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

%I #13 Jun 27 2021 07:52:56

%S 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 9,8,14,13,13,12,13,12,12,11,13,12,12,11,12,11,11,10,13,12,12,11,12,

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

%C More precisely, number of steps to reach 1 but passing through 7 first.

%C 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 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 a(n) = A023416(n+1) + floor(log_2(n+1)) + 4. - _Ralf Stephan_, Mar 03 2004

%o (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 Cf. A080791, A080800, A080801.

%K easy,nonn

%O 1,1

%A _Cino Hilliard_, Mar 25 2003