|
|
A000324
|
|
A nonlinear recurrence: a(0) = 1, a(1) = 5, a(n) = a(n-1)^2 - 4*a(n-1) + 4 for n>1.
(Formerly M3789 N1544)
|
|
9
|
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004
This is the special case k=4 of sequences with exact mutual k-residues. In general, a(1)=k+1 and a(n)=min{m | m>a(n-1), mod(m,a(i))=k, i=1,...,n-1}. k=1 gives Sylvester's sequence A000058 and k=2 Fermat sequence A000215. - Seppo Mustonen, Sep 04 2005
A000058, A000215, A000289 and this sequence here can be represented as values of polynomials defined via P_0(z)= 1+z, P_{n+1}(z) = z+ prod_{i=0..n} P_i(z), with recurrences P_{n+1}(z) = (P_n(z))^2 -z*P_n(z) +z, n>=0. - Vladimir Shevelev, Dec 08 2010
|
|
REFERENCES
|
Derek Jennings, Some reciprocal summation identities with applications to the Fibonacci and Lucas numbers, in: G. E. Bergum, Applications of Fibonacci Numbers, Vol. 7, Bergum G. E. et al. (eds.), Kluwer Academic Publishers, 1998, pp. 197-200.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
LINKS
|
|
|
FORMULA
|
a(n) = L(2^n)+2, if n>0 where L() is Lucas sequence.
a(n) = Lucas(2^(n-1))^2 for n > 1.
Sum_{n>=1} 4^n/a(n) = 4 (Jennings, 1998; Duverney, 2001). (End)
Product_{n>=1} (1 - 3/a(n)) = 1/4 (Duverney and Kurosawa, 2022). - Amiram Eldar, Jan 07 2023
|
|
MATHEMATICA
|
t = {1, 5}; Do[AppendTo[t, t[[-1]]^2 - 4*t[[-1]] + 4], {n, 11}] (* T. D. Noe, Jun 19 2012 *)
Join[{1}, RecurrenceTable[{a[n] == a[n-1]^2 - 4*a[n-1] + 4, a[1] == 5}, a, {n, 1, 8}]] (* Jean-François Alcover, Feb 07 2016 *)
Join[{1}, NestList[#^2-4#+4&, 5, 10]] (* Harvey P. Dale, Dec 11 2023 *)
|
|
PROG
|
(PARI) a(n)=if(n<2, max(0, 1+4*n), a(n-1)^2-4*a(n-1)+4)
(PARI) a(n)=if(n<1, n==0, n=2^n; fibonacci(n+1)+fibonacci(n-1)+2)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|