

A128932


Define the Fibonacci polynomials by F[1] = 1, F[2] = x; for n > 2, F[n] = x*F[n1] + F[n2] (cf. A049310, A053119). Swamy's inequality implies that F[n] <= F[n]^2 <= G[n] = (x^2 + 1)^2*(x^2 + 2)^(n3) for n >= 3 and x >= 1. The sequence gives a triangle of coefficients of G[n]  F[n] read by rows.


1



0, 0, 1, 0, 1, 2, 2, 5, 1, 4, 0, 1, 3, 0, 9, 0, 12, 0, 6, 0, 1, 8, 3, 28, 4, 38, 1, 25, 0, 8, 0, 1, 15, 0, 58, 0, 99, 0, 87, 0, 41, 0, 10, 0, 1, 32, 4, 144, 10, 272, 6, 280, 1, 170, 0, 61, 0, 12, 0, 1, 63, 0, 310, 0, 673, 0, 825, 0, 619, 0, 292, 0, 85, 0, 14, 0, 1
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OFFSET

3,6


COMMENTS

Swamy's (1966) inequality states that F[n]^2 <= G[n] for all real x and all integers n >= 3. Because F[n] >= 1 for all real x >= 1, we get F[n] <= G[n] for all integers n >= 3 and all real x >= 1.
Row n >= 3 of this irregular table gives the coefficients of the polynomial G[n]  F[n] (with exponents in increasing order). The degree of G[n]  F[n] is 2*n  2, so row n >= 3 contains 2*n  1 terms.
Guilfoyle (1967) notes that F[n] = det(A_n), where A_n is the (n1) X (n1) matrix [[x, 1, 0, 0, ..., 0, 0, 0], [1, x, 1, 0, ..., 0, 0, 0], [0, 1, x, 1, ..., 0, 0, 0], ..., [0, 0, 0, 0, ..., 1, x, 1], [0, 0, 0, 0, ..., 0, 1, x]], and Swamy's original inequality follows from Hadamard's inequality.
Koshy (2019) writes Swamy's original inequality in the form x^(n3)*F[n]^2 <= F[3]^2*F[4]^(n3) for x >= 1, and gives a counterpart inequality for Lucas polynomials. Notice, however, that the original form of Swamy's inequality is true for all real x. (End)


REFERENCES

Thomas Koshy, Fibonacci and Lucas numbers with Applications, Vol. 2, Wiley, 2019; see p. 33. [Vol. 1 was published in 2001.]
D. S. Mitrinovic, Analytic Inequalities, SpringerVerlag, 1970; p. 232, Sect. 3.3.38.


LINKS

M. N. S. Swamy and Joel Pitcain, Comment to Problem E1846, Amer. Math. Monthly, 75(3) (1968), 295. [It is pointed out that I^{n1}*F[n](x) = U_{n1}(I*x/2), where U_{n1}(cos(t)) = sin(n*t)/sin(t) and I = sqrt(1); Cf. A049310 and A053119, but with different notation.]


FORMULA

T(n,0) = 2^(n3)  (1  (1)^n)/2 = A166920(n3) for n >= 3.
Sum_{k=0}^{2*n2} T(n,k) = 4*3^(n3)  Fib(n) = A003946(n2)  A000045(n) for n >= 3. (End)


EXAMPLE

Triangle T(n,k) (with rows n >= 3 and columns k = 0..2*n2) begins:
0, 0, 1, 0, 1;
2, 2, 5, 1, 4, 0, 1;
3, 0, 9, 0, 12, 0, 6, 0, 1;
8, 3, 28, 4, 38, 1, 25, 0, 8, 0, 1;
15, 0, 58, 0, 99, 0, 87, 0, 41, 0, 10, 0, 1;
...


PROG

(PARI) lista(nn) = {my(f=vector(nn)); my(g=vector(nn)); my(h=vector(nn)); f[1]=1; f[2]=x; g[1]=0; g[2]=0; for(n=3, nn, g[n] = (x^2+1)^2*(x^2+2)^(n3)); for(n=3, nn, f[n] = x*f[n1]+f[n2]); for(n=1, nn, h[n] = g[n]f[n]); for(n=3, nn, for(k=0, 2*n2, print1(polcoef(h[n], k, x), ", ")); print(); ); } \\ Petros Hadjicostas, Jun 10 2020


CROSSREFS



KEYWORD

sign,tabf


AUTHOR



EXTENSIONS



STATUS

approved



