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A275931 a(n) = F(2*n+3)*F(2*n+1)^3, where F = Fibonacci (A000045). 1
2, 40, 1625, 74698, 3498056, 164257777, 7716095570, 362488657000, 17029226266313, 800010977816986, 37583485579350152, 1765623803357209825, 82946735218100281250, 3896730931076485826728, 183063407022834751530041, 8600083399124733831953002, 404020856351720236303380680 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
The right-hand side of Helmut Postl's identity F(2n+3) + F(n)*F(n+2)^3 = F(n+3)*F(n+1)^3, n even.
LINKS
FORMULA
From Colin Barker, Aug 31 2016: (Start)
a(n) = 55*a(n-1)-385*a(n-2)+385*a(n-3)-55*a(n-4)+a(n-5) for n>4.
G.f.: (2-70*x+195*x^2-47*x^3+x^4) / ((1-x)*(1-47*x+x^2)*(1-7*x+x^2)).
(End)
MATHEMATICA
Table[(Fibonacci[2 n + 3] Fibonacci[2 n + 1]^3), {n, 0, 20}] (* Vincenzo Librandi, Sep 02 2016 *)
LinearRecurrence[{55, -385, 385, -55, 1}, {2, 40, 1625, 74698, 3498056}, 20] (* Harvey P. Dale, Aug 09 2019 *)
PROG
(PARI) Vec((2-70*x+195*x^2-47*x^3+x^4)/((1-x)*(1-47*x+x^2)*(1-7*x+x^2)) + O(x^20)) \\ Colin Barker, Aug 31 2016
(Magma) [Fibonacci(2*n+3)*Fibonacci(2*n+1)^3: n in [0..25]]; // Vincenzo Librandi, Sep 02 2016
CROSSREFS
Cf. A000045.
Sequence in context: A012241 A341589 A286124 * A099707 A198248 A261732
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Aug 31 2016
STATUS
approved

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Last modified July 18 21:02 EDT 2024. Contains 374388 sequences. (Running on oeis4.)