A360745 a(n) is the maximum number of locations 1..n-1 which can be reached starting from a(1)=1, where jumps from location i to i +- a(i) are permitted (within 1..n-1). See example.

1, 1, 2, 3, 3, 4, 4, 4, 7, 7, 7, 8, 9, 9, 9, 9, 9, 9, 12, 12, 13, 13, 13, 13, 13, 14, 14, 17, 17, 17, 17, 17, 24, 24, 24, 25, 25, 26, 27, 27, 28, 29, 29, 29, 29, 29, 29, 29, 29, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 33, 33, 33, 34, 34, 37, 37, 37, 38, 38, 48, 48, 48, 48, 48, 49, 50, 51, 52, 53, 53, 53

Offset

1,3

Comments

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).

Links

Kevin Ryde, Table of n, a(n) for n = 1..10000

Kevin Ryde, PARI/GP Code

Neal Gersh Tolunsky, Graph of run lengths of a(n) for n = 1..50004

Example

a(13)=9 because we can reach 9 terms starting from a(1) as follows:
1, 1, 2, 3, 3, 4, 4, 4, 7, 7, 7, 8
1->1->2---->3------->4---------->8
1, 1, 2, 3, 3, 4, 4, 4, 7, 7, 7, 8
         3<----------------------8
1, 1, 2, 3, 3, 4, 4, 4, 7, 7, 7, 8
         3------->4---------->7
This is a total of 9 terms:
1, 1, 2, 3, 3, 4, 4, 4, 7, 7, 7, 8
1  1  2  3  3     4  4        7  8

Prog

(Python)
def A(lastn, mode=0):
  a, n, t=[1], 0, 1
  while n<lastn:
    d, g, r, rr=[[0]], 0, 0, [0]
    while len(d)>0:
      if not d[-1][-1] in rr:rr.append(d[-1][-1])
      if d[-1][-1]-a[d[-1][-1]]>=0:
        if d[-1].count(d[-1][-1]-a[d[-1][-1]])<t:g=1
      if d[-1][-1]+a[d[-1][-1]]<=n:
        if d[-1].count(d[-1][-1]+a[d[-1][-1]])<t:
          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(len(rr))
    n+=1
    print(n+1, a[n])
    if mode>0: print(a)
  return a  # S. Brunner, Feb 19 2023
(PARI) See links.

Crossrefs

Cf. A360744, A360746, A361383, A360593, A359005, A358838, A359008.

Sequence in context: A086333 A240792 A367129 * A217121 A368423 A253783

Adjacent sequences: A360742 A360743 A360744 * A360746 A360747 A360748

Keyword

nonn

Author

Neal Gersh Tolunsky, Feb 18 2023

Status

approved