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%I #58 Feb 28 2024 08:10:36
%S 1,1,2,2,3,3,3,8,6,4,5,15,20,10,5,8,30,45,40,15,6,13,56,105,105,70,21,
%T 7,21,104,224,280,210,112,28,8,34,189,468,672,630,378,168,36,9,55,340,
%U 945,1560,1680,1260,630,240,45,10,89,605,1870,3465,4290,3696,2310,990,330,55,11
%N Triangular array read by rows: T(n,k) = Fibonacci(n+1-k)*C(n,k-1), k = 1..n; n >= 1.
%C Row sums yield the even-subscripted Fibonacci numbers (A001906).
%C Row n shows the coefficients of the numerator of the n-th derivative of c(n)/(x^2+x-1), where c(n) = ((-1)^(n + 1))/n!; see the Mathematica program. - _Clark Kimberling_, Oct 22 2019
%H G. C. Greubel, <a href="/A094440/b094440.txt">Rows n = 1..100 of triangle, flattened</a>
%H M. Norfleet, <a href="https://www.fq.math.ca/Papers1/43-2/paper43-2-12.pdf">Characterization of second-order strong divisibility sequences of polynomials</a>, The Fibonacci Quarterly, 43(2) (2005), 166-169.
%F From _Peter Bala_, Aug 17 2007: (Start)
%F With an offset of 0, the row polynomials F(n,x) = Sum_{k = 0..n} C(n,k)* Fibonacci(n-k)*x^k satisfy F(n,x)*L(n,x) = F(2*n,x), where L(n,x) = Sum_{k = 0..n} C(n,k)*Lucas(n-k)*x^k.
%F Other identities and formulas include:
%F F(n+1,x)^2 - F(n,x)*F(n+2,x) = (x^2 + x - 1)^n;
%F Sum_{k = 0..n} C(n,k)*F(n-k,x)*L(k,x) = (2^n)*F(n,x);
%F F(n,2*x) = Sum_{k = 0..n} C(n,k)*F(n-k,x)*x^k;
%F F(n,3*x) = Sum_{k = 0..n} C(n,k)*F(n-k,2*x)*x^k, etc.
%F The sequence {F(n,r)}n>=1 gives the r-th binomial transform of the Fibonacci numbers: r = 1 gives A001906, r = 2 gives A030191, r = 3 gives A099453, r = 4 gives A081574, r = 5 gives A081575.
%F F(n,1/phi) = (-1)^(n-1)*F(n,-phi) = sqrt(5)^(n-1) for n >= 1, where phi = (1 + sqrt(5))/2.
%F The polynomials F(n,-x) satisfy a Riemann hypothesis: the zeros of F(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
%F G.f.: t/(1 - (2*x + 1)*t + (x^2 + x - 1)*t^2) = t + (1 + 2*x)*t^2 + (2 + 3*x + 3*x^2)*t^3 + (3 + 8*x + 6*x^2 + 4*x^3)*t^4 + ... . (End)
%F From _Peter Bala_, Jun 29 2016: (Start)
%F Working with an offset of 0, the n-th row polynomial F(n,x) = 1/sqrt(5)*( (x + phi)^n - (x - 1/phi)^n ), where phi = (1 + sqrt(5))/2.
%F d/dx(F(n,x)) = n*F(n-1,x).
%F F(-n,x) = -F(n,x)/(x^2 + x - 1)^n.
%F F(n,x - 1) = (-1)^(n-1)*F(n,-x).
%F F(n,x) is a divisibility sequence of polynomials, that is, if n divides m then F(n,x) divides F(m,x) in the polynomial ring Z[x]. (End)
%F From _G. C. Greubel_, Oct 30 2019: (Start)
%F Sum_{k = 1..n} T(n,k) = Fibonacci(2*n).
%F Sum_{k = 1..n} (-1)^k * T(n,k) = (-1)^n * Fibonacci(n). (End)
%F From _Clark Kimberling_, Oct 30 2019: (Start)
%F F(n,x) is a strong divisibility sequence of polynomials in Z[x]; that is,
%F gcd(F(x,h),F(x,k)) = F(x,gcd(h,k)) for h,k >= 1. Thus, if x is an integer, then F(n,x) is a strong divisibility sequence of integers; e.g., for x=3, we have A099453. (End)
%F Let p(n) denote the polynomial F(x,n). Then p(n) = k(b^n - c^n), where k = -1/sqrt(5), b = (1/2)(2x + 1 - sqrt(5)), c = (1/2)(2x + 1 + sqrt(5)), and for n >=3, p(n) = u*p(n - 1) + v*p(n - 2), where u = 1 + 2 x, v = 1 - x - x^2. - _Clark Kimberling_, Nov 11 2023
%e Triangle starts:
%e 1;
%e 1, 2;
%e 2, 3, 3;
%e 3, 8, 6, 4;
%e 5, 15, 20, 10, 5;
%e 8, 30, 45, 40, 15, 6;
%e 13, 56, 105, 105, 70, 21, 7;
%e ...
%e T(4,3) = F(2)*C(4,2) = 1*6 = 6.
%p with(combinat): T:=(n,k)->binomial(n,k-1)*fibonacci(n+1-k): for n from 1 to 11 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form # _Emeric Deutsch_
%t Table[Fibonacci[n+1-k]Binomial[n,k-1],{n,20},{k,n}]//Flatten (* _Harvey P. Dale_, Sep 14 2016 *)
%t (* Next program outputs polynomials having coefficients T(n,k) *)
%t g[x_, n_] := Numerator[(-1)^(n + 1) Factor[D[1/(1 - x - x^2), {x, n}]]]
%t Column[Expand[Table[g[x, n]/n!, {n, 0, 12}]]] (* _Clark Kimberling_, Oct 22 2019 *)
%o (Magma) /* As triangle */ [[Fibonacci(n+1-k)*Binomial(n,k-1): k in [1..n]]: n in [1.. 15]]; // _Vincenzo Librandi_, Aug 15 2017
%o (PARI) T(n,k) = binomial(n,k-1)*fibonacci(n-k+1);
%o for(n=1,12, for(k=1,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Oct 30 2019
%o (Sage) [[binomial(n,k-1)*fibonacci(n-k+1) for k in (1..n)] for n in (1..12)] # _G. C. Greubel_, Oct 30 2019
%o (GAP) Flat(List([1..12], n-> List([1..n], k-> Binomial(n,k-1)* Fibonacci(n-k+1) ))); # _G. C. Greubel_, Oct 30 2019
%Y Cf. A000032, A000045, A001906, A030191, A081574, A094435, A132148.
%Y Cf. A094435, A094436, A094437, A094438, A094439, A094441, A094442, A094443, A094444.
%Y Cf. A367208, A367209, A367210, A367211.
%K nonn,tabl
%O 1,3
%A _Clark Kimberling_, May 03 2004
%E Error in expansion of generating function corrected by _Peter Bala_, Sep 24 2008
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