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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
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
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
KEYWORD
nonn
AUTHOR
Amiram Eldar, Mar 30 2017
STATUS
approved