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 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 (list; graph; refs; listen; history; text; internal format)
 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) + (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). LINKS FORMULA 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. 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 2-1 = 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

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Last modified February 22 12:00 EST 2020. Contains 332135 sequences. (Running on oeis4.)