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A329934 a(1)=1, a(2)=1, a(n) = (number of times a(n-1) has appeared before) + (number of times a(n-2) has appeared before).

%I A329934 #18 Dec 14 2019 03:31:10
%S A329934 1,1,4,3,2,2,4,4,6,4,5,5,4,7,6,3,4,8,7,3,5,6,6,8,6,7,8,6,9,7,5,8,8,10,
%T A329934 6,8,13,7,6,13,10,4,9,9,6,12,10,4,11,9,5,9,10,9,10,11,7,8,13,10,9,13,
%U A329934 11,7,10,14,8,9,16,9,10,17,9,11,14,6,12,12,6,14,14
%N A329934 a(1)=1, a(2)=1, a(n) = (number of times a(n-1) has appeared before) + (number of times a(n-2) has appeared before).
%C A329934 Conjecture: This sequence grows logarithmically.
%H A329934 Joshua Oliver, <a href="/A329934/b329934.txt">Table of n, a(n) for n = 1..4000</a>
%H A329934 Rémy Sigrist, <a href="/A329934/a329934.png">Density plot of the first 10000000 terms</a>
%e A329934 a(n)=4 where n=3 because 1 (a(n-1)) has appeared twice before, and 1 (a(n-2)) has appeared twice before as well. 2+2 = 4.
%p A329934 b:= proc(n) option remember; `if`(n=0, 0, b(n-1)+x^a(n)) end:
%p A329934 a:= proc(n) option remember; `if`(n<3, 1, (p->
%p A329934       coeff(p, x, a(n-1))+coeff(p, x, a(n-2)))(b(n-1)))
%p A329934     end:
%p A329934 seq(a(n), n=1..120);  # _Alois P. Heinz_, Nov 24 2019
%t A329934 A={1,1};For[n=3,n<=81,n++,A=Append[A,Sum[Count[Table[Part[A,i],{i,1,n-1}],Part[A,n-k]],{k,2}]]];A
%o A329934 (PARI) o=vector(17); for (n=1, 81, print1 (v=if (n<3, 1, o[pp]+o[p]) ", "); o[v]++; [pp,p]=[p,v]) \\ _Rémy Sigrist_, Nov 27 2019
%Y A329934 Cf. A158416, A306246, A316774.
%K A329934 nonn
%O A329934 1,3
%A A329934 _Joshua Oliver_, Nov 24 2019