A360595 a(n) is the maximum number of locations 1..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 1..n-1) and a term can be visited up to three times.

0, 3, 1, 2, 2, 12, 1, 2, 2, 4, 2, 10, 15, 1, 2, 2, 4, 2, 10, 20, 1, 2, 2, 4, 2, 10, 13, 8, 2, 10, 2, 15, 7, 15, 25, 17, 53, 1, 2, 2, 4, 2, 10, 65, 1, 2, 2, 4, 2, 10, 13, 8, 2, 10, 2, 15, 7, 15, 72, 1, 2, 2, 4, 2, 10, 24, 18, 52

Offset: 1

Author

S. Brunner, Feb 14 2023

Keywords

• nonn

Comments

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

a(0)=0 is no terms before n=0 so an empty path.

Examples

			For n=6, the following is the longest chain of jumps starting from i = n-1 = 5,
  1  2  3  4  5   location number i
  0, 3, 1, 2, 2   a(i)
        1<----
         ->2
     3<----
      ------->2
        1<----
         ->2
     3<----
      ------->2
        1<----
         ->2
     3<----
It visited the terms 2,1,2,3 three times in a loop, which gives a total of 12 terms, so a(6)=12.
		

Crossrefs

Cf. A360593.

Programs

Python

def A(lastn,times=3,mode=0):
  a,n=[0],0
  while n0:
      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]])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+1,a[n])
    if mode>0:
      u,q=0,[]
      while u