%I #25 Jan 08 2023 22:47:06
%S 0,0,0,0,1,0,2,0,2,0,0,2,0,5,1,0,0,0,2,4,0,5,0,6,0,3,0,0,2,5,0,4,14,6,
%T 3,0,6,15,5,3,9,0,5,3,0,6,5,0,3,8,3,6,0,3,2,0,0,5,9,0,4,1,0,0,3,32,0,
%U 4,11,0,7,17,0,3,11,0,2,31,6,31,0,0,6,3,0,9,2,33,3,0,3,15,0,5
%N a(n) = b(nb(n)) where b is Van Eck's sequence A181391.
%C In Van Eck's sequence, b(n) is the distance between b(n1) and the previous occurrence of b(n1) there. Taking a(n) = b(nb(n)) is therefore the distance between the second and third last occurrence of b(n1) there.
%C If b(n1) has not yet occurred three times then the result is a(n) = 0 either by b(n)=0 when b(n1) has only occurred once, or b(nb(n)) = 0 when b(n1) has only occurred twice.
%e b(14) is 2, so we count back two steps to b(12), which is 5. Therefore a(14) = 5. As b(141) = b(13) = 0, the three occurrences of 0's are separated by b(14) and b(12) = a(14), that is, 2 and 5 steps:
%e .
%e n: ... 5 6 7 8 9 10 11 12 13 14 15 16 ...
%e .
%e <5><2>
%e   
%e   
%e b(n): ... 2, 0, 2, 2, 1, 6, 0, 5, 0, 2, 6, 5, ...
%e  
%e v 
%e a(14) = 5 <
%e two steps
%e back
%o (PARI) A181391_vec(N, a=0, i=Map())={vector(N, n, a=if(n>1, iferr(nmapget(i, a), E, 0)+mapput(i, a, n)))};
%o lista(nn) = my(v = A181391_vec(nn)); vector(#v, k, v[kv[k]]); \\ _Michel Marcus_, Dec 11 2022
%Y Cf. A181391.
%K nonn,easy
%O 1,7
%A _Tamas Sandor Nagy_, Dec 10 2022
