cp's OEIS Frontend

A080799 Number of divide by 2 and add 1 operations required to reach ...,7,8,4,2,1 when started at n.

%I A080799 #13 Jun 27 2021 07:52:56
%S A080799 6,5,8,7,7,6,10,9,9,8,9,8,8,7,12,11,11,10,11,10,10,9,11,10,10,9,10,9,
%T A080799 9,8,14,13,13,12,13,12,12,11,13,12,12,11,12,11,11,10,13,12,12,11,12,
%U A080799 11,11,10,12,11,11,10,11,10,10,9,16,15,15,14,15,14,14,13,15,14,14,13,14,13,13
%N A080799 Number of divide by 2 and add 1 operations required to reach ...,7,8,4,2,1 when started at n.
%C A080799 More precisely, number of steps to reach 1 but passing through 7 first.
%C A080799 A 3x+1 - type sequence cannot contain ..., 7, 8, 4, 2, 1 because 7 is odd and the recurrence will always yield 22 as the number that follows 7. So the x+1 conjecture has a property the 3x+1 conjecture does not have. The link will allow you to try very large numbers for these conjectures.
%H A080799 Cino Hilliard, <a href="http://groups.msn.com/BC2LCC/page.msnw?fc_p=%2Fkx%2Bp%20problems&amp;fc_a=0">The x+1 conjecture</a>
%F A080799 a(n) = A023416(n+1) + floor(log_2(n+1)) + 4. - _Ralf Stephan_, Mar 03 2004
%o A080799 (PARI) xpcount2(n,p) = { for(x=1,n, p1 = x; f=0; ct=0; while(p1>1, if(p1%2==0,p1/=2; ct++,p1 = p1*p+1; ct++); if(p1==7, p2=7; if(p2%2==0,p2/=2,p2 = p2*p+1); if(p2 ==8 && p1 ==7,f=1) ); ); if(f,print1(ct" ")) ) }
%Y A080799 Cf. A080791, A080800, A080801.
%K A080799 easy,nonn
%O A080799 1,1
%A A080799 _Cino Hilliard_, Mar 25 2003