OFFSET
1,2
COMMENTS
Conjecture 1: a(n) <= floor((4*n-1)/3).
Conjecture 2: The sequence L(k) = 9^(k-1) + 2 describes the number of consecutive terms that match the formula a(n) = floor((4*n-1)/3) : (Start)
[a(1)..a(3)] : L(1) = 9^(1-1) + 2 = 3
[a(8)..a(18)] : L(2) = 9^(2-1) + 2 = 11
[a(71)..a(153)] : L(3) = 9^(3-1) + 2 = 83
[a(908)..a(1638)] : L(4) = 9^(4-1) + 2 = 731
[a(5741)..a(12303)] : L(5) = 9^(5-1) + 2 = 6563
[a(51668)..a(110718)] : L(6) = 9^(6-1) + 2 = 59051
[a(465011)..a(996453)] : L(7) = 9^(7-1) + 2 = 531443
... (End).
PROG
(PARI)
a_seq(N) = {my(a = [1], used = Set([1]), forb = Set([]), maxv = 2); while(#a < N, my(cand = a[#a] + 1); while(1, if( setsearch(used, cand)==0 && setsearch(forb, cand)==0, break); cand++; ); a = concat(a, cand); used = setunion(used, [cand]); my(m = #a); if(m >= 3, my(s = a[m-2] + a[m-1] + a[m]); forb = setunion(forb, [s]); ); if(m >= 2, my(s = a[m-1] + a[m] + (m+1 <= N)); ); if(m >= 1 && m+1 <= N && m+2 <= N, 0); ); a; };
A394576 = a_seq(1000);
vector(100, n, A394576[n]) \\ Hoang Xuan Thanh, May 14 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Hoang Xuan Thanh, May 11 2026
STATUS
approved
