cp's OEIS Frontend

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.

%I A087656 #16 Feb 15 2015 14:49:54
%S A087656 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,
%T A087656 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,
%U A087656 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
%N 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.
%C A087656 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) + (p-1) and clearly A059975(n) = A059975(n/p) + (p-1). The natural start for the induction is A059975(1) = a(2) = 0 (one place before the currently listed sequences start).
%F A087656 If p is prime a(p+1)=p-1; it appears that a(n)=(n-1)/2 iff n is in A079148 or in A053177.
%e A087656 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 2-1 = 1 hence a(6) = 4.
%o A087656 (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)
%Y A087656 Same as A059975 apart from offset.
%K A087656 nonn
%O A087656 3,2
%A A087656 _Benoit Cloitre_, Oct 04 2003