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 A128932 Define the Fibonacci polynomials by F[1] = 1, F[2] = x; for n > 2, F[n] = x*F[n-1] + F[n-2] (cf. A049310, A053119). Swamy's inequality implies that F[n] <= F[n]^2 <= G[n] = (x^2 + 1)^2*(x^2 + 2)^(n-3) 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 3,6 COMMENTS From Petros Hadjicostas, Jun 10 2020: (Start) 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 (n-1) X (n-1) 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^(n-3)*F[n]^2 <= F[3]^2*F[4]^(n-3) 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, Springer-Verlag, 1970; p. 232, Sect. 3.3.38. LINKS Table of n, a(n) for n=3..79. Richard Guilfoyle, Comment to the solution of Problem E1846, Amer. Math. Monthly, 74(5), 1967, 593. [It is pointed out that the inequality is a special case of Hadamard's inequality.] M. N. S. Swamy, Problem E1846 proposed for solution, Amer. Math. Monthly, 73(1) (1966), 81. M. N. S. Swamy and R. E. Giudici, Solution to Problem E1846, Amer. Math. Monthly, 74(5), 1967, 592-593. M. N. S. Swamy and Joel Pitcain, Comment to Problem E1846, Amer. Math. Monthly, 75(3) (1968), 295. [It is pointed out that I^{n-1}*F[n](x) = U_{n-1}(I*x/2), where U_{n-1}(cos(t)) = sin(n*t)/sin(t) and I = sqrt(-1); Cf. A049310 and A053119, but with different notation.] Wikipedia, Fibonacci polynomials. Wikipedia, Hadamard's inequality. FORMULA From Petros Hadjicostas, Jun 10 2020: (Start) T(n,0) = 2^(n-3) - (1 - (-1)^n)/2 = A166920(n-3) for n >= 3. Sum_{k=0}^{2*n-2} T(n,k) = 4*3^(n-3) - Fib(n) = A003946(n-2) - A000045(n) for n >= 3. (End) EXAMPLE Triangle T(n,k) (with rows n >= 3 and columns k = 0..2*n-2) 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)^(n-3)); for(n=3, nn, f[n] = x*f[n-1]+f[n-2]); for(n=1, nn, h[n] = g[n]-f[n]); for(n=3, nn, for(k=0, 2*n-2, print1(polcoef(h[n], k, x), ", ")); print(); ); } \\ Petros Hadjicostas, Jun 10 2020 CROSSREFS Cf. A000045, A003946, A049310, A053119, A166920. Sequence in context: A073690 A079301 A079300 * A286150 A071950 A274847 Adjacent sequences: A128929 A128930 A128931 * A128933 A128934 A128935 KEYWORD sign,tabf AUTHOR N. J. A. Sloane, Apr 28 2007 EXTENSIONS Name edited by Petros Hadjicostas, Jun 10 2020 STATUS approved

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