0,2

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

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).

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, alternative link.

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.

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 + ....

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)

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 *)

(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)

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

Cf. A000032, A000058, A000215, A000289.

Sequence in context: A328333 A173776 A289909 * A123817 A124421 A262918

Adjacent sequences:  A000321 A000322 A000323 * A000325 A000326 A000327

nonn,easy,changed

N. J. A. Sloane

approved