cp's OEIS Frontend

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.

%I A360595 #28 Jul 20 2023 07:23:28
%S A360595 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,
%T A360595 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,
%U A360595 72,1,2,2,4,2,10,24,18,52
%N 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.
%C A360595 When a location is visited more than once, each such visit counts in a(n).
%C A360595 a(0)=0 is no terms before n=0 so an empty path.
%e A360595 For n=6, the following is the longest chain of jumps starting from i = n-1 = 5,
%e A360595   1  2  3  4  5   location number i
%e A360595   0, 3, 1, 2, 2   a(i)
%e A360595         1<----
%e A360595          ->2
%e A360595      3<----
%e A360595       ------->2
%e A360595         1<----
%e A360595          ->2
%e A360595      3<----
%e A360595       ------->2
%e A360595         1<----
%e A360595          ->2
%e A360595      3<----
%e A360595 It visited the terms 2,1,2,3 three times in a loop, which gives a total of 12 terms, so a(6)=12.
%o A360595 (Python)
%o A360595 def A(lastn,times=3,mode=0):
%o A360595   a,n=[0],0
%o A360595   while n<lastn:
%o A360595     d,i,v,o,g,r=[[n]],0,1,[],0,0
%o A360595     while len(d)>0:
%o A360595       if len(d[-1])>v: v,o=len(d[-1]),d[-1][:]
%o A360595       if d[-1][-1]-a[d[-1][-1]]>=0:
%o A360595         if d[-1].count(d[-1][-1]-a[d[-1][-1]])<times:g=1
%o A360595       if d[-1][-1]+a[d[-1][-1]]<=n:
%o A360595         if d[-1].count(d[-1][-1]+a[d[-1][-1]])<times:
%o A360595           if g>0: d.append(d[-1][:])
%o A360595           d[-1].append(d[-1][-1]+a[d[-1][-1]])
%o A360595           r=1
%o A360595       if g>0:
%o A360595         if r>0: d[-2].append(d[-2][-1]-a[d[-2][-1]])
%o A360595         else: d[-1].append(d[-1][-1]-a[d[-1][-1]])
%o A360595         r=1
%o A360595       if r==0: d.pop()
%o A360595       r,g=0,0
%o A360595     a.append(v)
%o A360595     n+=1
%o A360595     if mode==0: print(n+1,a[n])
%o A360595     if mode>0:
%o A360595       u,q=0,[]
%o A360595       while u<len(o):
%o A360595         q.append(a[o[u]])
%o A360595         u+=1
%o A360595       print(n+1,a[n],q,o)
%o A360595   return a
%Y A360595 Cf. A360593.
%K A360595 nonn
%O A360595 1,2
%A A360595 _S. Brunner_, Feb 14 2023