A360594 -id:A360594 - cp's OEIS frontend

A360746 a(n) is the maximum number of locations 1..n-1 which can be reached starting from a(n-1), where jumps from location i to i +- a(i) are permitted (within 1..n-1); a(1)=1. See example.

1, 1, 2, 3, 4, 4, 5, 5, 5, 7, 8, 8, 8, 9, 9, 12, 10, 10, 12, 10, 12, 13, 13, 13, 16, 14, 14, 16, 17, 17, 17, 18, 18, 24, 25, 25, 25, 26, 27, 27, 27, 27, 28, 28, 30, 28, 33, 28, 29, 30, 30, 30, 33, 31, 31, 31, 32, 32, 33, 33, 31, 31, 32, 33, 33, 35, 33, 37

Offset: 1

Neal Gersh Tolunsky, Feb 18 2023

nonn

			a(7)=5 because we reach 5 terms starting from the most recent term a(6) (each line shows the next unvisited term(s) we can reach from the term(s) in the previous iteration):
1, 1, 2, 3, 4, 4
   1<----------4
1, 1, 2, 3, 4, 4
1<-1->2
1, 1, 2, 3, 4, 4
      2---->4
From the last iteration we can visit no new terms. We reached 5 terms, so a(7)=5:
1, 1, 2, 3, 4, 4
1  1  2     4  4

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program

Cf. A360744, A360745, A360593, A360594, A360595, A359005, A358838, A359008.

PARI
```
See Links section.
```

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 26 2023

A360593 Each term a(i) can reach a(i+a(i)) and a(i-a(i)) if these terms exist. a(n) is the greatest number of terms among a(1..n-1) that can be reached by starting at a(n-1) and visiting no term more than once; a(0)=0. See example.

Original entry on oeis.org

0, 1, 2, 2, 4, 2, 6, 2, 7, 5, 6, 6, 9, 10, 10, 6, 8, 7, 9, 8, 11, 8, 12, 14, 12, 14, 19, 16, 19, 14, 14, 16, 14, 21, 14, 16, 21, 14, 14, 16, 14, 18, 14, 16, 21, 21, 19, 21, 22, 22, 21, 23, 24, 24, 29, 29, 22, 26, 24, 28, 24, 26, 31, 24, 31, 34, 24, 30, 34, 29, 39

Offset: 0

S. Brunner, Feb 13 2023

nonn

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.

			This example shows the longest chain of jumps starting with a(7)=2:
0, 1, 2, 2, 4, 2, 6, 2
   1<----2<----2<----2
    ->2---->4
0<----------
It visited the 7 terms 2,2,2,1,2,4,0. So a(8)=7.

S. Brunner, Table of n, a(n) for n = 0..217

Cf. A360594, A360595, A360744, A360745, A360746.

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

A361383 a(n) is the number of locations 1..n-1 which can be reached starting from location i=a(n-1), where jumps from location i to i +- a(i) are permitted (within 1..n-1); a(1)=1. See example.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 4, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 15, 16, 15, 16, 15, 18, 17, 18, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 22, 24, 22, 24, 23, 24, 26, 26, 26, 26, 26, 26, 26, 26, 26, 29, 32, 33, 35, 32, 35, 32, 35, 32, 35, 32, 35, 32, 36, 35, 37

Offset: 1

Neal Gersh Tolunsky, Mar 09 2023

nonn

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.

			We find a(8)=4 by first looking at the previous term in the sequence so far (1,1,2,3,3,4,5), which is a(7)=5. This tells us to start at location i=5. Permitted steps can reach 4 locations as follows:
  1, 1, 2, 3, 3, 4, 5
     1<-------3
  1, 1, 2, 3, 3, 4, 5
  1<-1->2
Steps from each of these locations cannot reach anything new, so a(8)=4. The reachable terms are:
  1, 1, 2, 3, 3, 4, 5
  1  1  2     3

Samuel Harkness, Table of n, a(n) for n = 1..10000

Cf. A360744, A360745, A360746, A360593, A360594, A360595, A358838.

More terms from Samuel Harkness, Mar 10 2023

cp's OEIS Frontend

A360746 a(n) is the maximum number of locations 1..n-1 which can be reached starting from a(n-1), where jumps from location i to i +- a(i) are permitted (within 1..n-1); a(1)=1. See example.

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A360593 Each term a(i) can reach a(i+a(i)) and a(i-a(i)) if these terms exist. a(n) is the greatest number of terms among a(1..n-1) that can be reached by starting at a(n-1) and visiting no term more than once; a(0)=0. See example.

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A361383 a(n) is the number of locations 1..n-1 which can be reached starting from location i=a(n-1), where jumps from location i to i +- a(i) are permitted (within 1..n-1); a(1)=1. See example.

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