A000324 A nonlinear recurrence: a(0) = 1, a(1) = 5, a(n) = a(n-1)^2 - 4*a(n-1) + 4 for n>1.

1, 5, 9, 49, 2209, 4870849, 23725150497409, 562882766124611619513723649, 316837008400094222150776738483768236006420971486980609

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

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

• T. D. Noe, Table of n, a(n) for n = 0..12
• A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437.
• A. V. Aho and N. J. A. Sloane, Some doubly exponential sequences, Fibonacci Quarterly, Vol. 11, No. 4 (1973), pp. 429-437 (original plus references that F.Q. forgot to include - see last page!)
• Daniel Duverney, Irrationality of Fast Converging Series of Rational Numbers, Journal of Mathematical Sciences-University of Tokyo, Vol. 8, No. 2 (2001), pp. 275-316.
• Daniel Duverney and Takeshi Kurosawa, Transcendence of infinite products involving Fibonacci and Lucas numbers, Research in Number Theory, Vol. 8 (2002), Article 68.
• Solomon W. Golomb, On certain nonlinear recurring sequences, Amer. Math. Monthly, Vol. 70, No. 4 (1963), 403-405.
• Romeo Meštrović, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012. - _N. J. A. Sloane_, Jun 13 2012
• Seppo Mustonen, On integer sequences with mutual k-residues, 2005.
• Seppo Mustonen, On integer sequences with mutual k-residues, 2005. [Local copy]
• Index entries for sequences of form a(n+1)=a(n)^2 + ....

Crossrefs

a(n) = A001566(n-1)+2 (for n>0).

Cf. A000032, A000058, A000215, A000289.

Programs

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 *)

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)

Formula

a(n) = L(2^n)+2, if n>0 where L() is Lucas sequence.

For n>=1, a(n) = 4 + Product_{i=0..n-1} a(i). - Vladimir Shevelev, Dec 08 2010

From Amiram Eldar, Sep 10 2022: (Start)

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