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LaBar's conjecture (steps to return n to 1 after division by 3 and, if needed, multiplication by 2, addition of 1 or 2).
2

%I #17 Jul 18 2021 01:46:18

%S 18,44,4,216,34,50,181,68,13,126,125,228,278,256,49,364,82,68,180,575,

%T 202,1033,245,92,403,140,40,520,499,156,872,214,158,1400,221,264,399,

%U 368,317,1157,390,298,648,376,94,594,1155,412,1983,500,133,808,226,122

%N LaBar's conjecture (steps to return n to 1 after division by 3 and, if needed, multiplication by 2, addition of 1 or 2).

%H Enoch Haga, <a href="https://doi.org/10.1111/j.1949-8594.1983.tb10148.x">Problem</a>, School Science and Mathematics, Nov 1983, vol. 83, no 7, page 628.

%H Sean A. Irvine, <a href="https://github.com/archmageirvine/joeis/blob/master/src/irvine/oeis/a049/A049067.java">Java program</a> (github)

%H LaBar, <a href="https://doi.org/10.1111/j.1949-8594.1982.tb10084.x">Problem #3929</a>, School Science and Mathematics, Dec 1982, vol. 82 no 8, page 715.

%e Beginning at n=1, algorithm produces s+t+a=18.

%o (This was in the formula section but is clearly a program in some language. - _N. J. A. Sloane_, Apr 18 2017)

%o f=1\n=n+1:a=n\x=n/3\ (Case 1) if int(x)=x then o=o+1:n=x:s=s+n: if n=1 then print s+t+a:e=0:o=0:s=0:t=0:n=a:return to n=n+1\if n<>1 then return to x=n/3\ (Case 2) if int(x)<>x then e=e+1\if f=1 then y=n*2+2:f=0:n=y:t=t+n:else y=n*2+1:f=1\n=y:t=t+n\if n=1 then print s+t+a:return to n=n+1:else return to x=n/3

%Y Cf. A049074.

%K easy,nonn

%O 1,1

%A _Enoch Haga_