A360594 a(n) is the maximum number of locations 0..n-1 which can be visited in a single path starting from i=n-1, where jumps from location i to i +- a(i) are permitted (within 0..n-1) and each location can be visited up to 2 times.

0, 2, 1, 2, 4, 3, 8, 1, 2, 2, 4, 2, 8, 5, 4, 6, 13, 14, 14, 13, 13, 16, 22, 3, 17, 16, 20, 13, 13, 24, 22, 15, 24, 15, 14, 17, 14, 4, 15, 18, 23, 26, 28, 13, 16, 30, 28, 14, 15, 17, 16, 19, 16, 33, 18, 33, 32, 35, 39, 38, 40, 38, 39, 39, 36, 39, 38, 39, 41, 52

Offset

0,2

Comments

When a location is visited more than once, each such visit counts in a(n).

Links

Table of n, a(n) for n=0..69.

Example

For n=0 there are no locations 0..n-1, so the sole possible path is empty which is a(0) = 0 locations.
For n=1, the sole possible path is location 0 twice, 0 -> 0, which is 2 locations a(1) = 2.
For n=12, a path of a(12) = 8 locations visited starting from n-1 = 11 is:
  11 -> 9 -> 11 -> 9 -> 7 -> 8 -> 10 -> 6
Locations 11 and 9 are visited twice each and the others once.
  i    = 0  1  2  3  4  5  6  7  8  9 10 11
  a(i) = 0, 2, 1, 2, 4, 3, 8, 1, 2, 2, 4, 2
                                    2<----2
                                     ---->2
                              1<----2<----
                               ->2---->4
                           8<----------

Prog

(Python)
def A(lastn, times=2, mode=0):
  a, n=[0], 0
  while n<lastn:
    d, i, v, o, g, r=[[n]], 0, 1, [], 0, 0
    while len(d)>0:
      if len(d[-1])>v: v, o=len(d[-1]), d[-1][:]
      if d[-1][-1]-a[d[-1][-1]]>=0:
        if d[-1].count(d[-1][-1]-a[d[-1][-1]])<times:g=1
      if d[-1][-1]+a[d[-1][-1]]<=n:
        if d[-1].count(d[-1][-1]+a[d[-1][-1]])<times:
          if g>0: d.append(d[-1][:])
          d[-1].append(d[-1][-1]+a[d[-1][-1]])
          r=1
      if g>0:
        if r>0: d[-2].append(d[-2][-1]-a[d[-2][-1]])
        else: d[-1].append(d[-1][-1]-a[d[-1][-1]])
        r=1
      if r==0: d.pop()
      r, g=0, 0
    a.append(v)
    n+=1
    if mode==0: print(n, a[n])
    if mode>0:
      u, q=0, []
      while u<len(o):
        q.append(a[o[u]])
        u+=1
      print(n, a[n], q, o)
  return a

Crossrefs

Cf. A360593, A360595.

Sequence in context: A187500 A323460 A129144 * A295313 A105022 A263293

Adjacent sequences: A360591 A360592 A360593 * A360595 A360596 A360597

Keyword

nonn

Author

S. Brunner, Feb 13 2023

Status

approved