%I #11 Jun 05 2022 08:32:32
%S 4,7,11,6,18,9,10,29,12,15,16,14,47,17,20,24,19,26,22,23,76,25,28,32,
%T 27,39,30,31,42,33,36,37,35,123,38,41,45,40,52,43,44,63,46,49,50,48,
%U 68,51,54,58,53,60,56,57,199,59,62,66,61,73,64,65,84,67,70,71,69,102,72,75
%N a(n) is the value of the cell in the Wythoff array that lies in the next row and same column as the cell containing n.
%C From _Kevin Ryde_, Jun 05 2022: (Start)
%C a(n) is n with the "odd" part (A348853) of its Zeckendorf representation increased to the next greater "odd" number.
%C This increase is Zeckendorf digits +10 or +100 at the odd part, according to whether the final digits there are ..101 or ..001, respectively.
%C A354321(n) is the first of those three digits so that a(n) = n + Fibonacci(A035612(n) + 3 - A354321(n)).
%C (End)
%e The Wythoff array (A035513 or A083412) begins:
%e 1 2 3 5 8 ...
%e 4 7 11 18 29 ...
%e 6 10 16 26 42 ...
%e ...
%e so a(1) = 4, a(2) = 7, a(3) = 11, a(4) = 6, ...
%o (PARI) T(n,k) = (n+sqrtint(5*n^2))\2*fibonacci(k+1) + (n-1)*fibonacci(k); \\ A035513
%o cell(n) = for (r=1, oo, for (c=1, oo, if (T(r,c) == n, return([r, c])); if (T(r,c) > n, break););); \\ see A003603 and A035612
%o a(n) = {my(pos = cell(n)); T(pos[1]+1, pos[2]);}
%o (PARI) { my(phi=quadgen(5),s=phi-1,c=2*phi-3);
%o a(n) = my(t=n,k=3,r);
%o until(r<s, [t,r]=divrem(t+1,phi); k++);
%o n + fibonacci(k - (r<c)); }
%Y Cf. A035513 and A083412 (Wythoff array), A003603 (row number), A035612 (column number).
%Y Cf. A348853 (odd part), A354321 (above 01), A000045 (Fibonacci numbers).
%Y Cf. A022342 (same row, next column).
%Y Cf. A349102 (binary increase odd).
%K nonn,easy
%O 1,1
%A _Michel Marcus_, Mar 21 2022