A361101 a(n) is the smallest positive number not among the terms in a(1..n-1) with index a(n-1)*k for any integer k; a(1)=1.

1, 2, 1, 3, 2, 1, 4, 1, 5, 1, 6, 2, 4, 4, 4, 4, 5, 3, 6, 4, 5, 3, 6, 4, 5, 3, 6, 5, 3, 7, 1, 8, 2, 6, 5, 3, 8, 2, 9, 1, 10, 2, 9, 1, 11, 4, 6, 5, 3, 8, 2, 9, 1, 12, 1, 13, 1, 14, 1, 15, 1, 16, 1, 17, 1, 18, 1, 19, 3, 8, 2, 10, 2, 11, 4, 6, 6, 6, 8, 2, 20, 3, 8, 3, 8, 3

Offset

1,2

Comments

From Samuel Harkness, Mar 02 2023: (Start)

This sequence is partly defined by the following cases for a(353) onwards. But what is the pattern to the 1s and 2s?

For k >= 0:

a(353+4k) = 1 or 2;

a(354+2k) = (354+2k)/2 - 63;

a(355+4k) = 1.

Of the 14912 n equal to 1 (mod 4) between 352 and 60000, 4983 (33.416%) equal 1 and 9929 (66.584%) equal 2. From observation it appears that 1/3 of a(353+4k); k>=0 equal 1 and 2/3 equal 2 (see Figure).

(End)

Links

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..8000

Samuel Harkness, Scatterplot for the first 600 terms

Samuel Harkness, Percentage of 1 mod 3 between 353 and 60000 equal to 1

Example

To find a(13), we look at the last term in the sequence thus far: (1, 2, 1, 3, 2, 1, 4, 1, 5, 1, 6, 2). Since it is a 2, the next term will be the smallest not among the even-indexed terms of the sequence thus far, which are (2, 3, 1, 1, 1, 2). 4 is the smallest missing number, so a(13)=4.

Mathematica

K = {1}; While[Length@K <= 85, A = {}; For[q = Last@K, q <= Length@K, q += Last@K, AppendTo[A, K[[q]]]]; k = 1; While[MemberQ[A, k], k++]; AppendTo[K, k]]; Print[K]  (* Samuel Harkness, Mar 06 2023 *)

Prog

(PARI) { p = 1; for (n = 1, #a = vector(86), x = 2^0; forstep (k = p, n-1, p, x = bitor(x, 2^a[k]); ); print1 (p = a[n] = valuation(1+x, 2)", "); ); } \\ Rémy Sigrist, Mar 02 2023

Crossrefs

Cf. A358921, A361172.

Sequence in context: A128260 A083368 A112379 * A308780 A246700 A073932

Adjacent sequences: A361098 A361099 A361100 * A361102 A361103 A361104

Keyword

nonn

Author

Neal Gersh Tolunsky, Mar 02 2023

Status

approved