OFFSET
1,1
COMMENTS
From Kevin Ryde, Jun 05 2022: (Start)
a(n) is n with the "odd" part (A348853) of its Zeckendorf representation increased to the next greater "odd" number.
This increase is Zeckendorf digits +10 or +100 at the odd part, according to whether the final digits there are ..101 or ..001, respectively.
A354321(n) is the first of those three digits so that a(n) = n + Fibonacci(A035612(n) + 3 - A354321(n)).
(End)
EXAMPLE
PROG
(PARI) T(n, k) = (n+sqrtint(5*n^2))\2*fibonacci(k+1) + (n-1)*fibonacci(k); \\ A035513
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
a(n) = {my(pos = cell(n)); T(pos[1]+1, pos[2]); }
(PARI) { my(phi=quadgen(5), s=phi-1, c=2*phi-3);
a(n) = my(t=n, k=3, r);
until(r<s, [t, r]=divrem(t+1, phi); k++);
n + fibonacci(k - (r<c)); }
KEYWORD
nonn,easy
AUTHOR
Michel Marcus, Mar 21 2022
STATUS
approved