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%I #21 Dec 03 2014 00:57:31
%S 1,17,62,118,251,250,433,1052,407,1280,2002,1968,793,3052,5614,1456,
%T 5337,5939,2413,5615,20217,11648,11080,31024,5425,1737,28027,26846,
%U 17451,33535,10261,64019,23552,44937,30086,84870,17353,55243,48824,108936,58618,87977
%N Maximum fixed points under iteration of sum of cubes of digits in base n.
%C 1 is considered a fixed point in all bases, 0 is not.
%C a(n)=1 iff A194025(n)=1.
%C In order for a number with d digits in base n to be a fixed point, it must satisfy the condition d*(n-1)^3<n^d, which can only occur when d<=4 for n>2. Because all binary numbers are "happy" (become 1 under iteration), there are no fixed points with more than 4 digits in any base.
%C Furthermore, 4-digit solutions of the form x0mm or xmmm (where m is n-1) represent extreme values of sum of cubed digits, and so 4-digit numbers can only be solutions if xn^3+n^2-1<=2n^3+x^3. For x=2 this reduces to n<=3, so any 4-digit solution must begin with 1 in bases above 3.
%H Christian N. K. Anderson, <a href="/A226026/b226026.txt">Table of n, a(n) for n = 2..1000</a>
%H Christian N. K. Anderson, <a href="/A226026/a226026_1.txt">Table of base, maximum fixed point, number of fixed points, and all fixed points</a> for base 2 to 1000.
%e In base 5, the numbers 1, 28 and 118 are written as 1, 103, and 433. The sum of the cubes of their digits are 1, 1+0^3+3^3=28, and 4^3+3^3+3^3=118. There are no other solutions, so a(5)=118.
%o (R) inbase=function(n,b) { x=c(); while(n>=b) { x=c(n%%b,x); n=floor(n/b) }; c(n,x) }
%o yfp=vector("list",100)
%o for(b in 2:100) { fp=c()
%o for(w in 0:1) for(x in 1:b-1) for(y in 1:b-1) if((u1=w^3+x^3+y^3)<=(u2=w*b^3+x*b^2+y*b) & u1+b^3>u2+b-1)
%o if(length((z=which((1:b-1)*((1:b-1)^2-1)==u2-u1)-1))) fp=c(fp,u2+z)
%o yfp[[b]]=fp[-1]
%o cat("Base",b,":",fp,"\n")
%o }
%Y Number of fixed points in base n: A194025.
%Y All fixed points in base 10: A046197.
%Y Cf. A193583, A209242.
%K nonn,base
%O 2,2
%A _Kevin L. Schwartz_ and _Christian N. K. Anderson_, May 23 2013