A305876 a(n) = Fibbinary(2^n).

1, 2, 5, 16, 36, 84, 273, 648, 2114, 4757, 16516, 37161, 87045, 282896, 673924, 2184233, 5263877, 17107472, 38830244, 134554132, 303080705, 707272770, 2300725397, 5457925252, 17805431433, 42970665029, 139654661284, 314223120404, 1099646108737, 2474203744786

Offset: 0

Author

Alois P. Heinz, Jun 12 2018

Keywords

• nonn
• base

Examples

			a(6) = A003714(2^6) = A003714(64) = 273 = 100010001_2 because F(0+2) + F(4+2) + F(8+2) = 1 + 8 + 55 = 64, where 0, 4, 8 are the indices of 1 bits in 100010001_2.  A014417(64) = 100010001 = A007088(273).
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..2306

Crossrefs

Cf. A000045, A000079, A003714 (Fibbinary), A007088, A014417, A305380.

Programs

Maple

F:= proc(n) F(n):= `if`(n<2, n, F(n-1)+F(n-2)) end:
b:= proc(n) local j;
      if n=0 then 0
    else for j from 2 while F(j+1)<=n do od;
         b(n-F(j))+2^(j-2)
      fi
    end:
a:= n-> b(2^n):
seq(a(n), n=0..35);

Python

def A305876(n):
    m, tlist, s = 2**n, [1,2], 0
    while tlist[-1]+tlist[-2] <= m:
        tlist.append(tlist[-1]+tlist[-2])
    for d in tlist[::-1]:
        s *= 2
        if d <= m:
            s += 1
            m -= d
    return s # Chai Wah Wu, Jun 14 2018

Formula

a(n) = A003714(2^n).

A014417(2^n) = A007088(a(n)).