320, 257, 192, 136, 91, 58, 32, 17, 585, 530, 428, 323, 232, 160, 105, 66, 36, 19
Offset: 1
Examples
Northwest corner (the array is read by falling antidiagonals): 1....4....10...21...40....72....125 3....8....18...35...64....112...191 5....12...26...49...88....152...257 7....16...34...63...112...192...323 9....20...42...77...136...232...389 11...24...50...91...160...272...455
Links
- Clark Kimberling, Antidiagonals n=1..80, flattened
Programs
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Mathematica
b[n_] := Fibonacci[n]; 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}] (* A213768 *) Table[t[n, n], {n, 1, 40}] (* A213769 *) s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}] Table[s[n], {n, 1, 50}] (* A213770 *)
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*n - 3)*x) and g(x) = (1 - x - x^2)(1 - x )^2.
T(n,k) = 2*n*Fibonacci(k+2) + Lucas(k+2) - 2*(k+n) - 3. - Ehren Metcalfe, Jul 08 2019
Comments