A284637 Discriminants of polynomials having Fibonacci numbers (A000045) for coefficients, P_n(x) = Sum_{k=1..n} F(k)*x^(2n-1-k) + Sum_{k=1..(n-1)} (-1)^k*F(n-k)*x^(n-k-1); a(1) = 1.

1, 5, 900, 2592000, 152587890625, 88060251340800000, 608462684559542896890625, 39491298245528363382865920000000, 24652445390187744298440793976121600000000, 136940866302168849110603332519531250000000000000000

Offset

1,2

Comments

D. H. Lehmer and E. Lehmer showed that the roots of these polynomials can be explicitly given, and that a(n) is divisible by 5^(n-1)*n^(2n-4).

The quotients a(n)/(5^(n-1)*n^(2n-4)) are 1, 1, 4, 81, 15625, 16777216, 137858491849, 7355827511386641, 2758702310349224820736, 7011372354671045074462890625, ...

Links

Amiram Eldar, Table of n, a(n) for n = 1..46

D. H. Lehmer and E. Lehmer, Properties of polynomials having Fibonacci numbers for coefficients, Fibonacci Quarterly, Vol 21, No. 1 (1983), pp. 62-64.

Formula

a(n) ~ 5 * n^(2*n - 4) * phi^(2*n*(n-2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Mar 02 2023

Example

The first 5 polynomials are:
P_1(x) = 1
P_2(x) = x^2 + x - 1
P_3(x) = x^4 + x^3 + 2x^2 - x + 1
P_4(x) = x^6 + x^5 + 2x^4 + 3x^3 - 2x^2 + x - 1
P_5(x) = x^8 + x^7 + 2x^6 + 3x^5 + 5x^4 - 3x^3 + 2x^2 - x + 1
The discriminant of P_2(x), for example, is a(2) = 1^2 - 4*1*(-1) = 5.

Mathematica

a={}; n=0; While[Length[a]<10, n++; f:=Fibonacci[Range[n]]; c = Join[Drop[Reverse[-(-1)^Range[n]]*f, -1], Reverse[f]]; p=x^Range[0, 2n-2].c; d=Discriminant[p, x]; AppendTo[a, d]]; a

Prog

(PARI) a(n) = if (n==1, 1, poldisc(sum(k=1, n, fibonacci(k)*x^(2*n-1-k)) + sum(k=1, n-1, (-1)^k*fibonacci(n-k)*x^(n-k-1)))); \\ Michel Marcus, Mar 02 2023

Crossrefs

Cf. A000045.

Sequence in context: A299837 A299726 A094630 * A173914 A015940 A093853

Adjacent sequences: A284634 A284635 A284636 * A284638 A284639 A284640

Keyword

nonn

Author

Amiram Eldar, Mar 30 2017

Status

approved