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A136496 Solution of the complementary equation b(n)=a(a(n))+n; this is sequence b; sequence a is A136495.

%I A136496 #35 Mar 09 2025 12:51:36
%S A136496 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 A136496 79,81,85,87,90,94,96,99,103,107,109,113,115,118,122,124,127,131,135,
%U A136496 137,140,144,148,150,154,156,159,163,165,168,172,176,178,181,185,189
%N A136496 Solution of the complementary equation b(n)=a(a(n))+n; this is sequence b; sequence a is A136495.
%C A136496 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).
%C A136496 Position of the n-th occurrence of the digit 2 in A105083(n-1) for n>=1. - _Jeffrey Shallit_, Mar 08 2025
%D A136496 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.
%H A136496 Jeffrey Shallit, <a href="https://arxiv.org/abs/2503.01026">The Narayana Morphism and Related Words</a>, arXiv:2503.01026 [math.CO], 2025.
%F A136496 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 A136496 a(n) = A136495(n) + A005374(n-1) + n. - _Alan Michael Gómez Calderón_, Dec 23 2024
%e A136496 b(1) = a(a(1))+1 = a(1)+1 = 1+1 = 2;
%e A136496 b(2) = a(a(2))+2 = a(3)+2 = 4+2 = 6;
%e A136496 b(3) = a(a(3))+3 = a(4)+3 = 5+3 = 8;
%e A136496 b(4) = a(a(4))+4 = a(5)+4 = 7+4 = 11.
%Y A136496 Cf. A020942, A035513, A105083, A136189, A136495.
%K A136496 nonn
%O A136496 1,1
%A A136496 _Clark Kimberling_, Jan 01 2008