%I #23 Jan 06 2025 08:59:21
%S 2,6,8,11,15,19,21,25,27,30,34,36,39,43,47,49,52,56,60,62,66,68,71,75,
%T 79,81,85,87,90,94,96,99,103,107,109,113,115,118,122,124,127,131,135,
%U 137,140,144,148,150,154,156,159,163,165,168,172,176,178,181,185,189
%N Solution of the complementary equation b(n)=a(a(n))+n; this is sequence b; sequence a is A136495.
%C b = 1 + (column 1 of Z) = 1 + A020942. The pair (a,b) also satisfy the following complementary equations: b(n)=a(a(a(n)))+1; a(b(n))=a(n)+b(n); b(a(n))=a(n)+b(n)-1; (and others).
%D Clark Kimberling and Peter J. C. Moses, Complementary equations and Zeckendorf arrays, in Applications of Fibonacci Numbers, vol.10, Proceedings of the Thirteenth International Conference on Fibonacci Numbers and Their Applications, William Webb, editor, Congressus Numerantium, Winnipeg, Manitoba 201 (2010) 161-178.
%F Let Z = (3rd order Zeckendorf array) = A136189. Then a = ordered union of columns 1,3,4,6,7,9,10,12,13,... of Z, b = ordered union of columns 2,5,8,11,14,... of Z.
%F a(n) = A136495(n) + A005374(n-1) + n. - _Alan Michael Gómez Calderón_, Dec 23 2024
%e b(1) = a(a(1))+1 = a(1)+1 = 1+1 = 2;
%e b(2) = a(a(2))+2 = a(3)+2 = 4+2 = 6;
%e b(3) = a(a(3))+3 = a(4)+3 = 5+3 = 8;
%e b(4) = a(a(4))+4 = a(5)+4 = 7+4 = 11.
%Y Cf. A020942, A035513, A136189, A136495.
%K nonn
%O 1,1
%A _Clark Kimberling_, Jan 01 2008