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A163205
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The non-repetitive Kaprekar binary numbers in decimal.
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40
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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
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1,2
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COMMENTS
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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.
From Joseph Myers, Aug 29 2009: (Start)
Note that all base-2 cycles are fixed points.
Initial terms in base 2: 0, 1001, 10101, 101101, 110001, 1011101, 1101001, 10111101, 11011001, 11100001. (End)
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REFERENCES
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M. Charosh, Some Applications of Casting Out 999...'s, Journal of Recreational Mathematics 14, 1981-82, pp. 111-118.
D. R. Kaprekar, On Kaprekar numbers, J. Rec. Math., 13 (1980-1981), pp. 81-82.
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LINKS
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Joseph Myers, Table of n, a(n) for n = 1..9802 [From Joseph Myers, Aug 29 2009]
Juergen Koeller, The Kaprekar Number
Wikipedia, Kaprekar Number
Index entries for the Kaprekar map
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FORMULA
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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 1-3 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.
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EXAMPLE
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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.
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MATHEMATICA
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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]& (* Jean-François Alcover, Apr 23 2017 *)
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PROG
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(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); } } } }}
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CROSSREFS
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Cf. A160761, A143088, A164884, A164885, A164886, A164887.
In other bases: A164997 (base 3), A165016 (base 4), A165036 (base 5), A165055 (base 6), A165075 (base 7), A165094 (base 8), A165114 (base 9), A099009 (base 10).
Sequence in context: A242990 A053476 A110680 * A154862 A350632 A020137
Adjacent sequences: A163202 A163203 A163204 * A163206 A163207 A163208
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KEYWORD
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nonn,base
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AUTHOR
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Damir Olejar, Jul 23 2009
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EXTENSIONS
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Initial zero added for consistency with other bases by Joseph Myers, Aug 29 2009
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STATUS
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approved
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