generate a Series / return nth term in the series

Question

I need to generate a sequence such that its members contain only 1 , 2 , 3 digits. For example, 1 2 3 11 12 13 21 22 23 31 32 33 111 .... and so on up to 10^18th term. I'm not able to deduce any pattern for this. It seems impossible to write a code up to 10^18 numbers of terms in the series.

1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233, 311, 312, 313, 321, 322, 323, 331, 332, 333, 1111, 1112, 1113, 1121, 1122, 1123, 1131, 1132, 1133, 1211, 1212, 1213, 1221 ...

I expect to find the given n-th term in the series. It's a number system which contains only 1 , 2 , 3 or is combinations of these digits as a number as explained in the sequence just like our normal number system.

Answer 1

This is just a base-3 numbering system only the digits go from 1 to 3 instead of 0 to 2. The math works out the same way:

1 = *3^0 * 3 ^ 0
2 = *3^0 * 3 ^ 0
3 = *3^0 * 3 ^ 0
4 = *3^1 + *3^0 * 3 ^ 1 + * 3 ^ 0
5 = *3^1 + *3^0 * 3 ^ 1 + * 3 ^ 0
6 = *3^1 + *3^0 * 3 ^ 1 + * 3 ^ 0
7 = *3^1 + *3^0 * 3 ^ 1 + * 3 ^ 0
...
19 = *3^2 + *3^1 + *3^0 * 3 ^ 2 + * 3 ^ 1 + * 3 ^ 0

Write two methods:

1. digit(n): computes the right-most digit for a given n . Some test cases: digit(4) = 1, digit(5)=2, digit(15)=3.
2. leftover(n): computes the number which represents n but with the right-most digit chopped off. Some test cases: leftover(4) = 1, leftover(15) = 4, leftover(23) = 7.

Now combine the two methods into the solution to your problem which repeatedly chops off the right most digit until there's nothing left. You might find it easier to do this recursively.

Answer 2

The sequence you've mentioned already is known as Numbers that contain only 1's, 2's and 3's . It is formulated by Hieronymus Fischer .

a(n) = sum_{j=0..m-1} (1 + b(j) mod 3)*10^j,

where m = floor(log_3(2*n+1)), b(j) = floor((2*n+1-3^m)/(2*3^j)).

You can examine explanation of the formula on the aforementioned link above. I've written so far basic level of it using long . To reach 10^18th term, you need to use BigInteger class of Java.

class SequenceGeneratorWith123 {

    // Written by Soner

    private static double logOfBase(long base, long num) {
        return Math.log(num) / Math.log(base);
    }

    private static int mfunc(long n) {
        return (int) Math.floor(logOfBase(3, 2 * n + 1));
    }

    private static int b(int j, double m, long n) {
        return (int) Math.floor((2 * n + 1 - Math.pow(3, m)) / (2 * Math.pow(3, j)));
    }

    public static void main(String[] args) {

        for (int i = 0; i < 9; i++) {
            long n = (long) Math.pow(10, i);
            int m = mfunc(n);
            long sum = 0;

            for (int j = 0; j < m ; j++) {
                sum += ((1 + b(j, m, n) % 3) * Math.pow(10, j));
            }
            System.out.printf("a(10^%d) = %d\n", i, sum);
        }

        System.out.println("After the point, overflow will occur " +
                        "because of long type.");    
    }
}

Output:

a(10^0) = 1
a(10^1) = 31
a(10^2) = 3131
a(10^3) = 323231
a(10^4) = 111123331
a(10^5) = 11231311131
a(10^6) = 1212133131231
a(10^7) = 123133223331331
a(10^8) = 13221311111312132
After the point, overflow will occur because of long type.

---

You just need to play with the code, that is, we can obtain your wish by merely changing main() somewhat.

long n = 1;
// How many terms you need you can alter it by pow() method.
// In this example 10^2 = 100 terms will be obtained.
int term = (int)Math.pow(10, 2);
for (int i = 0; i < term; i++) {

    int m = mfunc(n);
    long sum = 0;

    for (int j = 0; j < m ; j++) {
        sum += ((1 + b(j, m, n) % 3) * Math.pow(10, j));
    }
    System.out.printf("%d. term = %d\n", i + 1, sum);
    n++;
}

Output:

1. term = 1
2. term = 2
3. term = 3
4. term = 11
5. term = 12
6. term = 13
7. term = 21
8. term = 22
9. term = 23
10. term = 31
11. term = 32
12. term = 33
13. term = 111
14. term = 112
15. term = 113
16. term = 121
17. term = 122
18. term = 123
19. term = 131
20. term = 132
21. term = 133
22. term = 211
23. term = 212
24. term = 213
25. term = 221
26. term = 222
27. term = 223
28. term = 231
29. term = 232
30. term = 233
31. term = 311
32. term = 312
33. term = 313
34. term = 321
35. term = 322
36. term = 323
37. term = 331
38. term = 332
39. term = 333
40. term = 1111
41. term = 1112
42. term = 1113
43. term = 1121
44. term = 1122
45. term = 1123
46. term = 1131
47. term = 1132
48. term = 1133
49. term = 1211
50. term = 1212
51. term = 1213
52. term = 1221
53. term = 1222
54. term = 1223
55. term = 1231
56. term = 1232
57. term = 1233
58. term = 1311
59. term = 1312
60. term = 1313
61. term = 1321
62. term = 1322
63. term = 1323
64. term = 1331
65. term = 1332
66. term = 1333
67. term = 2111
68. term = 2112
69. term = 2113
70. term = 2121
71. term = 2122
72. term = 2123
73. term = 2131
74. term = 2132
75. term = 2133
76. term = 2211
77. term = 2212
78. term = 2213
79. term = 2221
80. term = 2222
81. term = 2223
82. term = 2231
83. term = 2232
84. term = 2233
85. term = 2311
86. term = 2312
87. term = 2313
88. term = 2321
89. term = 2322
90. term = 2323
91. term = 2331
92. term = 2332
93. term = 2333
94. term = 3111
95. term = 3112
96. term = 3113
97. term = 3121
98. term = 3122
99. term = 3123
100. term = 3131