cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Damir Olejar

Damir Olejar's wiki page.

Damir Olejar has authored 3 sequences.

A160761 The Kaprekar binary numbers in decimal.

Original entry on oeis.org

9, 9, 9, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 45, 45, 49, 45, 49, 49, 45, 45, 49, 49, 45, 49, 45, 45, 45, 49, 49, 45, 49, 45, 45, 49, 45, 45, 45, 93, 93, 105, 93, 105, 105, 105, 93, 105, 105, 105, 105, 105, 105, 93, 93, 105, 105, 105, 105, 105, 105, 93, 105, 105, 105
Offset: 1

Views

Author

Damir Olejar, May 25 2009

Keywords

Examples

			The number 9 is 1001 in binary. The maximum number using the same number of 0's and one'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 always gives us the same number and pattern of 0's and 1's. Therefore 9 is one of the Kaprekar numbers. Numbers that end the procedure in 0 are excluded since they are not Kaprekar numbers.

References

  • 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), 81-82.

Links

Crossrefs

Cf. A099009, A099010, A090429, A069746, A163205.

Programs

  • Java
    class pattern { public static void main(String args[]) { int mem1 = 0; int mem2 =1; for (int i = 1; i<3000; 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
  • Mathematica
    nmax = 100; 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, 1, nmax}]][[2, 1]] (* Jean-François Alcover, Apr 23 2017 *)

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 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.

A163205 The non-repetitive Kaprekar binary numbers in decimal.

Original entry on oeis.org

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
Offset: 1

Views

Author

Damir Olejar, Jul 23 2009

Keywords

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.
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)

Examples

			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 occurred for 1001, the number 9 occurring for 1100 is ignored to avoid repetition.

References

  • 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.

Links

Crossrefs

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).

Programs

  • Java
    import java.util.*; class pattern { public static void main(String args[]) { int mem1 = 0; int mem2 =1; ArrayList memory = new ArrayList(); for (int i = 1; i
  • 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]& (* Jean-François Alcover, Apr 23 2017 *)

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 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.

Extensions

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

A049341 a(n+1) = sum of digits of a(n) + a(n-1).

Original entry on oeis.org

3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3
Offset: 0

Views

Author

Damir Olejar

Keywords

Comments

a(n+1) = A007953(a(n) + a(n-1)) for n > 0.

Examples

			After 6,9 we get 6+9 = 15 -> 1+5 = 6.

Links

Crossrefs

Cf. A030132, A030133, A049342.

Programs

  • Haskell
    a049341 n =  a030132_list !! n
a049341_list =
   3 : 6 : map a007953 (zipWith (+) a049341_list $ tail a049341_list)
-- Reinhard Zumkeller, Aug 20 2011
  • Mathematica
    LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 1},{3, 6, 9, 6, 6, 3, 9, 3},112] (* Ray Chandler, Aug 27 2015 *)

Formula

Period 8.

Extensions

Definition improved by Reinhard Zumkeller, Aug 20 2011

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