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A213587 Rectangular array: (row n) = b**c, where b(h) = F(h+1), c(h) = F(n+h), F = A000045 (Fibonacci numbers), n>=1, h>=1, and ** = convolution. 7

%I #14 Sep 08 2022 08:46:02

%S 1,4,2,10,7,3,22,17,11,5,45,37,27,18,8,88,75,59,44,29,13,167,146,120,

%T 96,71,47,21,310,276,234,195,155,115,76,34,566,511,443,380,315,251,

%U 186,123,55,1020,931,821,719,614,510,406,301,199,89,1819,1675,1497,1332,1162,994,825,657,487,322,144

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

%C Principal diagonal: A213588.

%C Antidiagonal sums: A213589.

%C Row 1, (1,2,3,5,...)**(1,2,3,5,...): A004798.

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

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

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

%H Clark Kimberling, <a href="/A213587/b213587.txt">Antidiagonals n = 1..60, flattened</a>

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

%F Columns: T(n,k) = T(n-1,k) + T(n-2,k).

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

%F T{n, k) = (k*Lucas(n+k+2) - Fibonacci(k)*Lucas(n-1))/5. - _G. C. Greubel_, Jul 08 2019

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

%e 1....4....10....22....45....88....167

%e 2....7....17....37....75....146...276

%e 3....11...27....59....120...234...443

%e 5....18...44....96....195...380...719

%e 8....29...71....155...315...614...1162

%e 12...47...115...251...510...994...1881

%t (* First program *)

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

%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}]] (* A213587 *)

%t r[n_]:= Table[T[n, k], {k, 40}] (* columns of antidiagonal triangle *)

%t Table[T[n, n], {n, 1, 40}] (* A213588 *)

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

%t Table[s[n], {n, 1, 50}] (* A213589 *)

%t (* Second program *)

%t Table[((n-k+1)*LucasL[n+3] - Fibonacci[n-k+1]*LucasL[k-1])/5, {n, 12}, {k, n}]//Flatten (* _G. C. Greubel_, Jul 08 2019 *)

%o (PARI) lucas(n) = fibonacci(n+1) + fibonacci(n-1);

%o t(n,k) = ((n-k+1)*lucas(n+3) - fibonacci(n-k+1)*lucas(k-1))/5;

%o for(n=1,12, for(k=1,n, print1(t(n,k), ", "))) \\ _G. C. Greubel_, Jul 08 2019

%o (Magma) [[((n-k+1)*Lucas(n+3) - Fibonacci(n-k+1)*Lucas(k-1))/5: k in [1..n]]: n in [1..12]]; // _G. C. Greubel_, Jul 08 2019

%o (Sage) [[((n-k+1)*lucas_number2(n+3,1,-1) - fibonacci(n-k+1)* lucas_number2(k-1, 1,-1))/5 for k in (1..n)] for n in (1..12)] # _G. C. Greubel_, Jul 08 2019

%o (GAP) Flat( List([1..12], n-> List([1..n], k-> ((n-k+1)*Lucas(1,-1, n+3)[2] - Fibonacci(n-k+1)*Lucas(1,-1,k-1)[2])/5 ))) # _G. C. Greubel_, Jul 08 2019

%Y Cf. A213500.

%K nonn,tabl,easy

%O 1,2

%A _Clark Kimberling_, Jun 19 2012

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