cp's OEIS Frontend

For clarification:

The terms of the sequence so far are written as a one-dimensional grid, and from every term a(i) you can jump back or forth a(i) terms, without i getting < 0 or > n, as these terms don't exist. Then search for the longest possible chain of jumps you can do starting at a(n) and visiting no term more than once. The number of targets visited by this chain is the next term of the sequence.

A chain of jumps C always starts at n, with C1 = a(n-1). Then the first jump always has to go back C1 terms, so C2 = a(n-1-C1) = a(n-1-a(n-1)), and then it continues with jumping back or forth C2 terms, whichever produces the longest chain of jumps:

C3 = a(n-1-C1-C2) or a(n-1-C1+C2),

C4 = a(n-1-C1{-/+}C2{-/+}C3), etc.

The first numbers which appear to be missing from this sequence are:

3, 13, 15, 17, 20, 25, 27, 32, 33, 36, 43, 44, 48, 50, 51, 62, 63, 69, 70, 71, 75, 77, 78, 80, etc.

a(10)=7 is the earliest term whose solution cannot be represented by a single path in which each index is visited once.

a(n) is also the smallest number of terms you can reach from any starting term in the sequence so far. This is true because every term leads back to a(1)=1.

Note that each location can visit up to two terms (doesn't have to be a path), although in this case the example sections shows a path.

a(21)=13 is the earliest term whose solution cannot be represented by a single path in which each index is visited once (found by Kevin Ryde).

Slabs on the sidewalk are numbered n = 0, 1, 2,... and each has a label L(n) = 1, 1, 2, 2, 3, 3,...

At a given slab, a jump can be taken forward or backward by L(n) places (but not back before slab 0).

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For every n >= 0,

let L(n) = 1 + floor(n/2) -- the label on slab n,

let forward(n) = n + L(n) -- jumping forward,

if n > 0, let backward(n) = n - L(n) -- jumping backward,

let lambda(n) = floor((2/3)*n),

let mu(n) = 1 + 2*n,

let nu(n) = 2 + 2*n.

Observe that given n >= 0, there are exactly two ways of landing onto slab n with a direct jump backwards:

backward-jumping from mu(n) to n, and

backward-jumping from nu(n) to n.

If n is a multiple of 3, there is no other ways of jumping onto slab n. But if n is not a multiple of 3, there is one additional way:

forward-jumping from lambda(n) to n.

(Note that L is A008619, forward(n) == A006999(n+1), lambda is A004523, mu is A005408, nu is A299174.)

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Every slab n > 0 is reachable from slab 0, since there always exists some slab s < n which reaches n by one or more jumps:

if n != 0 (mod 3), then s = lambda(n) = floor((2/3)*n) takes one forward jump to n,

if n == 0 (mod 3) but n != 0 (mod 9), then s = lambda o lambda o mu(n) = floor((8/9)*n) takes two forward jumps and one backward jump to n,

if n == 0 (mod 9), then s = lambda o lambda o nu(n) = floor((8/9)*n + 6/9) takes two forward jumps and one backward jump to n.

This demonstrates that the sequence never stops.

This also gives the following bounds:

a(n) <= 1 + (4/3)*n,

a(n) <= 6*log(n)/log(9/8).

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The sequence is a surjective mapping N -> N, since given any n >= 0:

a(forward^n(0)) == n.

Note that location n-1 itself is counted as a term which can reach i=n-1.

Conjecture: a(n) is also the largest number such that starting point i=n can reach every previous location (with a(1)=1 and the same rule for jumps as in the current name).

A047619 appears to be the indices of 1's in this sequence.

A023758 appears to be the indices of terms for which a(n)=n-1.

A089633 appears to be the distinct values of the sequence (and its complement A158582 the missing values).

The sequence appears to consist of monotonically increasing runs of length 4.

It appears that a(A004767(n))=A100892(n) and a(A016825(n))=A100892(n)-1.

The sequence is 1244 initial terms followed by a repeating block of 4925 terms so that a(n) = a(n-4925) for n >= 6170. - Kevin Ryde, Jul 31 2023

For clarification: We start at the term with index a(n-1). From each term at index i, we can jump to up to two locations, which are a(i) terms away in either direction. We continue this process from the terms we have reached until we have visited all possible terms.

The radius of the sequence's digraph is the smallest eccentricity of any vertex (location) in the graph. The eccentricity of a location i means the largest number of jumps in the shortest path from location i to any other location.

The diameter of the sequence's digraph is the largest eccentricity of any vertex (location) in the graph. The eccentricity of a location i means the largest number of jumps in the shortest path from location i to any other location.