A325722 -id:A325722 - cp's OEIS frontend

A308232 Start the sequence with a(1) = 1 and read the digits one by one from there. The sequence is always extended with the concatenation kd, d being the digit that was read and k the number of d's present so far in the sequence.

1, 11, 21, 31, 12, 41, 13, 51, 61, 22, 14, 71, 81, 23, 15, 91, 16, 101, 32, 42, 111, 24, 17, 121, 18, 131, 52, 33, 141, 25, 19, 151, 161, 26, 171, 10, 181, 43, 62, 34, 72, 191, 201, 211, 82, 44, 221, 27, 231, 92, 241, 251, 28, 261, 53, 271, 35, 102, 63, 73, 281, 54, 291, 112, 45, 301, 29, 311, 55, 321, 331, 36, 341, 122, 46, 351

Offset: 1

Eric Angelini and Carole Dubois, May 16 2019

All integers > 0 will appear exactly once, except 2, 3, 4, 5, 6, 7, 8 and 9 which will never appear.

			The sequence starts with a(1) = 1.
We read this 1, see that there is only one digit 1 so far in the sequence, thus k = 1; we have then [kd] = 11 and this 11 becomes a(2);
We read now the first digit of a(2) = 11, which is 1; as this 1 is the 2nd occurrence of 1 so far in the sequence, we have k = 2 and [kd] = 21; this 21 becomes a(3);
We read now the second digit of a(2) = 11, which is 1; as this 1 is the 3rd occurrence of 1 so far in the sequence, we have k = 3 and [kd] = 31; this 31 becomes a(4);
We read now the first digit of a(3) = 21, which is 2; as this 2 is the 1st occurrence of 2 so far in the sequence, we have k = 1 and [kd] = 12; this 12 becomes a(5);
We read now the second digit of a(3) = 21, which is 1; as this 1 is the 4th occurrence of 1 so far in the sequence, we have k = 4 and [kd] = 41; this 41 becomes a(6); etc.

Carole Dubois, Table of n, a(n) for n = 1..5001
Carole Dubois, Digit count for three sequences (see Xrefs)
Carole Dubois, Digit-count for this sequence and two others visible in the Xref section

Cf. A325721 and A325722 where the same idea is developed (addition and multiplication instead of concatenation).

A325721 Start the sequence with a(1) = 1 and read the digits one by one from there. The sequence is always extended with the sum d + k, d being the digit read and k the number of d digits present so far in the sequence.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 3, 1, 5, 4, 7, 6, 9, 8, 11, 10, 5, 6, 7, 2, 8, 9, 10, 4, 11, 12, 8, 3, 7, 9, 10, 11, 5, 12, 6, 11, 13, 12, 4, 13, 14, 9, 15, 6, 10, 16, 17, 18, 7, 19, 7, 8, 20, 8, 21, 9, 14, 22, 10, 11, 23, 5, 24, 12, 25, 12, 26, 13, 13, 27, 15, 14, 14, 8, 6, 15, 9, 28, 16, 29, 10, 10, 11, 30, 7, 31, 32, 12, 9, 11, 13, 11

Offset: 1

Eric Angelini and Carole Dubois, May 16 2019

			The sequence starts with a(1) = 1.
We read this 1, see that there is only one digit 1 so far in the sequence, thus k = 1; we have then d + k = (1 + 1) = 2 and this 2 becomes a(2);
We read this 2, see that there is only one digit 2 so far in the sequence, thus k = 1; we have then d + k = (2 + 1) = 3 and this 3 becomes a(3);
We read this 3, see that there is only one digit 3 so far in the sequence, thus k = 1; we have then d + k = (3 + 1) = 4 and this 4 becomes a(4);
...
We now read the first digit of a(10) = 10, see that this 1 is the 2nd digit 1 so far in the sequence, thus k = 2; we have then d + k = (1 + 2) = 3 and this 3 becomes a(11);
We now read the second digit of a(10) = 10, see that this 0 is the 1st digit 0 so far in the sequence, thus k = 1; we have then d + k = 1 and this 1 becomes a(12);
We now read the single digit of a(11) = 3; we see that this 3 is the 2nd digit 3 so far in the sequence, thus k = 2; we have then d + k = 5 and this 5 becomes a(13); etc.

Carole Dubois, Table of n, a(n) for n = 1..5001
Carole Dubois, Digit-count for this sequence and two others visible in the Xref section

Cf. A325722 where the same idea is developed, but with d*k instead of d+k. See also A308232 for the concatenation kd.

Cf. A322182.

cp's OEIS Frontend

A308232 Start the sequence with a(1) = 1 and read the digits one by one from there. The sequence is always extended with the concatenation kd, d being the digit that was read and k the number of d's present so far in the sequence.

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