A003482 a(n) = 7*a(n-1) - a(n-2) + 4, with a(0) = 0, a(1) = 5.
0, 5, 39, 272, 1869, 12815, 87840, 602069, 4126647, 28284464, 193864605, 1328767775, 9107509824, 62423800997, 427859097159, 2932589879120, 20100270056685, 137769300517679, 944284833567072, 6472224534451829, 44361286907595735, 304056783818718320
Offset: 0
Examples
G.f. = 5*x + 39*x^2 + 272*x^3 + 1869*x^4 + 12815*x^5 + 87840*x^6 + ... - _Michael Somos_, Jun 26 2018
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Heiko Harborth, Fermat-like binomial equations, Applications of Fibonacci numbers, Proc. 2nd Int. Conf., San Jose, California, August 1986, 1-5 (1988).
- Sébastien Labbé and Jana Lepšová, A Fibonacci's complement numeration system, arXiv:2205.02574 [cs.FL], 2022.
- Sébastien Labbé and Jana Lepšová, A Fibonacci analogue of the two's complement numeration system, RAIRO-Theor. Inf. Appl. (2023) Vol. 57, No. 12. See p. 16.
- D. A. Lind, The quadratic field Q(sqrt(5)) and a certain diophantine equation, Fibonacci Quart. 6(3) (1968), 86-93.
- Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
- Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
- John Riordan and N. J. A. Sloane, Correspondence, 1974
- David Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quart. 13 (1973), 295-298.
- S. M. Tanny and M. Zuker, On a unimodal sequence of binomial coefficients, Discrete Math. 9 (1974), 79-89.
- Index entries for linear recurrences with constant coefficients, signature (8,-8,1).
Programs
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Maple
A003482:=z*(-5+z)/(z-1)/(z**2-7*z+1); # conjectured by Simon Plouffe in his 1992 dissertation
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Mathematica
LinearRecurrence[{8,-8,1},{0,5,39},30] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *) nxt[{a_,b_}]:={b,7b-a+4}; NestList[nxt,{0,5},30][[;;,1]] (* Harvey P. Dale, Jun 09 2025 *)
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PARI
a(n)=fibonacci(2*n)*fibonacci(2*n+3) \\ Charles R Greathouse IV, May 29 2013
Formula
a(n) = Fibonacci(2*n) * Fibonacci(2*n+3).
a(n) = Fibonacci(2*n+2)^2 - Fibonacci(2*n+1)^2. - Gary Detlefs, Oct 12 2011
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3). - Vladimir Joseph Stephan Orlovsky and Vincenzo Librandi, Jan 22 2012
a(n) = -4/5 + (sqrt(5)/5 + 2/5)*(7/2 + 3*sqrt(5)/2)^n - (sqrt(5)/5 - 2/5)*(7/2 - 3*sqrt(5)/2)^n. - Antonio Alberto Olivares, May 29 2013
a(n) = -A206351(-n) for all n in Z. - Michael Somos, Jun 26 2018
From Sébastien Labbé, May 06 2022: (Start)
a(n) = Sum_{k=2..2*n+1} Fibonacci(k)^2.
a(n) = A001654(2*n+1)-1. (End)
Comments