A082385 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.

1, 8, 512, 134, 92, 737, 713, 371, 27, 351, 153, 64, 280, 520, 133, 55, 125, 134, 92, 737, 713, 371, 216, 225, 141, 66, 432, 99, 1458, 702, 351, 153, 343, 118, 514, 190, 730, 370, 512, 134, 92, 737, 713, 371, 729, 1080, 513, 153, 1, 2, 8, 512, 134

Offset: 1

Author

Cino Hilliard, Apr 13 2003

Keywords

• base
• easy
• nonn

Comments

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

Crossrefs

Cf. A046156, A055012.

Programs

Mathematica

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

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;) }

Extensions

Edited by Stefan Steinerberger, Sep 05 2007