

A087656


Let f be defined on the rationals by f(p/q) =(p+1)/(q+1)=p_{1}/q_{1} where (p_{1},q_{1})=1. Let f^k(p/q)=p_{k}/q_{k} where (p_{k},q_{k})=1. Sequence gives least k such that p_{k}q_{k} = 1 starting at n.


2



1, 2, 2, 4, 3, 6, 3, 4, 5, 10, 4, 12, 7, 6, 4, 16, 5, 18, 6, 8, 11, 22, 5, 8, 13, 6, 8, 28, 7, 30, 5, 12, 17, 10, 6, 36, 19, 14, 7, 40, 9, 42, 12, 8, 23, 46, 6, 12, 9, 18, 14, 52, 7, 14, 9, 20, 29, 58, 8, 60, 31, 10, 6, 16, 13, 66, 18, 24, 11, 70, 7, 72, 37, 10, 20, 16, 15, 78, 8, 8, 41, 82
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OFFSET

3,2


COMMENTS

Proof that this is the same as A059975 except for offset, from Joseph Myers, Feb 21 2004. Claim: a(n+1) = A059975(n). If p is the least prime factor of n then the rule here gives (n+1)/1 > (n+2)/2 > ... > (n+p)/p = (n/p + 1)/1 so a(n+1) = a(n/p + 1) + (p1) and clearly A059975(n) = A059975(n/p) + (p1). The natural start for the induction is A059975(1) = a(2) = 0 (one place before the currently listed sequences start).


LINKS

Table of n, a(n) for n=3..84.


FORMULA

If p is prime a(p+1)=p1; it appears that a(n)=(n1)/2 iff n is in A079148 or in A053177.


EXAMPLE

6 > (6+1)/(1+1) = 7/2 > (7+1)/(2+1) = 8/3 > (8+1)/(3+1) = 9/4 > (9+1)/(4+1) = 2/1 and 21 = 1 hence a(6) = 4.


PROG

(PARI) a(x)=if(x<0, 0, c=0; while(abs(numerator(x)denominator(x)1)>0, x=(numerator(x)+1)/(denominator(x)+1); c++); c)


CROSSREFS

Same as A059975 apart from offset.
Sequence in context: A257010 A156864 A059975 * A122811 A089173 A126090
Adjacent sequences: A087653 A087654 A087655 * A087657 A087658 A087659


KEYWORD

nonn


AUTHOR

Benoit Cloitre, Oct 04 2003


STATUS

approved



