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A384662 Solution of the complementary equation b(n)=a(a(n))+a(n)+2 with a(1)=1; this is the sequence b.

%I A384662 #7 Jun 16 2025 00:00:27
%S A384662 4,6,8,14,19,23,25,28,30,32,37,39,41,44,49,52,55,60,64,67,73,78,82,84,
%T A384662 87,89,94,99,103,106,110,113,115,118,122,124,129,131,135,138,140,142,
%U A384662 148,150,153,158,160,165,167,169,171,174,178,181,183,186,190,193
%N A384662 Solution of the complementary equation b(n)=a(a(n))+a(n)+2 with a(1)=1; this is the sequence b.
%C A384662 Sequence a is A384661.
%H A384662 Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. Article 07.1.4 (2007), 1-13.
%F A384662 {b(n)-b(n-1) : n>=2} = {2, 3, 4, 5, 6}.
%e A384662 b(1) = a(a(1))+a(1)+2 = 1+1+2 = 4;
%e A384662 b(2) = a(a(2))+a(2)+2 = 2+2+2 = 6;
%e A384662 b(3) = a(a(3))+a(3)+2 = 3+3+2 = 8;
%e A384662 b(4) = a(a(4))+a(4)+2 = 5+7+2 = 14.
%Y A384662 Cf. A136498, A136499, A136500, A384661, A384663, A384664.
%K A384662 nonn
%O A384662 1,1
%A A384662 _Clark Kimberling_, Jun 09 2025