A322341 - OEIS

%I #8 Dec 04 2018 13:22:10

%S 1,2,3,4,5,6,7,8,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,

%T 45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75,77,79,81,83,85,87,89,

%U 91,93,95,97,99,102,26,107,110,14,38,58,119,122,62,18,16,10,12,135,138,20,143,146,74,30,76,22,52,159,162,82,167,170,86

%N The number of digits used in the sequence up to a(n) [a(n) included] is a multiple of a(n). The sequence starts with a(1) = 1 and is always extended with the smallest integer not yet present in the sequence that does not lead to a contradiction.

%C The sequence begins like A080676 but diverges from the 56th term on. The sequence is conjectured to be a permutation of A000027 (the positive integers): after computing 20000 terms, we see that the five smallest missing numbers are 2550, 2625, 3365, 3385 and 3399

%H Jean-Marc Falcoz, <a href="/A322341/b322341.txt">Table of n, a(n) for n = 1..20001</a>

%e a(1) = 1 as 1 digit is used so far [1] and a(1) is a multiple of 1;

%e a(2) = 2 as 2 digits are used so far [1 and 2] and a(2) is a multiple of 2;

%e a(3) = 3 as 3 digits are used so far [1, 2 and 3] and a(3) is a multiple of 3;

%e ...

%e a(10) is not equal to 10 as 11 digits would be used so far and 10 is not a multiple of 11;

%e a(10) = 11 as 11 digits are used so far [1, 2, 3, 4, 5, 6, 7, 8, 9, 1 and 1] and a(10) is a multiple of 11;

%e ...

%e a(56) = 26 as 104 digits are used so far [including a(56)] and 104 is a multiple of 26 (indeed 26*4 = 104);

%o (PARI) u=1; s=0; w=0; for (n=1, 84, for (v=u, oo, if (!bittest(s,v) && (w+(x=#digits(v)))%v==0, print1 (v ", "); s+=2^v; while (bittest(s,u), u++)

%o ; w+=x; break))) \\ _Rémy Sigrist_, Dec 04 2018

%Y Cf. A080676.

%K base,nonn,look

%O 1,2

%A _Eric Angelini_ and _Jean-Marc Falcoz_, Dec 04 2018