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A226026
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Maximum fixed points under iteration of sum of cubes of digits in base n.
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2
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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
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OFFSET
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2,2
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COMMENTS
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1 is considered a fixed point in all bases, 0 is not.
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.
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.
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LINKS
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EXAMPLE
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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.
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PROG
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(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")
}
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CROSSREFS
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Number of fixed points in base n: A194025.
All fixed points in base 10: A046197.
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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