A290864
Numbers k such that the k-th Fibonacci polynomial evaluated at k is prime.
Original entry on oeis.org
5 is in the sequence because A117715(5,5) = 701 is prime.
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select(t -> isprime(combinat:-fibonacci(t,t)), [2,seq(seq(6*i+j,j=[1,5]),i=0..100)]); # Robert Israel, Aug 13 2017
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Select[Range[100], PrimeQ@ Fibonacci[#, #] &] (* Giovanni Resta, Aug 13 2017 *)
A320570
a(n) = L_n(n), where L_n(x) is the Lucas polynomial.
Original entry on oeis.org
2, 1, 6, 36, 322, 3775, 54758, 946043, 18957314, 432083484, 11035502502, 312119004989, 9682664443202, 326872340718053, 11928306344169798, 467875943531657100, 19629328849962024962, 877095358067166709187, 41583555684469161804998, 2084882704791413248133431
Offset: 0
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[2] cat [(&+[(n/(n-j))*(Factorial(n-j)*n^(n-2*j)/(Factorial(j)*Factorial(n-2*j))): j in [0..Floor(n/2)]]): n in [1..20]]; // G. C. Greubel, Oct 15 2018
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Table[LucasL[n, n], {n, 0, 19}] (* or *)
Table[Round[((n + Sqrt[n^2 + 4])^n + (n - Sqrt[n^2 + 4])^n)/2^n], {n, 0, 19}] (* Round is equivalent to FullSimplify here *)
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for(n=0,20, print1(if(n==0,2, sum(j=0,floor(n/2), (n/(n-j))*((n-j)!*n^(n-2*j)/(j!*(n-2*j)!)))), ", ")) \\ G. C. Greubel, Oct 15 2018
A352798
a(n) = 1/(cf[0;n,n,n,...,n] - cf[0;n,n,...,n]) where the first continued fraction has n+1 terms and the second has n terms.
Original entry on oeis.org
1, -10, 330, -21960, 2551640, -461930274, 120572270007, -42930583856160, 20008932768992430, -11825788272679695050, 8643081649999714376976, -7654102744143874729100040, 8076084821027629176909996013, -10010473694454865001226770534530, 14402393216408406872433735669683370
Offset: 1
a(2) = -10 because the two continued fractions are cf[0;2,2] = 0 + 1/(2 + 1/2) = 2/5 and cf[0;2] = 0 + 1/2 = 1/2 and the reciprocal of their difference is 1/(2/5 - 1/2) = -10.
a(3) = 330 because the two continued fractions are cf[0;3,3,3] = 0 + 1/(3 + 1/(3 + 1/3)) = 10/33 and cf[0;3,3] = 0 + 1/(3 + 1/3) = 3/10, and 1/(10/33 - 3/10) = 330.
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a:= n-> (f-> -(-1)^n*f(n,n)*f(n+1,n))(combinat[fibonacci]):
seq(a(n), n=1..15); # Alois P. Heinz, Jul 06 2022
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a(n) = (-1)^(n+1) * vecprod(Vec(lift(Mod('x,'x^2-n*'x-1)^(n+1)))); \\ Kevin Ryde, Apr 18 2022
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from sympy.ntheory.continued_fraction import continued_fraction_reduce
def A352798(n): return int(1/(continued_fraction_reduce([0]+[n]*n)-continued_fraction_reduce([0]+[n]*(n-1)))) # Chai Wah Wu, Jul 06 2022
A320565
a(n) = ((1 + sqrt(4*n^2 + 1))^n - (1 - sqrt(4*n^2 + 1))^n)/(2^n * sqrt(4*n^2 + 1)).
Original entry on oeis.org
0, 1, 1, 10, 33, 701, 4033, 132301, 1089921, 48460114, 520210801, 29215223489, 386721507745, 26250621340841, 413242502386337, 32899021525375426, 600383148312628737, 54846079150716441949, 1138470675779123657425, 117372939125452004885621
Offset: 0
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[2^(1-n)*(&+[Binomial(n,2*k+1)*(4*n^2+1)^k: k in [0..Floor(n/2)]]): n in [0..20]]; // G. C. Greubel, Oct 15 2018
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a[0] = Limit[n^(n - 1) Fibonacci[n, 1/n], n -> 0]; (* a[0] = 0 *)
a[n_] := a[n] = n^(n - 1) Fibonacci[n, 1/n];
Table[a[n], {n, 0, 19}]
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for(n=0,20, print1( 2^(1-n)*sum(k=0,floor(n/2), binomial(n,2*k+1)*(4*n^2+1)^k) , ", ")) \\ G. C. Greubel, Oct 15 2018
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