A227145 Numbers satisfying an infinite nested recurrence relation.

0, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 9, 9, 9, 9, 10, 10, 11, 11, 12, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 16, 17, 17, 17, 17, 18, 18, 19, 19, 20, 20, 20, 21, 21, 22, 22, 22, 22, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 27, 27

Offset: 1

Author

Frank Ruskey, Jul 04 2013

Keywords

• nonn

Comments

Conjecture: a(F_n) = F_{n-2} for n>1, where F_n is the n-th Fibonacci number.

Conjecture: a(n) ~ n*(3-sqrt(5))/2. -Jeffrey Shallit, Oct 12 2022

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000
• Marcel Celaya and Frank Ruskey, Morphic words and nested recurrence relations, arxiv 1307.0153 (Jun 29 2013), [math.CO] (see page 11).

Crossrefs

Cf. A060144.

Programs

Maple

a:= proc(n) option remember; local i, r, s;
      if n<2 then 0 else r, s:= n, 1;
         for i while s>0 do r, s:= r-s, (a@@i)(n-i) od: r
      fi
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jul 04 2013

Mathematica

a[n_] := a[n]= Which[n <= 1, 0,True, n - 1 -Sum[Nest[a, n - i, i], {i,1,n}]]; Table[a[i], {i, 0, 30}] (* José María Grau Ribas, Jul 10 2013 *)

Formula

a(n) = n - 1 - a(n-1) - a(a(n-2)) - a(a(a(n-3))) - a(a(a(a(n-4)))) - ... with a(n) = 0 if n <= 1.