A226026 Maximum fixed points under iteration of sum of cubes of digits in base n.

1, 17, 62, 118, 251, 250, 433, 1052, 407, 1280, 2002, 1968, 793, 3052, 5614, 1456, 5337, 5939, 2413, 5615, 20217, 11648, 11080, 31024, 5425, 1737, 28027, 26846, 17451, 33535, 10261, 64019, 23552, 44937, 30086, 84870, 17353, 55243, 48824, 108936, 58618, 87977

Offset: 2

Author

Kevin L. Schwartz and Christian N. K. Anderson, May 23 2013

Keywords

• nonn
• base

Comments

1 is considered a fixed point in all bases, 0 is not.

a(n)=1 iff A194025(n)=1.

In order for a number with d digits in base n to be a fixed point, it must satisfy the condition d*(n-1)^32. Because all binary numbers are "happy" (become 1 under iteration), there are no fixed points with more than 4 digits in any base.

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.

Examples

			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.
		

Links

• Christian N. K. Anderson, Table of n, a(n) for n = 2..1000
• Christian N. K. Anderson, Table of base, maximum fixed point, number of fixed points, and all fixed points for base 2 to 1000.

Crossrefs

Number of fixed points in base n: A194025.

All fixed points in base 10: A046197.

Cf. A193583, A209242.

Programs

R

inbase=function(n,b) { x=c(); while(n>=b) { x=c(n%%b,x); n=floor(n/b) }; c(n,x) }
yfp=vector("list",100)
for(b in 2:100) { fp=c()
    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)
        if(length((z=which((1:b-1)*((1:b-1)^2-1)==u2-u1)-1))) fp=c(fp,u2+z)
    yfp[[b]]=fp[-1]
    cat("Base",b,":",fp,"\n")
}