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.

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

Paolo P. Lava, Table of n, a(n) for n = 1..3735

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

Cf. A260275, A260385, A260386.

Sequence in context: A030299 A352991 A268532 * A001292 A292523 A162391

Adjacent sequences: A260272 A260273 A260274 * A260276 A260277 A260278

Keyword

nonn,base,fini

Author

Paolo P. Lava, Jul 24 2015

Status

approved