A325721 - OEIS

%I #38 May 25 2019 12:12:10

%S 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,

%T 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,

%U 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

%N 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.

%H Carole Dubois, <a href="/A325721/b325721.txt">Table of n, a(n) for n = 1..5001</a>

%H Carole Dubois, <a href="/A325721/a325721.png">Digit-count for this sequence and two others visible in the Xref section</a>

%e The sequence starts with a(1) = 1.

%e 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);

%e 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);

%e 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);

%e ...

%e 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);

%e 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);

%e 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.

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

%Y Cf. A322182.

%K base,nonn

%O 1,2

%A _Eric Angelini_ and _Carole Dubois_, May 16 2019