A206351 a(n) = 7*a(n-1) - a(n-2) - 4 with a(1)=1, a(2)=3.
1, 3, 16, 105, 715, 4896, 33553, 229971, 1576240, 10803705, 74049691, 507544128, 3478759201, 23843770275, 163427632720, 1120149658761, 7677619978603, 52623190191456, 360684711361585, 2472169789339635
Offset: 1
Examples
G.f. = x + 3*x^2 + 16*x^3 + 105*x^4 + 715*x^5 + 4896*x^6 + 33553*x^7 + ... - _Michael Somos_, Jun 26 2018
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
- Pridon Davlianidze, Problem B-1279, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 4 (2020), p. 368; An Unusual Generalization, Solution to Problem B-1279 by J. N. Senadheera, ibid., Vol. 59, No. 4 (2021), pp. 370-371.
- Index entries for linear recurrences with constant coefficients, signature (8,-8,1).
Programs
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Haskell
a206351 n = a206351_list !! (n-1) a206351_list = 1 : 3 : map (subtract 4) (zipWith (-) (map (* 7) (tail a206351_list)) a206351_list) -- Reinhard Zumkeller, Feb 08 2012
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Magma
[Fibonacci(2*n)*Fibonacci(2*n-3): n in [1..30]]; // G. C. Greubel, Aug 12 2018
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Maple
genZ := proc(n) local start; option remember; start := [1, 3]; if n < 3 then start[n] else 7*genZ(n - 1) - genZ(n - 2) - 4 end if end proc: seq(genZ(n),n=1..20);
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Mathematica
LinearRecurrence[{8, -8, 1}, {1, 3, 16}, 50] (* Charles R Greathouse IV, Feb 07 2012 *) RecurrenceTable[{a[1] == 1, a[2] == 3, a[n] == 7 a[n - 1] - a[n - 2] - 4}, a, {n, 20}] (* Bruno Berselli, Feb 07 2012 *) a[ n_] := Fibonacci[2 n] Fibonacci[2 n - 3]; (* Michael Somos, Jun 26 2018 *) nxt[{a_,b_}]:={b,7b-a-4}; NestList[nxt,{1,3},20][[;;,1]] (* Harvey P. Dale, Aug 29 2024 *)
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PARI
Vec((1-5*x)/(1-8*x+8*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Feb 07 2012
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PARI
{a(n) = fibonacci(2*n) * fibonacci(2*n - 3)}; /* Michael Somos, Jun 26 2018 */
Formula
From Bruno Berselli, Feb 07 2012: (Start)
G.f.: x*(1-5*x)/(1-8*x+8*x^2-x^3).
a(n) = A081018(n-1) + 1. (End)
a(n) = -A003482(-n) = Fibonacci(2*n)*Fibonacci(2*n-3). - Michael Somos, Jun 26 2018
a(n) = A089508(n-1) + 2 for n>1. - Bruno Berselli, Jun 20 2019 [Formula found by Umberto Cerruti]
Product_{n>=2} (1 - 1/a(n)) = 1/phi (A094214) (Davlianidze, 2020). - Amiram Eldar, Nov 30 2021
a(n) = (Fibonacci(2*n-2) + 1/Lucas(2*n-2))*(Fibonacci(2*n-1) + 1/Lucas(2*n-1)). - Peter Bala, Sep 03 2022
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