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A317127 a(0) = a(1) = a(2) = 1; for n >= 3, a(n) = freq(a(n-1),n) + freq(a(n-3),n) where freq(i, j) is the number of times i appears in the terms a(0) .. a(j-1).

%I A317127 #29 Jul 16 2024 04:19:42
%S A317127 1,1,1,6,4,4,3,3,4,5,3,6,3,8,3,10,2,6,4,5,5,7,4,8,3,11,3,14,2,9,2,6,5,
%T A317127 7,6,9,4,11,4,14,4,16,3,16,4,17,3,18,2,13,2,10,3,15,3,22,2,17,3,18,4,
%U A317127 22,4,22,6,17,6,14,6,16,6,18,6,20,4,22,5,17,8,8,8,10,8,12,4,19,2,20,3,20,6,24,4
%N A317127 a(0) = a(1) = a(2) = 1; for n >= 3, a(n) = freq(a(n-1),n) + freq(a(n-3),n) where freq(i, j) is the number of times i appears in the terms a(0) .. a(j-1).
%C A317127 Inspired by A316774.
%C A317127 In this sequence, it is obvious that we have exactly three 1’s that are a(0) = a(1) = a(2) = 1. Can we determine the frequency characteristics of some other positive integers? For example, are there infinitely many 2's in this sequence?
%H A317127 Antti Karttunen, <a href="/A317127/b317127.txt">Table of n, a(n) for n = 0..65537</a>
%H A317127 Altug Alkan, <a href="/A317127/a317127.png">A line graph of a(n) for n <= 500</a>
%p A317127 b:= proc() 0 end:
%p A317127 a:= proc(n) option remember; local t;
%p A317127       t:= `if`(n<3, 1, b(a(n-1))+b(a(n-3)));
%p A317127       b(t):= b(t)+1; t
%p A317127     end:
%p A317127 seq(a(n), n=0..100); # after _Alois P. Heinz_ at A316774
%t A317127 c = <||>; f[n_] := If[KeyExistsQ[c,n],c[n],0]; a[n_] := a[n] = Block[{v}, v = If[n<3, 1, f[a[n-1]] + f[a[n-3]]]; If[f[v]>0, c[v] = c[v]+1, c[v]=1]; v]; Array[a, 93, 0] (* _Giovanni Resta_, Jul 24 2018 *)
%o A317127 (PARI)
%o A317127 up_to = 5000;
%o A317127 listA317127off1(up_to) = { my(v = vector(up_to),c); v[1] = v[2] = v[3] = 1; for(n=4,up_to, c=0; for(k=1,(n-1), c += ((v[k]==v[n-1])+(v[k]==v[n-3]))); v[n] = c); (v); };
%o A317127 listA317127off1(up_to) = { my(v = vector(up_to), m = Map(), c); v[1] = v[2] = v[3] = 1; mapput(m, 1, 3); for(n=4,up_to, c = (mapget(m, v[n-1])+mapget(m,v[n-3])); v[n] = c; mapput(m, c, if(!mapisdefined(m, c), 1, 1+mapget(m, c)))); (v); }; \\ Faster!
%o A317127 v317127 = listA317127off1(1+up_to);
%o A317127 A317127(n) = v317127[1+n]; \\ _Antti Karttunen_, Jul 23 2018
%Y A317127 Cf. A316774.
%K A317127 nonn,look
%O A317127 0,4
%A A317127 _Altug Alkan_, Jul 21 2018