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A127595
a(n) = F(4n) - 2F(2n) where F(n) = Fibonacci numbers A000045.
5
0, 1, 15, 128, 945, 6655, 46080, 317057, 2176335, 14925184, 102320625, 701373311, 4807434240, 32951037313, 225850798095, 1548007091840, 10610205501105, 72723448842367, 498453982018560, 3416454544730369, 23416728143799375
OFFSET
0,3
COMMENTS
a(n) is a divisibility sequence; that is, if h|k then a(h)|a(k).
LINKS
E. L. Roettger and H. C. Williams, Appearance of Primes in Fourth-Order Odd Divisibility Sequences, J. Int. Seq., Vol. 24 (2021), Article 21.7.5.
Hugh Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory vol. 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Odd and even linear divisibility sequences of order 4, INTEGERS, 2015, #A33.
FORMULA
a(n) = F(2n)*(L(2n)-2) = A001906(n)*A004146(n), where L(n) are the Lucas numbers A000032.
a(2n) = 5*(F(2n))^3*L(2n), a(2n+1) = F(2n+1)*L(2n+1)^3.
a(n) = [(Phi^(2n))-1]^2*[(Phi^(4n))-1]/[sqrt(5)*(Phi^(4n))].
G.f.: A(x)=x*(1+(r+2)*x+x^2)/((1-r*x+x^2)*(1-(r^2-2)*x+x^2)) at r=3. The case r=2 is A000578.
a(n) = -a(-n) for all n in Z. - Michael Somos, Dec 30 2022
EXAMPLE
G.f. = x + 15*x^2 + 128*x^3 + 945*x^4 + 6655*x^5 + ... - Michael Somos, Dec 30 2022
MATHEMATICA
With[{r = 3}, CoefficientList[Series[x (1 + (r + 2) x + x^2)/((1 - r x + x^2)*(1 - (r^2 - 2)*x + x^2)), {x, 0, 20}], x]] (* Michael De Vlieger, Nov 09 2021 *)
PROG
(PARI) {a(n) = my(w = quadgen(5)^(2*n)); imag(w^2 - 2*w)}; /* Michael Somos, Dec 30 2022 */
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Peter Bala, Apr 10 2007
STATUS
approved