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A260275
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Fixed points of the function A260529(n) = concatenation of the positions of digits 9, 8,..., 0 in the decimal representation of n, using 1 for the rightmost digit etc., skipping digits which don't occur.
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6
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1, 12, 21, 123, 231, 312, 321, 1234, 1324, 2143, 2341, 3412, 3421, 4123, 4231, 4312, 4321, 12345, 13425, 14235, 14325, 21354, 23451, 24153, 24351, 31524, 32541, 34512, 34521, 45123, 45231, 45312, 45321, 51234, 51324, 52143, 52341, 53412, 53421, 54123, 54231, 54312
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OFFSET
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1,2
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COMMENTS
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Given a number n with k digits, label the positions of the digits starting from LSD = 1 to MSD = k. Then concatenate in ascending order the positions of the maximum digit in n. Repeat the same process for all the different digits, in descending order. Sequence lists the fixed points of this transform.
If we consider the numbers that under this transform produce a multiple of the number itself, for n<= 10^9 we should add only 11780892. This has digit 9 is in position 2, 8 in positions 3 and 5, 7 in position 6, 2 in position 1, 1 in positions 7 and 8, 0 in position 4. Finally, 23561784 / 11780892 = 2.
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LINKS
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EXAMPLE
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In 2341 digit 4 is in position 2, 3 in position 3, 2 in position 4, 1 in position 1. Therefore concat(2,3,4,1) = 2341 that is a fixed point.
In 53412 digit 5 is in position 5, 4 in position 3, 3 in position 4, 2 in position 1, 1 in position 2. Therefore concat(5,3,4,1,2) = 53412 that is a fixed point.
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MAPLE
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with(numtheory): P:=proc(q) local a, b, j, k, n;
for n from 1 to q do a:=convert(n, base, 10); b:=0;
for k from 9 by -1 to 0 do for j from 1 to nops(a) do
if a[j]=k then b:=b*10^(ilog10(j)+1)+j; fi; od;
od; if type(b/n, integer) then print(n); fi;
od; end: P(10^10);
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CROSSREFS
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KEYWORD
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nonn,base,fini
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AUTHOR
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STATUS
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approved
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