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%I
%S 1,5,900,2592000,152587890625,88060251340800000,
%T 608462684559542896890625,39491298245528363382865920000000,
%U 24652445390187744298440793976121600000000,136940866302168849110603332519531250000000000000000
%N 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.
%C 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).
%C The quotients a(n)/(5^(n-1)*n^(2n-4)) are 1, 1, 4, 81, 15625, 16777216, 137858491849, 7355827511386641, 2758702310349224820736, 7011372354671045074462890625, ...
%H Amiram Eldar, <a href="/A284637/b284637.txt">Table of n, a(n) for n = 1..46</a>
%H D. H. Lehmer and E. Lehmer, <a href="http://www.fq.math.ca/Scanned/21-1/lehmer.pdf">Properties of polynomials having Fibonacci numbers for coefficients</a>, Fibonacci Quarterly, Vol 21, No. 1 (1983), pp. 62-64.
%F a(n) ~ 5 * n^(2*n - 4) * phi^(2*n*(n-2)), where phi = A001622 is the golden ratio. - _Vaclav Kotesovec_, Mar 02 2023
%e The first 5 polynomials are:
%e P_1(x) = 1
%e P_2(x) = x^2 + x - 1
%e P_3(x) = x^4 + x^3 + 2x^2 - x + 1
%e P_4(x) = x^6 + x^5 + 2x^4 + 3x^3 - 2x^2 + x - 1
%e P_5(x) = x^8 + x^7 + 2x^6 + 3x^5 + 5x^4 - 3x^3 + 2x^2 - x + 1
%e The discriminant of P_2(x), for example, is a(2) = 1^2 - 4*1*(-1) = 5.
%t 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
%o (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
%Y Cf. A000045.
%K nonn
%O 1,2
%A _Amiram Eldar_, Mar 30 2017
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