A022290 Replace 2^k in binary expansion of n with Fibonacci(k+2).

0, 1, 2, 3, 3, 4, 5, 6, 5, 6, 7, 8, 8, 9, 10, 11, 8, 9, 10, 11, 11, 12, 13, 14, 13, 14, 15, 16, 16, 17, 18, 19, 13, 14, 15, 16, 16, 17, 18, 19, 18, 19, 20, 21, 21, 22, 23, 24, 21, 22, 23, 24, 24, 25, 26, 27, 26, 27, 28, 29, 29, 30, 31, 32, 21, 22, 23, 24, 24, 25, 26

Offset: 0

Author

Marc LeBrun

Keywords

• nonn
• tabf
• base

Examples

			n=4 = 2^2 is replaced by A000045(2+2) = 3. n=5 = 2^2 + 2^0 is replaced by A000045(2+2) + A000045(0+2) = 3+1 = 4. - _R. J. Mathar_, Jan 31 2015
From _Philippe Deléham_, Jun 05 2015: (Start)
This sequence regarded as a triangle with rows of lengths 1, 1, 2, 4, 8, 16, ...:
  0
  1
  2, 3
  3, 4, 5, 6
  5, 6, 7, 8, 8, 9, 10, 11
  8, 9, 10, 11, 11, 12, 13, 14, 13, 14, 15, 16, 16, 17, 18, 19
  ...
(End)
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

Other sequences that are built by replacing 2^k in the binary representation with other numbers: A029931 (naturals), A054204 (even-indexed Fibonacci numbers), A062877 (odd-indexed Fibonacci numbers), A059590 (factorials), A089625 (primes).

Cf. A003754, A022342, A026274.

Programs

Haskell

a022290 0 = 0
a022290 n = h n 0 $ drop 2 a000045_list where
   h 0 y _      = y
   h x y (f:fs) = h x' (y + f * r) fs where (x',r) = divMod x 2
-- Reinhard Zumkeller, Oct 03 2012

Maple

A022290 := proc(n)
    dgs := convert(n,base,2) ;
    add( op(i,dgs)*A000045(i+1),i=1..nops(dgs)) ;
end proc: # R. J. Mathar, Jan 31 2015
# second Maple program:
b:= (n, i, j)-> `if`(n=0, 0, j*irem(n, 2, 'q')+b(q, j, i+j)):
a:= n-> b(n, 1$2):
seq(a(n), n=0..127);  # Alois P. Heinz, Jan 26 2022

Mathematica

Table[Reverse[#].Fibonacci[1 + Range[Length[#]]] &@ IntegerDigits[n, 2], {n, 0, 54}] (* IWABUCHI Yu(u)ki, Aug 01 2012 *)

PARI

my(m=Mod('x,'x^2-'x-1)); a(n) = subst(lift(subst(Pol(binary(n)), 'x,m)), 'x,2); \\ Kevin Ryde, Sep 22 2020

Python

def A022290(n):
    a, b, s = 1,2,0
    for i in bin(n)[-1:1:-1]:
        s += int(i)*a
        a, b = b, a+b
    return s # Chai Wah Wu, Sep 10 2022

Formula

G.f.: (1/(1-x)) * Sum_{k>=0} F(k+2)*x^2^k/(1+x^2^k), where F = A000045.

a(n) = Sum_{k>=0} A030308(n,k)*A000045(k+2). - Philippe Deléham, Oct 15 2011

a(A003714(n)) = n. - R. J. Mathar, Jan 31 2015

a(A000225(n)) = A001911(n). - Philippe Deléham, Jun 05 2015

From Jeffrey Shallit, Jul 17 2018: (Start)

Can be computed from the recurrence:

a(4*k) = a(k) + a(2*k),

a(4*k+1) = a(k) + a(2*k+1),

a(4*k+2) = a(k) - a(2*k) + 2*a(2*k+1),

a(4*k+3) = a(k) - 2*a(2*k) + 3*a(2*k+1),

and the initial terms a(0) = 0, a(1) = 1. (End)

a(A003754(n)) = n-1. - Rémy Sigrist, Jan 28 2020

From Rémy Sigrist, Aug 04 2022: (Start)

Empirically:

- a(2*A003714(n)) = A022342(n+1),

- a(3*A003714(n)) = a(4*A003714(n)) = A026274(n) for n > 0.

(End)