

A135506


a(n) = x(n+1)/x(n)  1 where x(1)=1 and x(k) = x(k1) + lcm(x(k1),k).


20



2, 1, 2, 5, 1, 7, 1, 1, 5, 11, 1, 13, 1, 5, 1, 17, 1, 19, 1, 1, 11, 23, 1, 5, 13, 1, 1, 29, 1, 31, 1, 11, 17, 1, 1, 37, 1, 13, 1, 41, 1, 43, 1, 1, 23, 47, 1, 1, 1, 17, 13, 53, 1, 1, 1, 1, 29, 59, 1, 61, 1, 1, 1, 13, 1, 67, 1, 23, 1, 71, 1, 73, 1, 1, 1, 1, 13, 79, 1, 1, 41, 83, 1, 1, 43, 29, 1, 89
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

This sequence has properties related to primes. For instance: terms consist of 1's or primes only; if 3 never occurs, any prime p occurs finitely many times.
All prime numbers 'p' from the sequence A014963(n), which equals A003418(n+1)/A003418(n), are in a(n1) = p.  Eric Desbiaux, Jan 11 2015
For any prime p > 3, a(p1) = p. Also a(n) is not 3 for any n. All terms but a(1) and a(3) are odd, and probably all of them are not composite numbers; this is strongly related to a strong version of Linnik's Theorem (see RuizCabello link).  Serafín RuizCabello, Sep 30 2015


LINKS

Robert Israel, Table of n, a(n) for n = 1..10000
Serafín RuizCabello, On the use of the lowest common multiple to build a primegenerating recurrence, arXiv:1504.05041 [math.CO], 2015.


MAPLE

x[1]:= 1;
for n from 2 to 101 do
x[n]:= x[n1] + ilcm(x[n1], n);
a[n1]:= x[n]/x[n1]1;
od:
seq(a[n], n=1..100); # Robert Israel, Jan 11 2015


MATHEMATICA

a[n_] := x[n+1]/x[n]  1; x[1] = 1; x[k_] := x[k] = x[k1] + LCM[x[k1], k]; Table[a[n], {n, 1, 88}] (* JeanFrançois Alcover, Jan 08 2013 *)


PROG

(PARI) x1=1; for(n=2, 40, x2=x1+lcm(x1, n); t=x1; x1=x2; print1(x2/t1, ", "))


CROSSREFS

Cf. A106108.
Sequence in context: A242598 A225568 A291694 * A295516 A068822 A090079
Adjacent sequences: A135503 A135504 A135505 * A135507 A135508 A135509


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Feb 09 2008


STATUS

approved



