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A323190
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Integers k for which there exists an integer j such that s(k) + j + reversal(s(k) + j) = k where s(k) is the sum of digits of k.
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0
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0, 10, 11, 12, 14, 16, 18, 22, 33, 44, 55, 66, 77, 88, 99, 101, 110, 121, 132, 141, 143, 154, 161, 165, 176, 181, 187, 198, 201, 202, 221, 222, 241, 242, 261, 262, 281, 282, 302, 303, 322, 323, 342, 343, 362, 363, 382, 383, 403, 404, 423, 424, 443, 444, 463
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listen;
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text;
internal format)
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OFFSET
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1,2
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COMMENTS
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Theorem: all palindromes that have an even number of digits and all palindromes that have an odd number of digits and the digit in the middle is even are in this sequence.
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LINKS
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EXAMPLE
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k=10 is a term because a solution exists with j=4: s(10)=1, s(k) + j + reversal(s(k) + j) = 1 + 4 + reversal(1 + 4) = 10.
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MATHEMATICA
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ok[n_] := Block[{s = Total@ IntegerDigits@ n}, Select[Range[0, n], s + # + FromDigits@ Reverse@ IntegerDigits[s + #] == n &, 1] != {}]; Select[ Range[0, 1000], ok] (* Giovanni Resta, Feb 19 2019 *)
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PROG
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(Java)
package com.company;
public class Main {
public static void main(String args[]) {
int counter=1;
for (int i = 0; i < 10000; i++) {
for (int j = 0; j < 10000; j++) {
int sumPlus = sumDigits(i) + j;
int check = sumPlus + reverse(sumPlus);
if (check == i) {
System.out.println(String.format("n(%d)=%d, a=%d", counter, i, j));
counter++;
break;
}
}
}
System.out.println(String.format("%d", counter));
}
public static int reverse(int x) {
String s = String.valueOf(x);
StringBuilder sb = new StringBuilder();
sb.append(s);
sb.reverse();
return Integer.parseInt(sb.toString());
}
public static int sumDigits(int x) {
int result = 0;
while (x > 0) {
result += x % 10;
x = x / 10;
}
return result;
}
}
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CROSSREFS
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This sequence is a subsequence of A067030.
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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