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A384664 Solution of the complementary equation b(n)=a(a(n))+a(n)+3 with a(1)=1; this is the sequence b. Sequence a is A384663.

%I A384664 #5 Jun 22 2025 18:25:07
%S A384664 5,7,9,11,17,23,27,31,34,36,38,40,45,47,49,52,54,59,61,64,70,73,75,79,
%T A384664 82,88,93,98,102,105,108,110,115,119,124,126,132,136,139,141,144,148,
%U A384664 152,155,159,161,164,166,168,173,175,180,185,187,190,195,197,202
%N A384664 Solution of the complementary equation b(n)=a(a(n))+a(n)+3 with a(1)=1; this is the sequence b. Sequence a is A384663.
%H A384664 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 A384664 {b(n)-b(n-1) : n>=2} = {2, 3, 4, 5, 6}.
%e A384664 b(1) = a(a(1))+a(1)+3 = 1+1+3 = 5;
%e A384664 b(2) = a(a(2))+a(2)+3 = 2+2+3 = 7;
%e A384664 b(3) = a(a(3))+a(3)+3 = 3+3+3 = 9;
%e A384664 b(4) = a(a(4))+a(4)+3 = 4+4+3 = 11;
%e A384664 b(5) = a(a(5))+a(5)+3 = 6+8+3 = 17.
%Y A384664 Cf. A136499, A136500, A384661, A384662, A384663.
%K A384664 nonn
%O A384664 1,1
%A A384664 _Clark Kimberling_, Jun 16 2025