

A163205


The nonrepetitive Kaprekar binary numbers in decimal.


40



0, 9, 21, 45, 49, 93, 105, 189, 217, 225, 381, 441, 465, 765, 889, 945, 961, 1533, 1785, 1905, 1953, 3069, 3577, 3825, 3937, 3969, 6141, 7161, 7665, 7905, 8001, 12285, 14329, 15345, 15841, 16065, 16129, 24573, 28665, 30705, 31713, 32193, 32385
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OFFSET

1,2


COMMENTS

Same as A160761, but with no repetitions. The numbers also exist in A143088, except that every first and last number is omitted from A143088's pyramid.
Note that all base2 cycles are fixed points.
Initial terms in base 2: 0, 1001, 10101, 101101, 110001, 1011101, 1101001, 10111101, 11011001, 11100001. (End)


REFERENCES

M. Charosh, Some Applications of Casting Out 999...'s, Journal of Recreational Mathematics 14, 198182, pp. 111118.
D. R. Kaprekar, On Kaprekar numbers, J. Rec. Math., 13 (19801981), pp. 8182.


LINKS



FORMULA

1. Sort all integers from the number in descending order.
2. Sort all integers from the number in ascending order.
3. Subtract ascending from descending order to obtain a new number.
4. Repeat the steps 13 with a new number until a repetitive sequence is obtained or until a zero is obtained.
5. Call the repetitive sequence's number a Kaprekar number, ignore zeros and repetitions from the set of the final results.


EXAMPLE

The number 9 is 1001 in binary. The maximum number using the same number of 0's and 1's is found and the minimum number having the same number of 0's and 1's is found to obtain the equation such as 1100  0011 = 1001. Repeating the same procedure gives us the same number and pattern of 0's and 1's. Therefore 9 is one of the Kaprekar numbers. If 9 did not occur before, it is counted as a number that belongs to a sequence and added to a database to skip repetitions. Numbers that end the procedure in 0 are excluded since they are not Kaprekar numbers. A number 9 can also be obtained with, let's say, 1100. Since number 9 already occured for 1001, the number 9 occurring for 1100 is ignored to avoid repetition.


MATHEMATICA

nmax = 10^5; f[n_] := Module[{id, sid, min, max}, id = IntegerDigits[n, 2]; min = FromDigits[sid = Sort[id], 2]; max = FromDigits[Reverse[sid], 2]; max  min]; Reap[Do[If[(fpn = FixedPoint[f, n]) > 0, Sow[fpn]], {n, 0, nmax}]][[2, 1]] // Union // Prepend[#, 0]& (* JeanFrançois Alcover, Apr 23 2017 *)


PROG

(Java) import java.util.*; class pattern { public static void main(String args[]) { int mem1 = 0; int mem2 =1; ArrayList<Integer> memory = new ArrayList<Integer>(); for (int i = 1; i<Integer.MAX_VALUE; i++) { do { mem1 = mem2; String binaryi = Integer.toBinaryString(i); String binarysort = ""; String binaryminimum = ""; for (int n = 0; n< binaryi.length(); n++) { String g = binaryi.substring(n, n+1);
if (g.equals("0")) { binarysort = binarysort+"0"; } else { binarysort = "1"+binarysort; binaryminimum = binaryminimum + "1"; } } int binrev1 = Integer.parseInt(binarysort , 2); int binrev2 = Integer.parseInt(binaryminimum , 2); int diff = binrev1  binrev2; mem2 = diff; } while (mem2!=0 && mem2!=mem1); String memtobin = Integer.toBinaryString(mem1); int ones = 0; for (int t = 0; t<memtobin.length(); t++) { String o = memtobin.substring(t, t+1); if (o.equals("1")) ones++; } if (memtobin.length()!=ones) { if(!memory.contains (mem1)) {System.out.print(mem1+" "); memory.add(mem1); } } } }}


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS

Initial zero added for consistency with other bases by Joseph Myers, Aug 29 2009


STATUS

approved



