A376424 Nonnegative numbers m such that the run lengths in binary expansion of m, say (r_1, ..., r_k), satisfy r_1 + ... + r_i <> r_j + ... + r_k for any i in the interval 1..k-1 and j in the interval 2..k.

0, 1, 3, 4, 6, 7, 8, 14, 15, 16, 19, 23, 24, 25, 28, 29, 30, 31, 32, 35, 47, 48, 49, 60, 61, 62, 63, 64, 67, 68, 71, 76, 79, 80, 88, 95, 96, 97, 102, 103, 110, 111, 112, 113, 114, 115, 120, 121, 122, 123, 124, 125, 126, 127, 128, 131, 132, 135, 156, 159, 160

Offset

1,3

Comments

Visually, if we consider a row of bricks of widths r_1, ..., r_k (in that order) above a row of widths r_k, ..., r_1 (in that order), we never have 4 bricks whose corners meet.

There are A108411(k) terms with k binary digits (ignoring leading zeros).

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Index entries for sequences related to binary expansion of n

Example

The binary expansion of 35 is "100011", the corresponding run lengths are (1, 3, 2); the sums 1, 1+3 are distinct from the sums 3+2, 2, so 35 is a term. Visually, if we consider a row of bricks of widths 1, 3, 2 (in that order) above a row of widths 2, 3, 1 (in that order), we never have 4 bricks whose corners meet:
    .-.-----.---.
    | |     |   |
    .-.-.---.-.-.
    |   |     | |
    .---.-----.-.

Prog

(PARI) toruns(n) = { my (r = []); while (n, my (v = valuation(n+n%2, 2)); n \= 2^v; r = concat(v, r)); r }
is(n) = { if (n, my (r = toruns(n)); setintersect(vector(#r, k, vecsum(r[1..k])), vector(#r, k, vecsum(r[#r+1-k..#r])))==[vecsum(r)], 1); }

Crossrefs

Cf. A101211, A108411.

Sequence in context: A239458 A007370 A322376 * A044813 A154661 A275672

Adjacent sequences: A376421 A376422 A376423 * A376425 A376426 A376427

Keyword

nonn,base

Author

Rémy Sigrist, Sep 22 2024

Status

approved