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

Author

Neal Gersh Tolunsky, Feb 18 2023

Keywords

• nonn

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).

Examples

			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
		

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

Crossrefs

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

Programs

PARI

\\ See links.

Python

def A(lastn,mode=0):
  a,n,t=[1],0,1
  while n0:
      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]])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