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%I #7 Oct 01 2013 17:57:37
%S 1,8,512,134,92,737,713,371,27,351,153,64,280,520,133,55,125,134,92,
%T 737,713,371,216,225,141,66,432,99,1458,702,351,153,343,118,514,190,
%U 730,370,512,134,92,737,713,371,729,1080,513,153,1,2,8,512,134
%N For each n append T(n), T(T(n)), T^3(n), ..., T^r(n), where T(n) = A055012(n) and r is the smallest integer such that T^r(n) is one of the following numbers: 1, 55, 136, 153, 160, 370, 371, 407, 919.
%C Conjecture: The sequence always terminates with one of the following:(tested to n=1000000) 1,55,136,153,160,370,371,407,919 which eventually loop back to themselves. 1,153,370,371,407 loop back in 1 step and are the sum of the cubes of their digits. The others are 55,250,133,55. 136,244,136. 160,217,352,160. 919,1459,919. A046156, A046157 indicate this as a limit of possibilities of numbers that cubed digital roots roll back to the origional number. Proof? - Cino Hilliard, Apr 13 2003 Proof: In A055012 T.D.Noe notes that for n > 1999, A055012(n) < n. This means that by repeatedly applying A055012, we eventually reach a number smaller than 2000. As checked by Cino Hilliard, all numbers below 10^6 end in one of the listed cycles. - Stefan Steinerberger, Sep 05 2007
%t a = {}; For[n = 1, n < 9, n++, j = Plus @@ IntegerDigits[n]^3; AppendTo[a, j]; While[ !MemberQ[{1, 55, 136, 153, 160, 370, 371, 407, 919}, j], j = Plus @@ (IntegerDigits[j]^3); AppendTo[a, j]]]; a
%o (PARI) digitcube2(m) = {y=0; for(x=1,m, digitcube(x) ) } digitcube(n) = { while(1, s=0; while(n > 0, d=n%10; s = s+d*d*d; n=floor(n/10); ); print1(s" "); if(s==1 || s==55 || s==153 || s==160 || s==370 || s==371 || s==407 || s==919 || s==136,break); n=s;) }
%Y Cf. A046156, A055012.
%K base,easy,nonn
%O 1,2
%A _Cino Hilliard_, Apr 13 2003
%E Edited by _Stefan Steinerberger_, Sep 05 2007
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