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Square array of binomial transforms of Fibonacci numbers, read by ascending antidiagonals.
2

%I #19 Sep 08 2022 08:45:09

%S 1,1,1,1,2,2,1,3,5,3,1,4,10,13,5,1,5,17,35,34,8,1,6,26,75,125,89,13,1,

%T 7,37,139,338,450,233,21,1,8,50,233,757,1541,1625,610,34,1,9,65,363,

%U 1490,4172,7069,5875,1597,55,1,10,82,535,2669,9633,23165,32532,21250,4181,89

%N Square array of binomial transforms of Fibonacci numbers, read by ascending antidiagonals.

%C Array rows are solutions of the recurrence a(n) = (2*k+1)*a(n-1) - A028387(k-1)*a(n-2), where a(0) = 1 and a(1) = k+1.

%H G. C. Greubel, <a href="/A081572/b081572.txt">Antidiagonal rows n = 0..50, flattened</a>

%F Rows are successive binomial transforms of F(n+1).

%F T(n, k) = ((5+sqrt(5))/10)*( (2*n + 1 + sqrt(5))/2)^k + ((5-sqrt(5)/10)*( 2*n + 1 - sqrt(5))/2 )^k.

%F From _G. C. Greubel_, May 27 2021: (Start)

%F T(n, k) = Sum_{j=0..k} binomial(k,j)*n^(k-j)*Fibonacci(j+1) (square array).

%F T(n, k) = Sum_{j=0..k} binomial(k,j)*(n-k)^(k-j)*Fibonacci(j+1) (antidiagonal triangle). (End)

%e The array rows begins as:

%e 1, 1, 2, 3, 5, 8, 13, ... A000045;

%e 1, 2, 5, 13, 34, 89, 233, ... A001519;

%e 1, 3, 10, 35, 125, 450, 1625, ... A081567;

%e 1, 4, 17, 75, 338, 1541, 7069, ... A081568;

%e 1, 5, 26, 139, 757, 4172, 23165, ... A081569;

%e 1, 6, 37, 233, 1490, 9633, 62753, ... A081570;

%e 1, 7, 50, 363, 2669, 19814, 148153, ... A081571;

%e Antidiagonal triangle begins as:

%e 1;

%e 1, 1;

%e 1, 2, 2;

%e 1, 3, 5, 3;

%e 1, 4, 10, 13, 5;

%e 1, 5, 17, 35, 34, 8;

%e 1, 6, 26, 75, 125, 89, 13;

%e 1, 7, 37, 139, 338, 450, 233, 21;

%e 1, 8, 50, 233, 757, 1541, 1625, 610, 34;

%t T[n_, k_]:= If[n==0, Fibonacci[k+1], Sum[Binomial[k, j]*Fibonacci[j+1]*n^(k-j), {j, 0, k}]]; Table[T[n-k, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, May 26 2021 *)

%o (Magma)

%o A081572:= func< n,k | (&+[Binomial(k,j)*Fibonacci(j+1)*(n-k)^(k-j): j in [0..k]]) >;

%o [A081572(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, May 27 2021

%o (Sage)

%o def A081572(n,k): return sum( binomial(k,j)*fibonacci(j+1)*(n-k)^(k-j) for j in (0..k) )

%o flatten([[A081572(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, May 27 2021

%Y Array row n: A000045 (n=0), A001519 (n=1), A081567 (n=2), A081568 (n=3), A081569 (n=4), A081570 (n=5), A081571 (n=6).

%Y Array column k: A000027 (k=1), A002522 (k=2).

%Y Different from A073133.

%Y Cf. A028387.

%K easy,nonn,tabl

%O 0,5

%A _Paul Barry_, Mar 22 2003