A322341 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.

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

Offset

1,2

Comments

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

Links

Jean-Marc Falcoz, Table of n, a(n) for n = 1..20001

Example

a(1) = 1 as 1 digit is used so far [1] and a(1) is a multiple of 1;
a(2) = 2 as 2 digits are used so far [1 and 2] and a(2) is a multiple of 2;
a(3) = 3 as 3 digits are used so far [1, 2 and 3] and a(3) is a multiple of 3;
...
a(10) is not equal to 10 as 11 digits would be used so far and 10 is not a multiple of 11;
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;
...
a(56) = 26 as 104 digits are used so far [including a(56)] and 104 is a multiple of 26 (indeed 26*4 = 104);

Prog

(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++)
; w+=x; break))) \\ Rémy Sigrist, Dec 04 2018

Crossrefs

Cf. A080676.

Sequence in context: A225580 A071980 A058183 * A080676 A033061 A088380

Adjacent sequences: A322338 A322339 A322340 * A322342 A322343 A322344

Keyword

base,nonn,look

Author

Eric Angelini and Jean-Marc Falcoz, Dec 04 2018

Status

approved