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A260275
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.
6
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
OFFSET
1,2
COMMENTS
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.
LINKS
EXAMPLE
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.
MAPLE
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);
CROSSREFS
KEYWORD
nonn,base,fini
AUTHOR
Paolo P. Lava, Jul 24 2015
STATUS
approved